Solutions Manual to Optoelectronics and Photonics: Principles and Practices, Second Edition Safa Kasap

Solutions Manual to Optoelectronics and Photonics: Principles and Practices, Second Edition © 2013 Pearson Education

Safa Kasap

Revised: 11 December 2012 Check author's website for updates http://optoelectronics.usask.ca ISBN-10: 013308180X ISBN-13: 9780133081800 NOTE TO INSTRUCTORS If you are posting solutions on the internet, you must password the access and download so that only your students can download the solutions, no one else. Word format may be available from the author. Please check the above website. Report errors and corrections directly to the author at [email protected]. S.O. Kasap, Optoelectronics and Photonics: Principles and Practices, Second Edition, © 2013 Pearson Education © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Solutions Manual (Preliminary)

Chapter 1

1.2

11 December 2012

Preliminary Solutions to Problems and Questions Chapter 1 Note: Printing errors and corrections are indicated in dark red. See Question 1.47. These are correct in the e-version of the textbook 1.1

Maxwell's wave equation and plane waves

(a) Consider a traveling sinusoidal wave of the form Ex = Eo cos(tkz + o). The latter can also be written as Ex = Eo cos[k(vtz) + o], where v = /k is the velocity. Show that this wave satisfies Maxwell's wave equation, and show that v = (oor)1/2. (b) Consider a traveling function of any shape, even a very short delta pulse, of the form Ex = f[k(vtz)], where f is any function, which can be written is Ex = f(),  = k(vtz). Show that this traveling function satisfies Maxwell's wave equation. What is its velocity? What determines the form of the function f? Solution (a)

Ex = Eo cos(tkz + o)  2 Ex  0 x 2  2 Ex and 0 y 2  2 Ex and   k 2 E0 cos(t  kz  0 ) 2 z  2 Ex    2 E0 cos(t  kz  0 ) 2 t 2E 2E 2E 2E Substitute these into the wave equation        0 to find o r o x 2 y 2 z 2 t 2 k 2 E cos(t  kz  0 )   o r o   2 E0 cos(t  kz  0 )  0

2















 v  ( o r o )

(b)

Let

k

 k

2



1



 o r o

 ( o r o )

1

1

2

2

Ex  f [k (v t  z )]  f ( ) Take first and second derivatives with respect to x, y, z and t.



Chapter 1

1.3

11 December 2012

 2 Ex 0 x 2  2 Ex 0 y 2

Ex df  k z d 2  2 Ex 2 d f k  d 2 z 2

Ex df  kv t d 2  2 Ex 2 2 d f k v d 2 t 2

Substitute these into the wave equation

k2







2E 2E 2E 2E        0 to find o r o x 2 y 2 z 2 t 2

2 d2 f 2 2 d f     v 0 k o r o d 2 d 2

v2 

1

 o r  o

v  ( o r o )

1

2

1.2 Propagation in a medium of finite small conductivity An electromagnetic wave in an isotropic medium with a dielectric constant r and a finite conductivity  and traveling along z obeys the following equation for the variation of the electric field E perpendicular to z, d 2E 2E E       o o r o 2 2 dz t t





Show that one possible solution is a plane wave whose amplitude decays exponentially with propagation along z, that is E = Eoexp(z)exp[j(t – kz)]. Here exp(z) causes the envelope of the amplitude to decay with z (attenuation) and exp[j(t – kz)] is the traveling wave portion. Show that in a medium in whichis small, the wave velocity and the attenuation coefficient are given by v

 k



1

o o r

and



 2 o cn

where n is the refractive index (n = r1/2). (Metals with high conductivities are excluded.) Solution



Chapter 1

1.4

11 December 2012

We can write E = Eoexp(z)exp[j(t – kz)] as E = Eoexp[jt – j(k – j)z]. Substitute this into the wave resonance condition [– j(k – j)]2Eoexp[jt – j(k – j)z]  (joroEoexp[jt – j(k – j)z] =  jo Eoexp[jt – j(k – j)z]       (k – j)2  oro = jo  k2 jk – 2 + oro = jo Rearrange into real and imaginary parts and then equating the real parts and imaginary parts   k2 – 2 + oro jk = jo Real parts k2 – 2 + oro = 0 Imaginary parts k = o       c   o   o  o  Thus, 2k k 2 2n 2 o n where we have assumed /k = velocity = c/n (see below). From the imaginary part k 2   2  o o r   2 Consider the small  case (otherwise the wave is totally attenuated with very little propagation). Then k 2   2  o  o r and the velocity is  1 v  k  o o r 1.3 Point light source What is the irradiance measured at a distance of 1 m and 2 m from a 1 W light point source? Solution Then the irradiance I at a distance r from O is P 1W I  o2  = 8.0 W cm-2 2 4r 4 (1 m) which drops by a factor of 4 at r = 2 m to become 2.0 W cm-2 1.4 Gaussian beam A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a Gaussian beam, what is the divergence of the beam? What are its Rayleigh range and beam width at 10 m? Solution Using Eq. (1.1.7), we find, 4 4(633  109 m) 2   = 1.0110-3 rad = 0.058  (2wo )  (0.8 103 m) The Rayliegh range is  wo2  [ 12 (0.8 103 m)]2 zo   = 0.79 m  (633  109 m)



Chapter 1

1.5

11 December 2012

The beam width at a distance of 10 m is 2w = 2wo[1 + (z/zo)2]1/2 = (0.810-3 m){1 + [(10 m)/(0.79 m)]2}1/2 = 0.01016 m or 10.16 mm. 1.5 Gaussian beam in a cavity with spherical mirrors Consider an optical cavity formed by two aligned spherical mirrors facing each other as shown in Figure 1.54. Such an optical cavity is called a spherical mirror resonator, and is most commonly used in gas lasers. Sometimes, one of the reflectors is a plane mirror. The two spherical mirrors and the space between them form an optical resonator because only certain light waves with certain frequencies can exist in this optical cavity. The radiation inside a spherical mirror cavity is a Gaussian beam. The actual or particular Gaussian beam that fits into the cavity is that beam whose wavefronts at the mirrors match the curvature of the mirrors. Consider the symmetric resonator shown in Figure 1.54 in which the mirrors have the same radius of curvature R. When a wave starts at A, its wavefront is the same as the curvature of A. In the middle of the cavity it has the minimum width and at B the wave again has the same curvature as B. Such a wave in the cavity can replicate itself (and hence exist in the cavity) as it travels between the mirrors provided that it has right beam characteristics, that is the right curvature at the mirrors. The radius of curvature R of a Gaussian beam wavefront at a distance z along its axis is given by R(z) = z[1 + (zo/z)2] ; zo = wo2/ is the Rayleigh range Consider a confocal symmetric optical cavity in which the mirrors are separated by L = R. (a) Show that the cavity length L is 2zo, that is, it is the same as the Rayleigh range, which is the reason the latter is called the confocal length. (b) Show that the waist of the beam 2wo is fully determined only by the radius of curvature R of the mirrors, and given by 2wo = (2R/)1/2 (c) If the cavity length L = R = 50 cm, and = 633 nm, what is the waist of the beam at the center and also at the mirrors?

Figure 1.54 Two spherical mirrors reflect waves to and from each other. The optical cavity contains a Gaussian beam. This particular optical cavity is symmetric and confocal; the two focal points coincide at F. Solution (a) At z  R / 2 we have R( z )  R . Substitute these into R(z) = z[1 + (zo/z)2] to find R = (R/2)[1 + (2zo/R)2]



Chapter 1

1.6

11 December 2012 2

 



 (b)



 



Now use  



 2z  2  1  o   R   2 zo    1  R  L  2 zo R = (R/2)[1 + (2zo/R)2]  2z  2  1  o   R   2 zo    1  R  zo = wo2/,  2wo2    1   R  2 wo 

2

2 R



(c) Substitute  = 633 nm, L = R = 50 cm into the above equation to find 2wo = 449 m or 0.449 mm. At the mirror, z = R/2, and also zo = R/2 so that 1/ 2

  z 2  2 w  2 wo 1       zo  

1/ 2

  R / 2 2   2 wo 1       R / 2  

 2 wo ( 21 / 2 ) = 0.635 mm

1.6 Cauchy dispersion equation Using the Cauchy coefficients and the general Cauchy equation, calculate refractive index of a silicon crystal at 200 m and at 2 m, over two orders of magnitude wavelength change. What is your conclusion? Solution At  = 200µm, the photon energy is hc (6.62 1034 J s)(3  108 m s-1 ) 1   h     6.2062  103 eV 6 19 -1  (200 10 m) 1.6.  10 J eV Using the Cauchy dispersion relation for silicon with coefficients from Table 9.2, n = n-2(h + n0 + n2(h2 + n4(h4 = (2.0410-8)( 6.2062 103  + 3.4189+ (10-2)( 6.2062 103  + (10-2)( 6.2062 103     = 3.4184 At  = 2µm, the photon energy is hc (6.62 1034 J s)(3  108 m s -1 ) 1   h     0.6206eV 6 19  (2 10 m) 1.6 10 J eV -1 Using the Cauchy dispersion relation for silicon with coefficients from Table 9.2,



Chapter 1

1.7

11 December 2012





n = n-2(h + n0 + n2(h2 + n4(h4 = (2.0410-8)( 0.6206  + 3.4189+ (10-2)( 0.6206  + (10-2) ( 0.6206   = 3.4521

1.7 Sellmeier dispersion equation Using the Sellmeier equation and the coefficients, calculate the refractive index of fused silica (SiO2) and germania GeO2 at 1550 nm. Which is larger, and why? Solution The Sellmeier dispersion relation for fused silica is 0.696749 2 0.408218 2 0.890815 2 n2  1  2     0.06906602 μm 2  2  0.1156622μm 2  2  9.9005592μm 2 0.696749 (1550 nm ) 2 0.408218(1550 nm) 2 0.890815(1550 nm) 2 n  1   (1550 nm) 2  (69.0660 nm) 2 (1550 nm) 2  (115.662 nm) 2 (1550 nm) 2  (9900.559 nm) 2 so that 2

n = 1.4443 The Sellmeier dispersion relation for germania is 0.8068664 2 0.7181585 2 0.8541683 2 n2  1  2     (0.0689726μm) 2  2  (0.1539661μm) 2  2  (11.841931μm) 2

0.8068664 (1550 nm) 2 0.7181585(1550 nm) 2 0.8541683(1550 nm) 2 n  1   (1550 nm) 2  (68.9726 nm) 2 (1550 nm) 2  (153.9661nm) 2 (1550 nm) 2  (11841.931nm) 2 2

so that n = 1.5871 1.8 Sellmeier dispersion equation The Sellmeier dispersion coefficient for pure silica (SiO2) and 86.5%SiO2-13.5 mol.% GeO2 re given in Table 1.2 Write a program on your computer or calculator, or use a math software package or even a spread sheet program (e.g. Excel) to obtain the refractive index n as a function of  from 0.5 m to 1.8 m for both pure silica and 86.5%SiO2-13.5%GeO2. Obtain the group index, Ng, vs. wavelength for both materials and plot it on the same graph. Find the wavelength at which the material dispersion becomes zero in each material.



Chapter 1

1.8

11 December 2012

Solution Excel program to plot n and differentiate and find Ng



Chapter 1

1.9

11 December 2012

Figure 1Q8-1 Refractive index n and the group index Ng of pure SiO2 (silica) glass as a function of wavelength (Excel). The minimum in Ng is around 1.3 m. Note that the smooth line option used in Excel to pass a continuous smooth line through the data points. Data points are exactly on the line and are not shown for clarity.



Chapter 1

1.10

11 December 2012

Figure 1Q8-2 Refractive index n and the group index Ng of 86.5%SiO213.5%GeO as a function of wavelength (Excel). The minimum in Ng is around 1.4 m. Note that the smooth line option used in Excel to pass a continuous smooth line through the data points. Data points are exactly on the line and are not shown for clarity.

Material dispersion is proportional to derivative of group velocity over wavelength. The corresponding values are close to 1.3 and 1.4 m. 1.9 The Cauchy dispersion relation for zinc selenide ZnSe is a II-VI semiconductor and a very useful optical material used in various applications such as optical windows (especially high power laser windows), lenses, prisms etc. It transmits over 0.50 to 19 m. n in the 1 – 11 m range described by a Cauchy expression of the form 0.0485 0.0061 n  2.4365    0.0003λ 2 ZnSe dispersion relation λ2 λ4 in which  in m. What are the n-2, n0, n2 and n4 coefficients? What is ZnSe's refractive index n and group index Ng at 5 m? Solution hc h 



hc  (6.62 10 34 J s 

1 1.6  10

19

J eV

-1

)(3  108 m s -1 )  1.24 10 6 eVm

so that 0.0485 0.0061 (h ) 2  (h ) 4  0.0003(hc) 2 (h ) 2 2 4 (hc) (hc) Comparing with Cauchy dispersion equation in photon energy: n = n-2(h + n0 + n2(h2 + n4(h4 we have n  2.4365 

,



Chapter 1

1.11

11 December 2012

n0  2.4365 n2 

0.0485 0.0485   3.15 1010 eV -2 2 6 2 (hc) (1.24  10 )

n2  0.0003(hc)2  0.0003  (1.24 106 )2  4.62 1016 eV 2 n4 

and

0.0061 0.0061   2.58 1021 eV -4 4 (hc) (1.24  106 ) 4

At λ = 5 m 0.0485 0.0061   0.0003(5 m) 2 2 4 (5 m) (5 m) 0.0485 0.0061  2.4365    0.0003(25)  2.43 25 625

n  2.4365 

dn d 0.0485 0.0061 n  2.4365    0.0003λ 2 2 4 λ λ dn 2  0.0485 4 3  0.0061    2  0.0003λ d λ4 λ8 dn 0.097 0.0244    0.0006 λ d λ3 λ5 Ng  n  

Group index and   At λ = 5 m

 





dn 0.097 0.0244    0.0006  (5µm) d  (5µm)3 (5µm)5 dn  0.003783µm 1 d dn Ng  n    2.43  5µm  (0.003783µm 1 )  2.45 d

1.10 Refractive index, reflection and the Brewster angle (a) Consider light of free-space wavelength 1300 nm traveling in pure silica medium. Calculate the phase velocity and group velocity of light in this medium. Is the group velocity ever greater than the phase velocity? (b) What is the Brewster angle (the polarization angle p) and the critical angle (c) for total internal reflection when the light wave traveling in this silica medium is incident on a silica/air interface. What happens at the polarization angle? (c) What is the reflection coefficient and reflectance at normal incidence when the light beam traveling in the silica medium is incident on a silica/air interface?



Chapter 1

1.12

11 December 2012

(d) What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is incident on an air/silica interface? How do these compare with part (c) and what is your conclusion? Solution

Figure 1.8 Refractive index n and the group index Ng of pure SiO2 (silica) glass as a function of wavelength. (a) From Figure 1.8, at  = 1300 nm, n = 1.447, Ng = 1.462, so that The phase velocity is given by v = c/n = (3108 m s-1)/(1.447) = 2.073108 m s-1. The group velocity is given by vg = c/Ng = (3108 ms-1)/(1.462) = 2.052108 m s-1. The group velocity is about ~1% smaller than the phase velocity. (b)

The Brewster angle p is given by n 1 tan  p  2   0.691 n1 1.447



Chapter 1

1.13

11 December 2012

 p  tan 1 0.691  34.64



At the Brewster angle of incidence i = p, the reflected light contains only field oscillations normal to the plane of incidence (paper).

The critical angle is

sin  c 

c  sin 1 (0.691)  43.7

 (c)

n2 1   0.691 n1 1.447

Given n1  1.447 n2  1

n1  n2 1.447  1   0.1827 n1  n2 1.447  1



r/ /  r 

and

R  R/ /  r  r/ /  0.0333

(d)

2

2

Given n1  1

 and

n2  1.447 n  n 1  1.447 r/ /  r  1 2   0.1827 n1  n2 1.447  1

R  R/ /  r  r/ /  0.0333 2

2

Reflection coefficients are negative, which means that in external reflection at normal incidence there is a phase shift of 180°. 1.11 Snell's law and lateral beam displacement What is the displacement of a laser beam passing through a glass plate of thickness 2 mm and refractive index 1.570 if the angle of incidence is 40? (See Figure 1.14)



Chapter 1

1.14

11 December 2012

Figure 1.14 Lateral displacement of light passing obliquely through a transparent plate Solution The problem is sketched in Figure 1Q12-1

Figure 1Q12-1 Light beam deflection through a glass plate of thickness L = 2 mm. The angle of incidence is 40 and the glass has a refractive index of 1.570   d cos i  sin i 1   L ( n / no ) 2  sin 2 i  







 cos 40 d  sin 40 1  L  (1.570 /1) 2  sin 2 40

  

0.7660    0.6428 1    0.2986 2.46  0.4132  

d  0.2986 2 mm    d = 0.60 mm This is a significant displacement that can be easily measured by using a photodiode array. 







Chapter 1

1.15

11 December 2012

1.12 Snell's law and lateral beam displacement An engineer wants to design a refractometer (an instrument for measuring the refractive index) using the lateral displacement of light through a glass plate. His initial experiments involve using a plate of thickness L, and measuring the displacement of a laser beam when the angle of incidence i is changed, for example, by rotating (tilting) the sample. For i = 40, he measures a displacement of 0.60 mm, and when i = 80 he measures 1.69 mm. Find the refractive index of the plate and its thickness. (Note: You need to solve a nonlinear equation for n numerically.) Solution Figure 1.14 shows the lateral beam deflection through a transparent plate.

Figure 1.14 Lateral displacement of light passing obliquely through a transparent plate   cos i Apply d  L sin i 1   n 2  sin 2 i       cos 40 cos 80 and 0.60 mm  L sin 40 1  1.69 mm  L sin 801    n 2  sin 2 40  n 2  sin 2 80    Divide one by the other     cos 40 cos 40  1 1     0.60  sin 40   0.60  sin 40   n 2  sin 2 40  n 2  sin 2 40      0    1.69  sin 80   1.69  sin 80     cos 80 cos 80 1   1   2 2 2 2 n  sin 80  n  sin 80      cos 40 1   0.60  sin 40   x 2  sin 2 40  Define y    1.69  sin 80    cos 80 1   x 2  sin 2 80    0.76606  1   2 x  0.41318   = f(x) y  0.35503  0.6527   0.17365  1   x 2  0.96985  



Chapter 1

1.16

11 December 2012

We can plot y = f(x) vs. x, and find where f(x) cross the x-axis, which will give x = n 0.1 0.05 0

y

1.4

x

1.5

1.6

1.7

1.8

0.05

The above graph was generated in LiveMath (Theorist) (http://livemath.com)

Clearly, the x-axis is cut at n  1.575 Substitute n = 1.575 into one of the equations i.e.   cos 40 0.60 mm  L sin 401   1.5752  sin 2 40   Solving for L we find L  2.0 mm. 1.13 Snell's law and prisms Consider the quartz prism shown in Figure 1.55 that has an apex angle  = 60. The prism has a refractive index of n and it is in air. (a) What are Snell's law at interfaces at A (incidence and transmittance angles of i and t ) and B (incidence and transmittance angles of i and t)? (b) Total deflection  = 1 + 2 where 1 = i  t and 2 = t  i. Now,  + i  t = 180 and  +  = 180. Find the deflection of the beam for an incidence angle of 45 for the following three colors at which n is known: Blue, n = 1.4634 at  = 486.1 nm; yellow, n = 1.4587 at  = 589.2 nm; red, n = 1.4567 at  = 656.3 nm. What is the separation in distance between the rays if the rays are projected on a screen 1 m away.

Figure 1.55 A light beam is deflected by a prism through an angle . The angle of incidence is i. The apex angle of the prism is . Solution (a) Snell's law at interfaces at A: sin i n  sin t 1 Snell's law at interfaces at B:



Chapter 1

1.17

11 December 2012

sin i 1  sin t n (b) Consider the deflection angle   = 1 + 2 where 1 = i  t and 2 = t  i, i.e.  = i  t + t  i. Now,  sin  i   t  arcsin    n  and from Figure 1.55  sin i  i  180     t    t    arcsin   n  so that    sin i    t  arcsin n sin i  arcsin n sin    arcsin    n     and the deflection is,    sin  i   sin i     sin i    arcsin n sin    arcsin    i  arcsin      arcsin      n   n     n     so that finally,    sin  i     i    arcsin n sin    arcsin    n     Substituting the values, and keeping n as a variable, the deflection (n) as a function of n is 







 1      2n  

 (n )  (45  60 )  arcsin n sin 60  arcsin 

where sin(45) = 1/2. The separation L for two wavelengths1 and 2 corresponding to n1 and n2 at the screen at a distance L away is therefore L = L[(n1)  (n2)] where the deflections must be in radians. Consider the deflection of blue light 







  1    2 (1.4634)   

 blue  (45  60 )  arcsin (1.4634) sin 60  arcsin 

 blue = 34.115 Similarly, yellow = 33.709 The separation of blue and yellow beams at the screes is Lblue-yellow = L(blue  blue) = (1m)(/180)( 34.115 33.709) = 7.08 mm Table 1Q13-1 summarizes the results of the calculations for blue, yellow and red light. Table 1Q13-1 Deflection of blue, yellow and red light through a prism with apex angle 60. The angle of incidence is 45. Blue 486.1 nm

Yellow 589.2 nm

Red 656.3 nm



Chapter 1

1.18

11 December 2012

n

Deflection angle) L between colors L between blue and rede

1.4634 0.5954 rad 34.115

1.4587 0.5883 rad 33.709 7.08 mm

1.4567 5853 rad 33.537 3.00 mm

10.1 mm

1.14 Fermat's principle of least time Fermat's principle of least time in simple terms states that when light travels from one point to another it takes a path that has the shortest time. In going from a point A in some medium with a refractive index n1 to a point B in a neighboring medium with refractive index n2 as in Figure 1.56 the light path is AOB that involves refraction at O and satisfies Snell's law. The time it takes to travel from A to B is minimum only for the path AOB such that the incidence and refraction angles i and t satisfy Snell's law. Let's draw a straight line from A to B cutting the x-axes at O. The line AOB will be our reference line and we will place the origin of x and y coordinates at O. Without invoking Snell's law, we will vary point O along the x-axis (hence OO is a variable labeled x), until the time it takes to travel AOB is minimum, and thereby derive Snell's law. The time t it takes for light to travel from A to B through O is AO OB [( x1  x ) 2  y12 ]1 / 2 [( x 2  x ) 2  y 22 ]1 / 2 t    (1) c / n1 c / n 2 c / n1 c / n2 The incidence and transmittance angles are given by x1  x ( x2  x) sin  i  sin t  and (2) 2 2 1/ 2 [( x1  x )  y1 ] [( x2  x) 2  y22 ]1/2 Differentiate Eq. (1) with respect to x to find the condition for the "least time" and then use Eq. (2) in this condition to derive Snell's law.

Figure 1.56 Consider a light wave traveling from point A (x1, y2) to B (x1, y2) through an arbitrary point O at a distance x from O. The principle of least time from A to B requires that O is such that the incidence and refraction angles obey Snell's law.



Chapter 1

1.19

11 December 2012

Solution Differentiate t with respect to x dt 1/ 2  2( x1  x)[( x1  x) 2  y12 ]1/ 2 1/ 2  2( x2  x)[( x2  x) 2  y22 ]1/2   dx c / n1 c / n2 The time should be minimum so dt 0 condition for the "least time" dx ( x1  x)[( x1  x) 2  y12 ]1/2 ( x2  x)[( x2  x) 2  y22 ]1/2   0 c / n1 c / n2 ( x1  x) ( x2  x)   2 2 1/2 c / n1[( x1  x)  y1 ] c / n2 [( x1  x) 2  y12 ]1/2 Use Eq. (2) in the above expression to find n1 sin i  n2 sin  t 

sin i n2  sin t n1

Snell's law

1.15 Antireflection (AR) coating (a) Consider three dielectric media with flat and parallel boundaries with refractive indices n1, n2, and n3. Show that for normal incidence the reflection coefficient between layers 1 and 2 is the same as that between layers 2 and 3 if n2 = [n1n3]. What is the significance of this result? (b) Consider a Si photodiode that is designed for operation at 900 nm. Given a choice of two possible antireflection coatings, SiO2 with a refractive index of 1.5 and TiO2 with a refractive index of 2.3, which would you use and what would be the thickness of the antireflection coating? The refractive index of Si is 3.5. Explain your decision. (c) Consider a Ge photodiode that is designed for operation around 1200 nm. What are the best AR refractive index and coating thickness if the refractive index of Ge is about 4.0? Solution



Chapter 1

1.20

11 December 2012

(a)

Start with the reflection coefficient r12 between n1 and n2, n  n1n3 n1  n1n3 n n  n3 n n r12  1 2  1   1 3 n1  n2 n1  n1n3 n1  n1n3 n1n3  n3 

n1 n1n3  n1n3  n3 n1n3  n1n3 n1 n1n3  n1n3  n1n3  n3 n1n3



n1 n1n3  n3 n1n3 n1 n1n3  2n1n3  n3 n1n3

Now consider r23 between n2 and n3, n n  n3 n n  n3 n1  n1n3 n n r23  2 3  1 3  1 3  n2  n3 n1n3  n3 n1n3  n3 n1  n1n3 



n1 n1n3  n1n3  n1n3  n3 n1n3 n1 n1n3  n1n3  n1n3  n3 n1n3



n1 n1n3  n3 n1n3 n1 n1n3  2n1n3  n3 n1n3

r12  r23

To reduce the reflected light, waves A and B must interfere destructively. To obtain a good degree of destructive interference between waves A and B, the amplitudes of reflection coefficients must be comparable. When n2 = (n1n3)1/2, then the reflection coefficient between the air and coating is equal to that between the coating and the semiconductor. So the reflection is minimum. (b) We use n1 = 1 for air, n2 for the antireflection coefficient and n3 =3.5 for Si photodiode,

Rmin

 n2  n n    22 1 3   n2  n1n3 

2

For SiO2 n2 = 1.5 2

 1.52  1 3.5  Rmin (SiO 2 )   2   0.047  1.5  1 3.5  For TiO2

n2 = 2.3 2

 2.32  1 3.5  Rmin (TiO 2 )   2   0.041  2.3  1 3.5  Rmin (TiO 2 )  Rmin (SiO 2 )  So, TiO2 is a better choice



Chapter 1

1.21

11 December 2012

The thickness of the AR layer should be d = /(4n2) = (900 nm)/[4(2.3)] = 97.8 nm (c) Consider the Ge photodiode. Ge has n = 4.0. We use n1 = 1 for air, n2 for the antireflection coefficient and n3 = 4.0 for Ge photodiode, The ideal AR coating would have n2 = (n1n3)1/2 = 2.0 The thickness of the AR layer should be d = /(4n2) = (1200 mm)/[4(2)] = 150 nm 1.16 Single and double layer antireflection V-coating For a single layer AR coating of index n2 on a material with index n3 (> n2 > n1), as shown in Figure 1.57(a), the minimum reflectance at normal incidence is given by 2

n2  n n  R min   22 1 3  Single layer AR coating  n 2  n1n3  when the reflections A, B, … all interfere as destructively as possible. Rmin = 0 when n2 = (n1n3)1/2. The choice of materials may not always be the best for a single layer antireflection coating. Double layer AR coatings, as shown in Figure 1.57(b) can achieve lower and sharper reflectance at a specified wavelength as in Figure 1.57(c). To reduce the reflection of light at the n1/n4, interface, two layers n2 and n3, each quarter wavelength in the layer (/n2 and /n3) are interfaced between n1 and n4. The reflections A, B and C for normal incidence result in a minimum reflectance given by 2

 n 2 n  n4 n 22  R min   32 1 Double layer AR coating 2  n3 n1  n4 n2  Double layer reflectance vs. wavelength behavior usually has V-shape, and they are called V-coatings. (a) Show that double layer reflectance vanishes when Best double layer AR coating (n2/n3)2 = n1/n4 (b) Consider an InGaAs, a semiconductor crystal with an index 3.8, for use in a photodetector. What is the reflectance without any AR coating? (c) What is the reflectance when InGaAs is coated with a thin AR layer of Si3N4? Which material in the table would be ideal as an AR coating? (d) What two materials would you choose to obtain a V-coating? Note: The choice of an AR coating also depends on the technology involved in depositing the AR coating and its effects on the interface states between the AR layer and the semiconductor. Si1-xNx is a common AR coating on devices inasmuch as it is a good passive dielectric layer, its deposition technology is well established and changing its composition (x) changes its index. A B n1 n1

n2

n2

n3

n3

CBA

Reflectance Double layer Single layer

n4

 (a)

(b)

(c)

Figure 1.57(a) A single layer AR coating. (b) A double layer AR coating and its V-shaped reflectance spectrum over a wavelength range.



Chapter 1

1.22

11 December 2012

Solution (a) The minimum reflectance, 2







 n 2 n  n4 n22  Rmin   32 1 0 2 n n  n n 4 2   3 1 2 2 n3 n1  n4 n2  0







n32 n1  n4 n22







n1 n22  n4 n32







n1 n  ( 2 )2 n4 n3

(b)

Without an AR coating, the reflectance is R = [(n1  n3)/ (n1  n3)]2 = [(1  3.8)/ 1  3.8)]2 = 0.34 or 34%

(c)

Take

Best double layer AR coating

n3 = 3.8, n2 = 1.95, n1 = 1, and find the minimum reflectance from 2

R min

 n22  n1n3   2   n 2  n1n3 

Rmin

1.952  3.8  7    1.08 10 2 1.95  3.8 

2



For ideal an AR coating: n2  n1n3 

n1n3  1 3.8  1.9493

n2  1.9493  Looking at table, Si3N4 (n2 = 1.95) would be ideal. (d)

Two find 2 materials for a V-coating, consider first,

(n2 / n3 ) 2  n1 / n4

Best double layer AR coating

(n2 / n3 )  n1 / n4



Chapter 1

1.23

11 December 2012

( n2 / n3 )  1/ 3.8  0.51 This is the ratio we need. From the table MgF2 (n2 = 1.38) and CdS (n3 = 2.60) are the best two 1.38  0.53 . materials for V-coating: (n2 / n3 )  2.60 The minimum reflectance would be 2

Rmin

2

 n32 n1  n4 n22   2.602  3.8 1.382   2   0.001 2 2 2  n3 n1  n4 n2   2.60  3.8 1.38 

1.17 Single, double and triple layer antireflection coatings Figure shows the reflectance of an uncoated glass, and glass that has a single (1), double (2) and triple (3) layer antireflection coatings? The coating details are in the figure caption. Each layer in single and double layer AR coatings has a thickness of /4, where is the wavelength in the layer. The triple layer AR layer has three coatings with thicknesses /4, /2 and 4. Can you qualitatively explain the results by using interference? What applications would need single, double and triple layer coatings?

Solution Instructor’s choice of answers. Can be given out as a short project to students.

1.18 Reflection at glass-glass and air-glass interfaces A ray of light that is traveling in a glass medium of refractive index n1 = 1.460 becomes incident on a less dense glass medium of refractive index n2 = 1.430. Suppose that the free space wavelength of the light ray is 850nm. (a)

What should the minimum incidence angle for TIR be?

(b)

What is the phase change in the reflected wave when the angle of incidence i = 85° and when

i = 90°?



Chapter 1

1.24

11 December 2012

(c) What is the penetration depth of the evanescent wave into medium 2 when i = 85° and when i = 90°? (d) What is the reflection coefficient and reflectance at normal incidence (i = 0°) when the light beam traveling in the silica medium (n = 1.460) is incident on a silica/air interface? (e) What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is incident on an air/silica (n = 1.460) interface? How do these compare with part (d) and what is your conclusion? Solution (a) The critical angle c for TIR is given by n 1.430  c  sin 1 2  sin 1  78.36 1.460 n1 (b) Since the incidence angle i > c, there is a phase shift in the reflected wave. The phase change in Er, is given by . With n1 = 1.460, n2 = 1.430, and i = 85°, 1/2

sin 2 i  n 2   tan    cos i = 2.08675 1 tan 2   2.08675



1 2





2  2  1.430   sin (85)      1.460     cos(85)

1/2



  2 tan 1 (2.08673)=128.79 so that the phase change is 128.79 . For the Er,// component, the phase change is 1/2

sin 2 i  n 2  1 1 tan 2  / /  2   n 2 cos i



so that





1 tan n2



1 2

 

tan(1/2//  12  ) = (n1/n2)2tan(/2) = (1.460/1.430)2tan(1/2(128.79°)) = 2.17522

/ /  2 tan 1 (2.17522)-  49.38 which gives  We can repeat the calculation with i = 90°. The phase change in Er, is given by . With n1 = 1.460, n2 = 1.430, and i = 90°, 1/2

sin 2 i  n 2   cos i



 tan 



  2 tan 1 ()=180

tan

1 2

 

1 2

   

2  2  1.430   sin (90  )       1.460     cos(90)

1/2

so that the phase change is 180 . For the Er,// component, the phase change is 1/2

sin 2 i  n 2  1 1 tan 2  / /  2   n 2 cos i



so that





1 tan n2



1 2

 

tan(1/2//  12  )= 



Chapter 1

1.25

11 December 2012

which gives

// = 2tan-1()

(c) The amplitude of the evanescent wave as it penetrates into medium 2 is Et,(y,t)  Eto,exp(–2y) We ignore the z-dependence, expj(tkzz), as this represents propagation along z. The field strength drops to e-1 when y = 1/2 = , the penetration depth. The attenuation constant 2 for i = 85° is 2  2n2  n1  2    sin 2  i  1 o  n2   

1/ 2

1/ 2



2  2 (1.430)  1.460  2  2   sin (85 )  1  9  (850  10 )  1.430  

= 1.96106 m

.

so the penetration depth 1/2 = 1/(1.96106 m) = 5.110-7 m, or 0.51m. For 90°, repeating the calculation above, 1/ 2

2  2 (1.430)  1.460  2  2   sin (90 )  1  9  (850  10 )  1.430  

= 2.175106 m

so the penetration depth 1/2 = 1/(2.175106 m) = 4.510-7 m, or 0.45 m. Conclusion: The penetration depth increases as the angle of incidence decreases (d)

n1  n2 1.460  1  = 0.187 n1  n2 1.460  1 R// = R = r//2 = 0.035 or 3.5% r//  r 

and (e)

n1  n2 1  1.460  = 0.187 n1  n2 1  1.460 and R// = R = r//2 = 0.035 or 3.5% When r is a negative number, then there is a phase shift of 180° (or ) which is in agreement with part (b). r//  r 

1.19 Dielectric mirror Consider a dielectric mirror that is made up of quarter wave layers of GaAs with nH = 3.382 and AlAs with nL = 2.912, both around 1500 nm. The GaAs-AlAs dielectric mirror is inside a vertical cavity surface emitting laser diode operating at 1.5 m. The substrate is GaAs with n3 = nsubstrate = 3.382 . The light is incident on the mirror from another semiconductor that is GaAlAs with an index n0 = 3.40. Calculate how many pairs of layers N would be needed to get a reflectance above 95%. What would be the bandwidth? Solution



Chapter 1

1.26

11 December 2012 2

2   n1  2 N  n0    2 N  n0  2 N  n n     n    n    1 n3  2  2 3     RN      2N n0     n1    n12 N   n0 n22 N    n2    n3      n3  Insert RN = 0.95 and solve,  n 1  RN   ln 0   n 1 R  1  3 N  =14.58 i.e. 15 pairs are needed N 2 ln n1   n2   n  n2   4  =0.095167  arcsin 1 0   n1  n2 

 n1   n  2 

2N

  

n0

 1  RN n3  1  R N

But, 0 = 1500 nm,   = 142 nm 1.20 TIR and polarization at water-air interface (a) Given that the refractive index of water is about 1.33, what is the polarization angle for light traveling in air and reflected from the surface of the water? (b) Consider a diver in sea pointing a flashlight towards the surface of the water. What is the critical angle for the light beam to be reflected from the water surface? Solution (a)

Apply

n2 n1

n1  1

with

n2  1.33

 p  tan 1 (n2 / n1 )  tan 1 (1.33 / 1)  53.06

 (b)

tan  p 

Given

and

n1  1.33 n2  1

The critical angle is

 c  sin 1

n2 1  sin 1  48.75 1.33 n1

1.21 Reflection and transmission at a semiconductor-semiconductor interface A light wave with a free space wavelength of 890 nm (free space wavelength) that is propagating in GaAs becomes incident on AlGaAs. The refractive index of GaAs is 3.60, that of AlGaAs is 3.30. (a) Consider normal incidence. What are the reflection and transmission coefficients and the reflectance and transmittance? (From GaAs into AlGaAs)



Chapter 1

1.27

11 December 2012

(b) What is the Brewster angle (the polarization angle p) and the critical angle (c) for total internal reflection for the wave in (a); the wave that is traveling in GaAs and incident on the GaAs/AlGaAs interface. (c) What is the reflection coefficient and the phase change in the reflected wave when the angle of incidence i = 79? (d) What is the penetration depth of the evanescent wave into medium 2 when i = 79 and when i = 89? What is your conclusion? Solution (a) Given,

n1  3.60 n2  3.30

n1  n2 3.60  3.30   0.043 n1  n2 3.60  3.30

we have

r/ /  r 



R  R/ /  r  r/ /  0.0018 2

2n1 3.60  2  1.043 3.60  3.30 n1  n2 4n1n2 4  3.60  3.30 T = T  T/ /    0.998 2 (n1  n2 ) (3.60  3.30)2 t/ /  t 

and

 (b)

2

Apply

n2 n1 n sin  c  2 n1 n1  3.60

tan  p  and Take

n2  3.30

 



(c)

Take



 p  tan 1 (n2 / n1 )  tan 1 (3.30 / 3.60)  42.51 n 3.30  c  sin 1 2  sin 1  66.44 3.60 n1 n

n2 3.30   0.9166 n1 3.60

cos i  [n 2  sin 2 i ]1/2 r  cos  i  [n 2  sin 2  i ]1/ 2

then,

cos(79 )  [(0.9166) 2  sin 2 (79 )]1/ 2 cos(79 )  [(0.9166) 2  sin 2 (79 )]1/2 0.1908  0.3513 j  0.1908  0.3513 j 



Chapter 1

1.28

11 December 2012

r 

0.1908  0.3513 j 0.1908  0.3513 j  0.1908  0.3513 j 0.1908  0.3513 j 0.19082  0.35132  2 j 0.1908  0.3513 0.1598 0.087  0.1340 j  0.1598  0.5444  0.8385 j

and





r  0.5444  j (0.8385)

r  0.9997exp( j 237 )  We know that r | r | exp( j ) , thus

0.9997exp( j 57 )  0.9997exp( j )  

exp( j )  exp( j 237 )

  237   123 1/2

sin 2 i  n 2  or tan   cos i    122.98 = 123 Now the parallel component, r//,



1 2



1/2

sin 2 (79)  (0.9166) 2   cos(79)

[n 2  sin 2  i ]1/ 2  n 2 cos i r/ /  2 [n  sin 2 i ]1/2  n 2 cos i [(0.9166) 2  sin 2 (79 )]1/2  (0.9166) 2 cos(79 ) [(0.9166)2  sin 2 (79 )]1/2  (0.9166) 2 cos(79 ) 0.3513 j  0.1603  0.3513 j  0.1603 

r/ / 



 We know that

0.3513 j  0.1603 0.3513 j  0.1603  0.3513 j  0.1603 0.3513 j  0.1603 0.0977  j 0.1126  0.1491  0.6552  j 0.7552

r/ /  0.6552  j (0.7552) r/ /  1.088  exp( j 49.05 ) r/ / | r // | exp( j // ) , thus exp( j // )  exp( j 49.05 )



Chapter 1

1.29

11 December 2012



/ /  49.05

or,

  



 / /  12   





1 1 tan 12   tan 2 0.91662 n 1 1 ( / /   )  tan (2.1913)  65.47 2

tan

1 2



1 2



122.98  2.1913

/ /   )  130.94 / /  130.94  180  49.05 (

(d) The attenuation coefficient 2 for i =79° is 2  2n2  n1  2    sin 2  i  1 o  n2   

1/ 2

1/2

i.e.

2  2 (3.30)  3.60  2   2  sin (79 ) 1     (890 109 m)  3.30  

 8.92 106 m 1 .

so the penetration depth is 1/2 = 1/(8.92106 m-1) = 1.1210-7 m, or 0.112 µm For 89°, repeating the calculation 1/ 2

2  2 (3.30)  3.60  2   2  sin (89 ) 1     (890 109 m)  3.30  

 1.01107 m 1

So, the penetration depth is 1/2 = 1/( 1.01107 m1 ) = 9.910-8 m, or 990nm. The conclusion is that the penetration depth decreases as the incidence angle increases 1.22 Phase changes on TIR Consider a light wave of wavelength 870 nm traveling in a semiconductor medium (GaAs) of refractive index 3.60. It is incident on a different semiconductor medium (AlGaAs) of refractive index 3.40, and the angle of incidence is 80. Will this result in total internal reflection? Calculate the phase change in the parallel and perpendicular components of the reflected electric field. Solution

 c  sin 1

n2 3.40  sin 1  70.81 3.60 n1

i  80  c Clearly, So, this results in total internal reflection. n

n2 3.40   0.9444 n1 3.60



Chapter 1

1.30

11 December 2012

tan 

 

 

1/2

sin 2 i  n 2   cos i

1/2

sin 2 (80)  (0.9444)2   cos(80)

 1.6078

  2 tan 1 (1.6078)  116.24



 / /  12   





1 1 tan 12   tan 2 0.94442 n 1 1 ( / /   )  tan (1.8027)  60.98 2

tan 



1 2

1 2



1 2



116.24  1.8027

/ /   )  121.96 / /  121.96  180  58.04 (

1.23 Fresnel’s equation Fresnel's equations are sometimes given as follows: E n cos i  n 2 cos t r   r 0,  1 Ei 0, n1 cos i  n 2 cos t E n cos t  n 2 cos i r //  r 0,//  1 Ei 0,// n1 cos t  n 2 cos i E 2n1 cos i t   t 0,  Ei 0, n1 cos i  n 2 cos t E 2n1 cos i and t //  t 0,//  Ei 0,// n1 cos t  n 2 cos i Show that these reduce to Fresnel's equation given in Section 1.6. Using Fresnel's equations, find the reflection and transmission coefficients for normal incidence and show that and r //  nt //  1 r  1  t  where n = n2/n1. Solution

From Snell’s law



n2 / n1  n

sin i n2  n sin t n1 sin 2 t 

sin 2 i n2

Perpendicular component r 



Er 0, Ei 0,



n1 cos i  n2 cos t cos i  n2 / n1 cos t  n1 cos i  n2 cos t cos i  n2 / n1 cos t

cos i  n cos t cos i  n[1  sin 2 t ]1/2  r  cos i  n cos t cos i  n[1  sin 2 t ]1/ 2



Chapter 1

1.31

11 December 2012

sin 2 i 1/2 ] 2 n r  sin 2 i 1/2 cos i  n[1  ] n2 cos i  [n 2  sin 2 i ]1/2 r  cos i  [n 2  sin 2 i ]1/2 cos i  n[1 



 Parallel component

r/ / 



n1 cos t  n2 cos i cos t  n2 / n1 cos i cos t  n cos i   n1 cos t  n2 cos i cos t  n2 / n1 cos i cos t  n cos i

2n1 cos i 2 cos i 2 cos i   Ei 0,  n1 cos i  n2 cos t cos i  n2 / n1 cos t cos i  n cos t 2 cos i 2 cos i t   2 1/2 sin 2 i 1/2 cos i  n[1  sin t ] cos i  n[1  ] n2 2 cos i t  cos i  [n 2  sin 2 i ]1/ 2 E 2n1 cos i 2 cos i 2 cos i t / /  t 0,/ /    Ei 0,/ / n1 cos t  n2 cos i cos t  n2 / n1 cos i cos t  n cos i

t  









Ei 0,/ /



sin 2 i 1/2 [1  ]  n cos i [1  sin 2 t ]1/2  n cos i n2 r/ /   sin 2 i 1/2 [1  sin 2 t ]1/2  n cos i [1  ]  n cos i n2 [n 2  sin 2 i ]1/2  n 2 cos i r/ /  2 [n  sin 2 i ]1/2  n 2 cos i





Er 0,/ /





Et 0,



t// 

2 cos i  cos t  n cos i

2 cos i

t// 

2n cos i [n  sin 2 i ]1/2  n 2 cos i

[1 

sin i 1/2 ]  n cos i n2 2

2

cos i  [n 2  sin 2 i ]1/2 r  1  1 cos i  [n 2  sin 2 i ]1/2 cos i  [n 2  sin 2 i ]1/2  cos i  [n 2  sin 2 i ]1/2  cos i  [n 2  sin 2 i ]1/2



Chapter 1

1.32

11 December 2012

 

2 cos i cos i  [n 2  sin 2 i ]1/ 2 r  1  t r  1 

[n 2  sin 2 i ]1/2  n 2 cos i 2n cos i r/ /  nt / /  2 n 2 2 1/2 2 [n  sin i ]  n cos i [n  sin 2 i ]1/2  n 2 cos i   

[n 2  sin 2 i ]1/2  n 2 cos i  2n 2 cos i r/ /  nt / /  [n 2  sin 2 i ]1/ 2  n 2 cos i [n 2  sin 2 i ]1/2  n 2 cos i r/ /  nt / /  2 [n  sin 2 i ]1/2  n 2 cos i r/ /  nt / /  1

1.24 Fresnel's equations Consider a light wave traveling in a glass medium with an index n1 = 1.440 and it is incident on the glass-air interface. Using Fresnel equations only i.e. Eqs (6a) and 6(b) in §1.6, calculate the reflection coefficients r and r// and hence reflectances R and R// for (a) i = 25 and (b) i = 50. In the case of i = 50, find the phase change  and // from the reflection coefficients by writing r = |r|exp(j). Compare  and // from r and r// calculations with those calculated from Eqs (11) and (12). Solution

The above problem is solved using LiveMath and reproduced below. It should be relatively straightforward to follow.

na = n1; nb = n2; r = r//; r = r; subscript means ;  means //.


1.25 Goos-Haenchen phase shift A ray of light which is traveling in a glass medium (1) of refractive index n1 = 1.460 becomes incident on a less dense glass medium (2) of refractive index n2 = 1.430. Suppose that the free space wavelength of the light ray is 850nm. The angle of incidence i = 85. Estimate the lateral Goos-Haenchen shift in the reflected wave for the perpendicular field component. Recalculate the Goos-Haenchen shift if the second medium has n2 = 1 (air). What is your conclusion? Assume that the virtual reflection occurs from a virtual plane in medium B at a distance d that is the same as the penetration depth. Note that d actually depends on the polarization, the direction of the field, but we will ignore this dependence.



Chapter 1

1.34

Solution

Figure 1.20 The reflected light beam in total internal reflection appears to have been laterally shifted by an amount z at the interface. It appears as though it is reflected from a virtual plane at a depth d in the second medium from the interface.

The problem is shown in Figure 1.20. When i = 85, 1/ 2 2  2n2  n  1 2   sin  i  1 2  o  n 2   7 The penetration depth is  = 1/2 = 5.0910 m. As an estimate, we can assume that d ~  so that the Goose-Haenchen shift is z  2dtan = 2(5.0910-7 m)(tan85) = 11.610-6 m = 11.6 m We can repeat the calculation using n2 = 1 (air), then we find  = 1/2 = 1.2810-7 m, and z  2dtan = 2(1.2810-7 m)(tan85) = 2.9310-6 m = 2.93 m. The shift is small when the refractive index difference is large. The wave penetrates more into the second medium when the refractive index difference is smaller. Note: The use of d   is a rough approximation to estimate z. 1.26 Evanescent wave Total internal reflection (TIR) of a plane wave from a boundary between a more dense medium (1) n1 and a less dense medium (2) n2 is accompanied by an evanescent wave propagating in medium 2 near the boundary. Find the functional form of this wave and discuss how its magnitude varies with the distance into medium 2. Solution

The transmitted wave has the general form Et, = tEio,expj(tktr) in which t is the transmission coefficient. The dot product, examining ktr = yktcost + zkt sint.

However, from Snell's law, when i > c, sint = (n1/n2)sini > 1 and cost = [1 sin2t] = ±jA2 is a purely imaginary number. Thus, taking cost = jA2 Et, = tEio,expj(t – zktsint + jyktA2) = tEio,exp(yktA2)expj(t – zktsint)



Chapter 1

1.35

which has an amplitude that decays along y as exp(–2y) where 2 = ktA2. Note that jA2 is ignored because it implies a wave in medium 2 whose amplitude and hence intensity grows. Consider the traveling wave part expj(t – zktsint). Here, ktsint = kisini (by virtue of Snell's law). But kisini = kiz, which is the wave vector along z, that is, along the boundary. Thus the evanescent wave propagates along z at the same speed as the incident and reflected waves along z. Furthermore, for TIR we need sini > n2/n1. This means that the transmission coefficient,

t 

ni cos i 12

n 2  cos i   2   sin2  i  n1  

 t 0 exp j   

must be a complex number as indicated by texp(j) in which t is a real number and  is a phase change. Note that t does not, however, change the general behavior of propagation along z and the penetration along y. 1.27 TIR and FTIR (a) By considering the electric field component in medium in Figure 1.22(b), explain how you can adjust the amount of transmitted light. (b) What is the critical angle at the hypotenuse face of a beam splitter cube (Figure 1.22 (b)) made of glass with n1 = 1.6 and having a thin film of liquid with n2 =1.3. Can you use 45 prisms with normal incidence? (c) Explain how a light beam can propagate along a layer of material between two different media as shown in Figure 1.59 (a). Explain what the requirements are for the indices n1, n2, n3. Will there be any losses at the reflections? (d) Consider the prism coupler arrangement in Figure 1.59(b). Explain how this arrangement works for coupling an external light beam from a laser into a thin layer on the surface of a glass substrate. Light is then propagated inside the thin layer along the surface of the substrate. What is the purpose of the adjustable coupling gap?

Figure 1.22 (a) A light incident at the long face of a glass prism suffers TIR; the prism deflects the light. (b) Two prisms separated by a thin low refractive index film forming a beam-splitter cube. The incident beam is split into two beams by FTIR.



Chapter 1

1.36

Figure 1.59 (a) Light propagation along an optical guide. (b) Coupling of laser light into a thin layer - optical guide - using a prism. The light propagates along the thin layer.

Solution (a) Consider the prism A when the neighboring prism C in Figure 1.22 (b) in far away. When the light beam in prism A is incident on the A/B interface, hypotenuse face, it suffers TIR as i > c. There is however an evanescent wave whose field decays exponentially with distance in medium B. When we bring prism C close to A, the field in B will reach C and consequently penetrates C. (The tangential field must be continuous from B to C). One cannot just use the field expression for the evanescent wave because this was derived for a light beam incident at an interface between two media only; no third medium. The transmitted light intensity from A to C depends on the thickness of B. (b) For the prism A in Figure 1.22 (b), n1 = 1.6 and n2 = 1.3 so that the critical angle for TIR at the hypotenuse face is







c = arcsin(n2/n1) = arcsin(1.3/1.6) = 54.3

In this case, one cannot use a 45  prism. If the angle of incidence i at the n1/n2 layer is more than the critical angle c12 and if angle of (d) incidence i at the n1/n3 layer is more than the critical angle 13 then the light ray will travel by TIR, zigzagging between the boundaries as sketched in Figure 1.59(a). For example, suppose that n1 = 2 (thin layer); n2 = 1 (air) and n3 = 1.6 (glass),

c12 = arcsin(n2/n1) = arcsin(1/2) = 38.8,



Chapter 1

1.37

c13 = arcsin(n3/n1) = arcsin(1.6/2) = 53.1,

and

so that i > 53.1 will satisfy TIR. There is no loss in TIR as the magnitude of the amplitude of the reflected way is the same as that of the incident wave. Note: There is an additional requirement that the waves entering the thin film interfere constructively, otherwise the waves will interfere destructively to cancel each other. Thus there will be an additional requirement, called the waveguide condition, which is discussed in Chapter 2. (e) The light ray entering the prism is deflected towards the base of the prism as shown in Figure 1.59 (b). There is a small gap between the prism and the thin layer. Although the light arriving at the prism base/gap interface is reflected, because of the close proximity of the thin layer, some light is coupled into the thin layer per discussion in Part (a) due to frustrated TIR. This arrangement is a much more efficient way to couple the light into the thin layer because the incident light is received by the large hypotenuse face compared with coupling the light directly into the thin layer. 1.28 Complex refractive index and dielectric constant The complex refractive index  = n  jK can be defined in terms of the complex relative permittivity r = r1 jr2as  = n  jK =  r1/ 2 = r1  jr 2 where r1 and r2 are the real and imaginary parts of r. Show that

 ( 2   r22 )1 / 2   r1  n   r1  2  

1/ 2

 ( 2   2 )1/ 2   r1  K   r1 r 2  2  

and

1/ 2

Solution

Given

  n  jK   r   r1  j r 2

we have n 2  2 jnK  K 2   r1  j r 2  2nK   r 2 and n 2  K 2   r1  K   r 2 / 2n and substituting into the second equation above,

(1) (2)

2



  n   r 2    r1  2n  n 4  n 2 r1  14  r22  0



n2 

2

 r1   r21  4(  14  r22 ) 2



 r1   r21   r22 2

2

It is apparent that n has two solutions. The negative sign has to be excluded because this would make the numerator negative and lead to a complex number for n. By definition, n is a real number, and not imaginary. Thus, n2 

 r1   r21   r22 2



Chapter 1

1.38

1/ 2

 ( 2   2 )1 / 2   r1  n   r1 r 2  2  



From Eq. (1), n =r2/2K, Substitute this into Eq. (2), 2

 r2     K 2   r1 2 K   4 K  K 2 r1  14  r22  0



  r1   r21  4(  14  r22 )   r1   r21   r22 K   2 2



2

Where the negative sign is excluded as K cannot imaginary. Thus, 1/ 2

 ( 2   2 )1 / 2   r1  K   r1 r 2  2  



1.29 Complex refractive index Spectroscopic ellipsometry measurements on a germanium crystal at a photon energy of 1.5 eV show that the real and imaginary parts of the complex relative permittivity are 21.56 and 2.772 respectively. Find the complex refractive index. What is the reflectance and absorption coefficient at this wavelength? How do your calculations match with the experimental values of n = 4.653 and K = 0.298, R = 0.419 and  = 4.53  106 m-1? Solution From problem 1.28 we have 1/ 2

 ( 2   2 )1 / 2   r1  n   r1 r 2  2  

1/ 2

 ( 21.562  2.7722 )1 / 2  21.56    2  

 4.653

Similarly 1/ 2

 ( 2   2 )1 / 2   r1  K   r1 r 2  2  

1/ 2

 ( 21.562  2.7722 )1 / 2  21.56    2  

 0.298

Almost an exact agreement (not surprisingly). The reflectance R is given by ( n  1) 2  K 2 (4.653  1) 2  0.2982   0.42 or 42% R ( n  1) 2  K 2 (4.653  1) 2  0.2982 The absorption coefficient  is 2k as in Eq. (1.8.67) so that

 = 2k = 2koK = 2(2/o)K = 2(2/c)K 



2(2 )hK 2(2 )(1.5 eV)(0.298)  = 4.5310-6 m-1 15 8 -1 hc (4.136  10 eV s)(3  10 m s )

which agrees with the measurements. 1.30 Complex refractive index Figure 1.26 shows the infrared extinction coefficient K of CdTe . Calculate the absorption coefficient α and the reflectance R of CdTe at 60 μm and 80 μm.



Chapter 1

1.39

Solution

At 60 μm: K  1, and n  0.6, so that the corresponding free-space wave vector is ko = 2/ = 2/(6010-6 m) = 1.05105 m-1. The absorption coefficient  is 2k as in Eq. (1.8.6) so that

 = 2k = 2koK = 2(1.05105 m-1)(1) = 2.1105 m-1 which corresponds to an absorption depth 1/ of about 4.8 micron. The reflectance is R

( n  1) 2  K 2 (0.6  1) 2  12 = 0.32 or 32%  ( n  1) 2  K 2 (0.6  1)2  12

At 80 μm: K  0.27, and n  4.5, so that the corresponding free-space wave vector is ko = 2/ = 2/(8010-6 m) = 7.85104 m-1. The absorption coefficient  is 2k so that

 = 2k = 2koK = 2(7.85104 m-1)(0.27) = 4.2104 m-1 which corresponds to an absorption depth 1/ of about 24 micron. The reflectance is ( n  1) 2  K 2 (0.6  1) 2  12 = 0.41 or 41% R  ( n  1) 2  K 2 (0.6  1)2  12

1.31 Refractive index and attenuation in the infrared region - Reststrahlen absorption Figure 1.26 shows the refractive index n and the attenuation (absorption) coefficient K as a function of wavelength  in the infrared for a CdTe crystal due to lattice absorption, called Reststrahlen absorption. It results from the ionic polarization of the crystal induced by the optical field in the light wave. The relative permittivity r due to positive (Cd2+) and negative (Te) ions being made to oscillate by the optical field about their equilibrium positions is given in its simplest form by



 r   r  j r   rH 

Chapter 1

1.40

 rH   rL

(1)         1  j   T  T   T  where rL and r are the relative permittivity at low (L) and high (H) frequencies, well below and above the infrared peak,  is a loss coefficient characterizing the rate of energy transfer from the EM wave to lattice vibrations (phonons), and T is a transverse optical lattice vibration frequency that is related to the nature of bonding between the ions in the crystal. Table 1.3 provides some typical values for CdTe and GaAs. Eq. (1) can be used to obtain a reasonable approximation to the infrared refractive index n and absorption K due to Reststrahlen absorption. (a) Consider CdTe, and plot n and K vs.  from 40 m to 90 m and compare with the experimental results in Figure 1.26 in terms of the peak positions and the width of the absorption peak. (b) Consider GaAs, and plot n and K vs.  from 30 m to 50 m. (c) Calculate n and K for GaAs at  = 38.02 m and compare with the experimental values n = 7.55 and K = 0.629. 2

Figure 1.26 Optical properties of CdTe as a function of wavelength in the infrared region. (c) Calculate n and K for GaAs at  = 38.02 m and compare with the experimental values n = 7.55 and K = 0.629. Solution From Question 1.28



Chapter 1

1.41

1 2 2 K  n 2   r   r   r    r   2 which means that we can substitute the values of r and r from Eq. (1) into the above two equations and plot n and K as a function of wavelength. (a) CdTe n

(b) GaAs

More points can also be used (c) For GaAs at 38.02 m, the calculated values are n = 7.44 and K = 0.586, which compare reasonable well with experimental values of n = 7.55 and K = 0.629. 1.32 Coherence length A particular laser is operating in single mode and emitting a continuous wave lasing emission whose spectral width is 1 MHz. What is the coherence time and coherence length? Solution The spectral width in frequency  and the coherence time t are related by 1   t



Chapter 1

1.42

Thus, the coherence time is t  1/  = 1/(1106 Hz) =10-6 s or 1µs. The coherence length is lc = ct = (3 108 m s 1 ) 10-6 s = 300m 1.33 Spectral widths and coherence (a) Suppose that frequency spectrum of a radiation emitted from a source has a central frequency o and a spectral width . The spectrum of this radiation in terms of wavelength will have a central wavelength o and a spectral width . Clearly, o = c/o. Since  << o and  << o, using  = c/, show that the line width  and hence the coherence length lc are

  

o 2   o o c

and

l c  c t 

o2 





(b) Calculate  for a lasing emission from a He-Ne laser that has o = 632.8 nm and   1.5 GHz. Find its coherence time and length. Solution (a) See Example 1.9.3 (b) Consider the width in wavelength,  = (2/c) =(1.5109 s-1 (632/810-9 m)2/(3108 m s-1 = 3.1610-6 m. The coherence time is t  1/  = 1/(1.5109 Hz) = 0.66610-9 s The coherence length is lc = ct = (3108 m s-1)(0.66610-9 s) = 0.20 m = 20 cm 1.34 Coherence lengths Find the coherence length of the following light sources (a)

An LED emitting at 1550 nm with a spectral width 150 nm

(b)

A semiconductor laser diode emitting at 1550 nm with a spectral width 3 nm

(c)

A quantum well semiconductor laser diode emitting at 1550 nm with a spectral with of 0.1 nm

(d)

A multimode HeNe laser with a spectral frequency with of 1.5 GHz

(e)

A specially designed single mode and stabilized HeNe laser with a spectral width of 100 MHz

Solution  d c   2 (a)  d  so that  = (c/2) =(15010-9 m)(3108 m s-1(155010-9 m)2 = 1.8731013 Hz Thus, the coherence time is t  1/  = 1/(1.8731013 Hz) = 5.3410-14 s The coherence length is lc = ct = 1.610-5 m or 16m (b)   = (c/2) =(310-9 m)(3108 m s-1(155010-9 m)2 = 3.7461011 Hz



Chapter 1

1.43

The coherence time is

t  1/  = 1/(3.7461011Hz) = 2.6710-12 s or 2.67 fs The coherence length is lc = ct = 8.0110-4 m or 0.8 mm (c) Apply  = (c/2), that is,  = (c/2) =(0.110-9 m)(3108 m s-1(155010-9 m)2 = 1.2481010 Hz

The coherence time is

t  1/  = 1/(1.2481010 Hz) = 8.0110-11 s or 80.1 fs The coherence length is lc = ct = 2.410-2 m or 24 mm (d)

Apply

 

1 t

Thus, the coherence time is t  1/  = 1/(1.5109 Hz) =6.6610-10 s or 666 fs. The coherence length is lc = ct = (3 108 m s 1 ) 6.6610-10 s = 0.2 m = 20 cm (e) 1   t Thus, the coherence time is t  1/  = 1/(100106 Hz) =10-8 s The coherence length is lc = ct = (3 108 m s 1 ) 10-8 s = 3 m 1.35 Fabry-Perot optical cavity Consider an optical cavity formed between two identical mirrors. The cavity length is 50 cm and the refractive index of the medium is 1. The mirror reflectances are 0.97 each. What is the nearest mode number that corresponds to a radiation of wavelength 632.8 nm? What is the actual wavelength of the mode closest to 632.8 nm? What is the mode separation in frequency and wavelength? What are the finesse F and Q-factors for the cavity? Solution For = o = 632.8 nm, the corresponding mode number mo is, mo = 2L o = (20.5 m) / (632.810-9 m) = 1580278.1 and actual mo has to be the closest integer value to 1580278.1, that is 1580278 The actual wavelength of the mode closest to 632.8 nm is o = 2L  mo =(20.5 m) / (1580278) = 632.80005 nm The frequency separation m of two consecutive modes is c c c c c  m   m  1 – m      2L 2L  m 1  m 2L (m  1) m



Chapter 1

1.44

c 3  10 8 = 3108 Hz.  2L 2(0.5) The wavelength separation of two consecutive modes is 2 (632.8  10 9 ) 2 = 4.00410-13 m or 0.400 pm.  m  m  2L 2(0.5) Finesse is R 1 / 2 F =103.1 1 R Quality factor is Q = m0F =1580278103.14 = 1.63108  m 

or

1.36 Fabry-Perot optical cavity from a ruby crystal Consider a ruby crystal of diameter 1 cm and length 10 cm. The refractive index is 1.78. The ends have been silvered and the reflectances are 0.99 and 0.95 each. What is the nearest mode number that corresponds to a radiation of wavelength 694.3 nm? What is the actual wavelength of the mode closest to 694.3 nm? What is the mode separation in frequency and wavelength? What are the finesse F and Q-factors for the cavity? Solution Number mode nearest to the emission wavelength is 2L =(210 cm)(1.78)/(694.3 nm) = 512746.65 i.e. i.e. m0 = 512746. m  n The actual wavelength of the mode closest to 694.3 nm is 2 Ln 0  =(210 cm) (1.78)/ (512746) = 694.3008819 nm m0 The frequency separation m of two consecutive modes is c c c c c  m   m 1   m      = 8.43109 Hz. 2 Ln m 1 m 2 Ln 2 Ln (m  1) m The wavelength separation of two consecutive modes is

 

2m

= 0.00135408 nm = 1.35 pm 2 Ln Average geometric reflectance is R = (R1R2)1/2=0.96979 R 1 / 2 = 102.42 Finesse, F  1 R Quality factor, Q = m0F =1580278103.14= 5.25107 1.37 Fabry-Perot optical cavity spectral width Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use Eq. (1.11.3) to plot the peak closest to 632.8 nm for 4 values of R = 0.99, 0.90, 0.75 and 0.6. For each case find the spectral width m,, the finesse F and Q. How accurate is Eq.(1.11.5) in predicting m? (You may want to use a graphing software for this problem.) Solution



 c   m  m   m f ;  2L 

f = c/(2L)

Io (1  R )  4R sin 2 (kL) Io ; kmL = m  (1  R ) 2

I cavity  I max

Chapter 1

 m 

2

f

;

F

R 1 / 2

Cavity resonant frequencies

(2)

Cavity intensity

(3)

Maximum cavity intensity

(4)

Spectral width

(5)

1 R Resonant frequency υm Quality factor, Q    mF Spectral width υm F

1.45

(6)

Cavity fundamental mode is f = c/(2L) = 3.75108 Hz. The Graph below shows that the peak closest to 632.8 nm is 632.80025 nm which corresponds to m = 4.74083251014 Hz.

AU: Arbitrary Units



R

Chapter 1

1.46

, nm exp, MHz

0.99

0.9

0.75

0.6

1.5000E-06 1.12

1.6750E-05 12.55

4.6000E-05 34.46

8.3000E-05 62.18

calc, MHz

1.20

12.58

34.46

61.64

Fcalculation

312.58

29.80

10.88

6.08

Fexperiment

333.70

29.88

10.88

6.03

Q

4.219E+14

3.778E+13

1.376E+13

7.624E+12

Source From Graph From Graph

 m  F

f F

R 1 / 2 1 R

f  m υm Q υ m F

1.38 Diffraction Suppose that a collimated beam of light of wavelength 600 nm is incident on a circular aperture of diameter of 200 m. What is the divergence of the transmitted beam? What is the diameter at a distance 10 m? What would be the divergence if the aperture were a single slit of width 200 m? Solution (a)

sin  o  1.22

 D

 1.22

600 109  3.66 103 6 200 10

 o  0.209



The divergence angle is 2o  0.418 If R is the distance of the screen from the aperture, then the radius of the Airy disk, approximately b, can be calculated from b / R  tan o   o

b  R tan o  10  tan(0.209 )  0.036m  3.6cm Thus, the diameter is 2b = 0.072 m or 7.2 cm (b) Divergence from a single slit of width a is 2 2  600 109 360   2 o    6 103 rad  6 103   0.34 6 a 200 10 2 1.39 Diffraction Consider diffraction from a uniformly illuminated circular aperture of diameter D. The far field diffraction pattern is given by a Bessel function of the first kind and first order, J1, and the intensity at a point P on the angle  with respect to the central axis through aperture is 2

 2 J ( )  I ( )  I o  1     where Io is the maximum intensity,  = (1/2)kDsin is a variable quantity that represents the angular position  on the screen as well as the wavelength (k = 2) and the aperture diameter D. J1() can be calculated from



Chapter 1

J 1 ( ) 

1





0

1.47

cos(   sin  )d

Using numerical integration (or a suitable math software package), plot [J1()/ ] vs.  for  = 0 to 10 using suitable number of points, and then find the zeros. What are the first two that lead to dark rings? Solution 6.000E-01 5.000E-01

1

4.000E-01 3.000E-01

5

2.000E-01

J

1.000E-01 0.000E+00 -1.000E-01 -2.000E-01 -3.000E-01



-4.000E-01

0.00 2.00 4.00 6.00 8.00 10.00

There are two zeros at = 3.80 and 7.00

0.4 0.2

Y 0

0



2

4

6

8

The above is from Livemath (Theorist) The ratio of the intensity of first bright ring to the intensity at the center of the Airy disk 2

  J ( )    1        0  = = 0.017  J 1 (5.14)     5.14 

Additional Problem



Chapter 1

1.48

Consider an aperture that is 50 m in diameter and illuminated by a 550 nm green laser light beam. If the screen is 2 m away, what are the radius of the first dark ring?The first dark ring occurs when  so that  = (1/2)kDsin(1/2)(2)Dsin      = (1/2)kDsin(1/2)[2m(50 m) sin  gives  = 1.16o. Let R be the distance from the aperture to the screen. If the radius of the dark ring is r then r/R = tan. Thus substituting R = 2 m, and  = 1.16o we find r = 0.024 m and 2r = 0.048 m.

1.40 Bragg diffraction Suppose that parallel grooves are etched on the surface of a semiconductor to act as a reflection grating and that the periodicity (separation) of the grooves is 1 micron. If light of wavelength 1.3 m is incident at an angle 89 to the normal, find the diffracted beams. Solution When the incident beam is not normal to the diffraction grating, then the diffraction angle m for the m-th mode is given by, d(sinm  sini = m ; m = 0, 1, 2, 

so that for first order (1 m)(sinm  sin(89) = (+1)(1.3 m) and

(1 m)(sinm  sin(89) = (1)(1.3 m)

Solving these two equations, we find m = complex number for m = 1, and m = 17.5 for m = 1. m = 17.5 for m = 1, in fact, is the only solution.

Figure 1Q40-1 There is only one diffracted beam, which corresponds to m = 1. 1.41 Diffraction grating for WDM Consider a transmission diffraction grating. Suppose that we wish to use this grating to separate out different wavelengths of information in a WDM signal at 1550 nm. (WDM stands of wavelength division multiplexing.) Suppose that the diffraction grating has a periodicity of 2 m. The angle of incidence is 0 with respect to the normal to the diffraction grating. What is the angular separation of the two wavelength component s at 1.550 m and 1.540 m? How would you increase this separation?



Chapter 1

1.49

Solution Consider the transmission grating shown in Figure 1Q41-1 with normal incidence, i = 0

Figure 1Q41-1 Transmission gratings.

The grating equation for normal incidence with the grating in air is given by dsin = m ; m = 0, 1, 2,  in which we need to set d  2μm For   1.550μm 2μm  sin  m 1.550μm  sin  0.775m For m = 1   sin 1 (0.775)  50.08 For   1.540μm   For m = 1



2μm  sin  m 1.540μm sin  0.770 m

  sin 1 (0.770)  50.35 

  50.35  50.08  0.28



Chapter 1

1.50

1.42 A monochromator Consider an incident beam on a reflection diffraction grating as in Figure 1.60. Each incident wavelength will result in a diffracted wave with a different diffraction angle. We can place a small slit and allow only one diffracted wave m to pass through to the photodetector. The diffracted beam would consist of wavelengths in the incident beam separated (or fanned) out after diffraction. Only one wavelength m will be diffracted favourably to pass through the slit and reach the photodetector. Suppose that the slit width is s = 0.1 mm, and the slit is at a distance R = 5 cm from the grating. Suppose that the slit is placed so that it is at right angles to the incident beam: i + m = /2. The grating has a corrugation periodicity of 1 m. (a) What is the range of wavelengths that can be captured by the photodetector when we rotate the grating from i = 1 to 40? (b) Suppose that i = 15. What is the wavelength that will be detected? What is the resolution, that is, the range of wavelengths that will pass through the slit? How can you improve the resolution? What would be the advantage and disadvantage in decreasing the slit width s?

Figure 1.60 A monochromator based on using a diffraction grating Solution Grating equation is d sin( m )  sin( i )   m where m = 0, 1, 2,  Moreover, in this particular

case  m   i  90 0 . Therefore the grating equation may be transformed as

1  1  d sin(90 0   i )  sin( i )   2d sin (90 0   i   i ) cos (90 0   i   i )  2d sin 450   i   m or 2  2  2d  sin 450   i   m (a) The equation shows that the largest may be achieved for m = 1 which is usually done in monochromator. Substituting 400 and 10 into above formula we get the values of 123.3 nm and 982.4 nm that can be captured by the photodetector when we rotate the grating from i = 1 to 40. (b) At i=150 and m = 1 the above formula gives = 707.1 nm Spectral resolution may be found by differentiating the above formula s   2d cos45  i i  2d cos45  i  R which gives = 2.45 nm



Chapter 1

1.51

1.43 Thin film optics Consider light incident on a thin film on a substrate, and assume normal incidence for simplicity. (a) Consider a thin soap film in air, n1 = n2 = 1, n2 = 1.40. If the soap thickness d = 1 m, plot the reflectance vs. wavelengt from 0.35 to 0.75 m, which includes the visible range. What is your conclusion? (b) MgF2 thin films are used on glass plates for the reduction of glare. Given that n1 = 1, n2 = 1.38 and n3 = 1.70. (n for glass depends on the type of glass but 1.6 is a reasonable value), plot the reflectance as a function of wavelength from 0.35 to 0.75 m for a thin film of thickness 0.10 m. What is your conclusion? Solution Substitute = 2dn2(2 in r  r e  j r  r e  j ( 4n2 d /  ) r  1 2  j  1 2  j ( 4n2 d /  ) 1  r1r2 e 1  r1r2 e 2 and plot R = | r | as a function of wavelength from 0.35 to 0.75 m as in the figure. Clearly, certain wavelengths, in this case, violet, green, orange-red are reflected more than others (blue and yellow).

(a)

Figure: Reflectance vs wavelength in the visible range for a soap film (b)

Figure: Reflectance vs wavelength in the visible range for a MgF2 tin film coating on glass The reflectance is lowered substantially by the thin film coating, and remains low over the visible spectrum. Without the coating, the reflectance is 6.0%. With the coating, it is below 1%



Chapter 1

1.52

Authors comment: The above are for normal incidence. Obviously reflections at other angles will have R vs shifted in wavelength. This will affect the spectrum of the reflected light from the soap film but the MgF2 coating will still result in a relatively low reflectance over the visible because the minimum in the reflectance is very broad over the visible range. 1.44 Thin film optics Consider a glass substrate with n3 = 165 that has been coated with a transparent optical film (a dielectric film) with n2 = 2.50, n1 = 1 (air). If the film thickness is 500 nm, find the minimum and maximum reflectances and transmittances and their corresponding wavelengths in the visible range for normal incidence. (Assume normal incidence.) Note that the thin n2-film is not an AR coating, and for n1 < n3 < n2, 2

Rmax

 n2  n n  n n    22 1 3  and Rmin   3 1   n2  n1n3   n3  n1 

2

Solution Minimum reflectance Rmin occurs at = 2 or multiples of 2 , and maximum reflectance Rmax occurs at =  or an odd integer multiple of 2 . The corresponding equations to Eq. (1.11.8) are 2

Rmin

 n  n   1.65  1    3 1      n3  n1   1.65  1  2

2

= 0.060or 6.0 % 2

 n 2  n n   2.52  (1)(1.65)   = 0.34 or 34% Rmax   22 1 3    2 and  n2  n1n3   2.5  (1)(1.65)  Corresponding transmittances are, Tmax = 1 – Rmin = 0.94 or 94% and Tmin = 1 – Rmax = 0.66 or 66%. Since n2 is not an intermediate index between n1 and n3, the n2-film does not reduce the reflection that would have occurred at the n1-n3 interface had there been no n2-layer. Indeed R13 in the absence of n2, is the same as Rmin and the n2-layer increases reflection Since = 2dn2(2, and d = 500 nm, and the wavelengths for maximum reflectance are given by the condition  = (2m+1), m = 0, 1,2we can calculate the maximum reflectance wavelengths max = 4dn2/ 4dn2/(2m+1)]

 max

1 5000

3 1,667

5 1000

7 714

9 555

11 455

13 385

(nm)

1.45 Thin film optics Consider light incident on a thin film on a substrate, and assume normal incidence for simplicity. Plot the reflectance R and transmittance  as a function of the phase change  from = 4 to +4 for the following cases (a) Thin soap film in air, n1 = n2 = 1, n2 = 1.40. If the soap thickness d = 1 m, what are the maxima and minima in the reflectance in the visible range? (b) A thin film of MgF2 on a glass plate for the reduction of glare, where that n1 = 1, n2 = 1.38 and n3 = 1.70. (n for glass depends on the type of glass but 1.7 is a reasonable value). What should be the thickness od MgF2 to for minimum reflection at 550 nm? (c) A thin film of semiconductor on glass where n1 = 1, n2 = 3.5 and n3 = 1.55.



Chapter 1

1.53

Solution

(a) R and T vs  for a thin soap film in air, n1 = n3 = 1, n2 = 1.4 (LiveMath)

(b) R and T vs  for a thin film of MgF2 on glass n1 = 1, n2 = 1.38, n3 = 1.70 (LiveMath)

R and T vs  for a thin film of semiconductor on glass n1 = 1, n2 = 3.5, n3 = 1.55 (LiveMath)

1.46 Transmission through a plate Consider the transmittance of light through a partially transparent glass plate of index n2 in which light experiences attenuation (either by absorption or scattering). Suppose that the plate is in a medium of index n1,the reflectance at each n1-n2 interface is R and the attenuation coefficient is . (a)

Show that Tplate 

(1  R ) 2 e d (1  R 2 )e  2d

(b) If T is transmittance of a glass plate of refractive index n in a medium of index no show that, in the absence of any absorption in the glass plate, n/no= T-1 + (T-2 – 1)1/2 if we neglect any losses in the glass plate. (c) If the transmittance of a glass plate in air has been measured to be 89.96%. What is its refractive index? Do you think this is a good way to measure the refractive index? Solution



Chapter 1

1.54

Figure 1.47 Transmitted and reflected light through a slab of material in which there is no interference. (a) Consider a light beam of unit intensity that is passed through a thick plate of partially transparent material of index n2 in a medium of index n1 as in Figure 1.47. The first transmitted light intensity into the plate is (1R), and the first transmitted light out is (1R) (1R) e d = (1R)2 e d . However, there are internal reflections as shown, so that the second transmitted light is (1R) e d R e d R  e d  (1R) = R2(1R)2 e3 d so that the transmitted intensity through the plate is 3d

Tplate = (1R)2 e d + R2(1R)2 e 2d

R2 e

+ R4

5d

+ R4(1R)2 e

+ … = (1R)2 e d [1 +

e4d + ]

(1  R ) 2 e d 1  R 2 e2 d (b) For the transparent plate =0 and the transmittance of plate becomes 2 (1  R ) 2 e d 1  R  1  R  Tplate  = 1  R 2 e2 d 1 R2 1 R Tplate 

or

2

 n  n0   the equation above becomes Assuming that R    n  n0  2

T plate

 n  n0   1   2 2 2 2 n  n0    n  n0   n  n0  n / n0  1  n / n0  1 2(n / n0 )      2 2 2 2 2 n  n0   n  n0  n / n0  1  n / n0  1 (n / n0 ) 2  1  n  n0   1    n  n0 

which leads to quadratic equation (n / n0 ) 2  2 (n / n0 )  1  0 with the solution n/no= T-1 + (T-2 – 1)1/2 T . (c) The transmittance of 89.96% leads to refractive index of 1.5967. In practice, this is not very good method because it does not give sufficient precision.

1.47 Scattering Consider Rayleigh scattering. If the incident light is unpolarized, the intensity Is of the scattered light a point at a distance r at an angle  to the original light beam is given by



Is 

Chapter 1

1.55

1  cos2  r2

Plot a polar plot of the intensity Is at at a fixed distance r from the scatter as we change the angle  around the scattered. In a polar plot, the radial coordinate (OP in Figure 1.48 b)) is Is. Construct a contour plot in the xy plane in which a contour represents a constant intensity. You need to solve vary r and  or x and y such that Is remains constant. Note x = rcos and y = rsin ;  = arctan(y/x), r = (x2 + y2)1/2. Author's Note: There is a printing error. The minus sign should have been plus as in the above expression. This should have been obvious from Figure 1.48(b). The error will be corrected in the next reprint. The e-version of the book is correct. Solution (a) Polar plot Take I s  1  cos2  as we are interested in the angular dependence only (set r = 1). In the polar plot, the distance from the origin is the intensity. Do not confuse this with r. r is contant but in the polar plot the coordinate r is now the intensity. 2

 1 2 Intensity

3 6

4

Polar plot on LiveMath (Theorist) (b) Contour plot We can set the proportionality constant to 1, and write 1  cos2  Is  r2



r

1  cos2  Is

We can now plot the above on a polar plot in which the distance from the center is r, and r and  pairs of coordinates are such that they always yield a constant Is because we have set Is = constant. We can arbitrarily set I = 1, 2 or 3 to get 3 contour lines. Blue, Is = 1, black, Is = 2 and red, Is = 3 in AU (arbitrary units)



Chapter 1

1.56

2

 0.5 r 1

1.5 6

4

Contour plots on LiveMath (Theorist) Another interesting plot is the density plot in which the density represents the intensity Is. The brightness (density) represents the intensity at a point r,. 2

0 -2

x

0

2

y -2

Density plot of Rayleigh scattering. Brightness represents more light intensity at the point r,

1.48 One dimensional photonic crystal (a Bragg mirror) The 1D photonic crystal in Figure 1.50(a), which is essentially a Bragg reflector, has the dispersion behavior shown in Figure 1.51. The stopband  for normal incidence and for all polarizations of light is given by (R.H. Lipson and C. Lu, Eur. J. Phys. 30, S33, 2009) n n    2(2 /  ) arcsin  2 1  o  n2  n1 

where  is the stop-band, o is the center frequency defined in Figure 1.50(a) and n2 and n1 are the high and low refractive indices. Calculate the lowest stop band in terms of photon energy in eV, and wavelength (nm) for a structure in which n2 = 4 and n1 = 1.5, and n1d1 = n2d2 = /4 and d1 = 2 m. Solution



Chapter 1

1.57

 n n    o 2(2 /  ) arcsin  2 1   n2  n1  n2 = 4 and n1 = 1.5, d1 = 2 m n1d1  n2 d 2   / 4

1.5  2  m  4d 2   / 4 d 2  0.75 m 0  12 m c 0  2  1.57 1014

0

 4  1.5  15   1.57 1014  2(2 /  ) arcsin    5.4110  4  1.5  8 2  3 10   2 c /    3.48 107 m  348nm 5.411015 1 5.411015 E  h / 2  6.62 1034 J s  )  3.56eV 1.6 1019 J eV -1 2 1.49 Photonic crystals Concepts have been borrowed from crystallography, such as a unit cell, to define a photonic crystal. What is the difference between a unit cell used in a photonic crystal and that used in a real crystal? What is the size limit on the unit cell of a photonic crystal? Is the refractive index a microscopic or a macroscopic concept? What is the assumption on the refractive index? Solution The size limit on the unit cell of a photonic crystal is that it must be longer than the wavelength scale. Refractive index is a macroscopic concept.



Chapter 1

1.58

NOTES FROM THE AUTHOR Some of the problems have been solved by using LiveMath (previously Mathview and Theorist). http://livemath.com

LiveMath interpretation This is a comment, and is not used in calculations A square represents a mathematical statement

The first line with a square is a mathematical statement. The second line with a triangle isolates c. It is a mathematical conclusion from the first line. The dot at the center of the triangle represents a working conclusion, something that will be used elsewhere in calculations. The third line with a triangle is a mathematical conclusion from the second line. It calculates c Italic text next to mathematical conclusions explains the mathematical operation e.g. isolate



Chapter 2

2.2

3 February 2013

Preliminary Solutions to Problems and Question Chapter 2 Note: Printing errors and corrections are indicated in dark red. Currently none reported. 2.1 Symmetric dielectric slab waveguide Consider two rays such as 1 and 2 interfering at point P in Figure 2.4 Both are moving with the same incidence angle but have different m wavectors just before point P. In addition, there is a phase difference between the two due to the different paths taken to reach point P. We can represent the two waves as E1(y,z,t) = Eocos(tmymz + ) and E2(y,z,t) = Eocos(tmymz) where the my terms have opposite signs indicating that the waves are traveling in opposite directions.  has been used to indicate that the waves have a phase difference and travel different optical paths to reach point P. We also know that m =k1cosm and m = k1sinm, and obviously have the waveguide condition already incorporated into them through m. Show that the superposition of E1 and E2 at P is given by E ( y, z , t )  2 Eo cos( m y  12  ) cos(t   m z  12  ) What do the two cosine terms represent? The planar waveguide is symmetric, which means that the intensity, E2, must be either maximum (even m) or minimum (odd m) at the center of the guide. Choose suitable  values and plot the relative magnitude of the electric field across the guide for m = 0, 1 and 2 for the following symmetric dielectric planar guide : n1 = 1.4550, n2 = 1.4400, a = 10 m, = 1.5 m (free space), the first three modes have 1 = 88.84, 2 = 87.673 = 86.51. Scale the field values so that the maximum field is unity for m = 0 at the center of the guide. (Note: Alternatively, you can choose  so that intensity (E2) is the same at the boundaries at y = a and y = a; it would give the same distribution.) Solution E ( y )  Eo cos(t   m z   m z   )  Eo cos(t   m z   m z ) Use the appropriate trigonometric identity (see Appendix D) for cosA + cosB to express it as a product of cosines 2cos[(A+B)/2]cos[(AB)/2], E ( y, z , t )  2 Eo cos( m y  12  ) cos(t   m z  12  ) The first cosine term represents the field distribution along y and the second term is the propagation of the field long the waveguide in the z-direction. Thus, the amplitude is Amplitude = 2 Eo cos( m y  12  ) The intensity is maximum or minimum at the center. We can choose  = 0 ( m = 0),  =  ( m = 1),  = 2 ( m = 2), which would result in maximum or minimum intensity at the center. (In fact,  = m). The field distributions are shown in Figure 2Q1-1.



Chapter 2

2.3

3 February 2013

Figure 2Q1-1 Amplitude of the electric field across the planar dielectric waveguide. Red, m = 0; blue, m = 1; black, m = 2.

2.2 Standing waves inside the core of a symmetric slab waveguide Consider a symmetric planar dielectric waveguide. Allowed upward and downward traveling waves inside the core of the planar waveguide set-up a standing wave along y. The standing wave can only exist if the wave can be replicated after it has traveled along the y-direction over one round trip. Put differently, a wave starting at A in Figure 2.51 and traveling towards the upper face will travel along y, be reflected at B, travel down, become reflected again at A, and then it would be traveling in the same direction as it started. At this point, it must have an identical phase to its starting phase so that it can replicate itself and not destroy itself. Given that the wavevector along y is m, derive the waveguide condition.

Figure 2.51 Upward and downward traveling waves along y set-up a standing wave. The condition for setting-up a standing wave is that the wave must be identical, able to replicate itself, after one round trip along y.

Solution From Figure 2.51 it can be seen that the optical path is AB  BA  4a

With the ray under going a phase change  with each reflection the total phase change is   4a m  2



Chapter 2

2.4

3 February 2013

The wave will replicate itself, is the phase is same after the one round-trip, thus   4a m  2  2m and since  m  k1 cos m 

2n1 (2a)



2n1



cos m we get

cos m  m  m

as required. 2.3

Dielectric slab waveguide

(a) Consider the two parallel rays 1 and 2 in Figure 2.52. Show that when they meet at C at a distance y above the guide center, the phase difference is 

m = k12(a  y)cosmm





(b)

Using the waveguide condition, show that

 m   m ( y )  m  (c)

y (m   m ) a

The two waves interfering at C can be most simply and conveniently represented as E ( y )  A cos(t )  A cos[t   m ( y )]

Hence find the amplitude of the field variation along y, across the guide. What is your conclusion?

Figure 2.52 Rays 1 and 2 are initially in phase as they belong to the same wavefront. Ray 1 experiences total internal reflection at A. 1 and 2 interfere at C. There is a phase difference between the two waves.

Solution (a)

From the geometry we have the following: (a y)/AC = cos

and



C/AC = cos(2)

The phase difference between the waves meeting at C is  = kAC kAC = k1AC k1ACcos(2) 



Chapter 2

2.5

3 February 2013







= k1AC[1  cos(2)] k1AC[1 + cos(2)] 







= k1(a y)/cos][ 1 + 2cos2  1]







=k1(a y)/cos][2cos2]







= k1(a y)cos

(b)

 2 (2a )n1  Given,   cos  m  m  m  

 (m  m ) m  m  2n1 (2a) k1 (2a)



cos  m 

Then,

 m  2k1 (a  y ) cos  m  m  2k1 (a  y )



y y   m  1  m  m   m  m  (m  m ) a  a

 m  ( y )  m  (c)

m  m  m k1 (2a)

y y (m  m )  m   m ( y )  m  ( m  m ) a a

The two waves interfering at C are out phase by , E ( y )  A cos(t )  A cos[t   m ( y )]

where A is an arbitrary amplitude. Thus, E  2 A cos[t  12  m ( y )] cos12  m ( y ) or

E  2 A cos12  m ( y )cos(t   ) = Eocos(t + )

in which m/2, and cos(t +) is the time dependent part that represents the wave phenomenon, and the curly brackets contain the effective amplitude. Thus, the amplitude Eo is y  m  Eo  2 A cos   (m  m ) a 2 2   To plot Eo as a function of y, we need to find m for m = 0, 1 , 2…The variation of the field is a truncated) cosine function with its maximum at the center of the guide. See Figure 2Q1-1. 2.4 TE field pattern in slab waveguide Consider two parallel rays 1 and 2 interfering in the guide as in Figure 2.52. Given the phase difference

 m   m ( y )  m 

y (m  m ) a

between the waves at C, distance y above the guide center, find the electric field pattern E (y) in the guide. Recall that the field at C can be written as E ( y )  A cos(t )  A cos[t   m ( y )] . Plot the field pattern for the first three modes taking a planar dielectric guide with a core thickness 20 m, n1 = 1.455 n2 = 1.440, light wavelength of 1.3 m.



Chapter 2

2.6

3 February 2013

Figure 2.52 Rays 1 and 2 are initially in phase as they belong to the same wavefront. Ray 1 experiences total internal reflection at A. 1 and 2 interfere at C. There is a phase difference between the two

Solution

The two waves interfering at C are out phase by , E ( y )  A cos(t )  A cos[t   m ( y )] where A is an arbitrary amplitude. Thus, 1   1  E  2 A cos t   m ( y ) cos   m ( y ) 2   2  or

 1  E  2 A cos   m ( y )  cost   = Eocos(t + ) 2  

in which m/2, and cos(t +) is the time dependent part that represents the wave phenomenon, and the curly brackets contain the effective amplitude. Thus, the amplitude Eo is y  m  Eo  2 A cos   (m  m ) 2a  2  To plot Eo as a function of y, we need to find m for m = 0, 1 and 2 , the first three modes. From Example 2.1.1 in the textbook, the waveguide condition is (2a)k1 cos  m  m  m we can now substitute for m which has different forms for TE and TM waves to find,

TE waves

2   n2   2 sin  m      n1      tan ak1 cos  m  m    2 cos  m 

1/ 2

 f TE ( m )



Chapter 2

2.7

3 February 2013

TM waves

2   n2   2 sin  m      n1      tan ak1 cos  m  m    2 2   n2    cos  m  n1 

1/ 2

 f TM ( m )

The above two equations can be solved graphically as in Example 2.1.1 to find m for each choice of m. Alternatively one can use a computer program for finding the roots of a function. The above equations are functions of m only for each m. Using a = 10 m,  = 1.3 m, n1 = 1.455 n2 = 1.440, the results are: TE Modes

m=0

m=1

m=2

m (degrees)

88.84 

 



m (degrees)

163.75

147.02

129.69

TM Modes

m=0

m=1

m=2

m (degrees)

88.84 

 



m (degrees)

164.08

147.66

130.60

There is no significant difference between the TE and TM modes (the reason is that n1 and n2 are very close).

Figure 2Q4-1 Field distribution across the core of a planar dielectric waveguide

We can set A = 1 and plot Eo vs. y using y  m   (m  m ) Eo  2 cos  2a  2  with the m and m values in the table above. This is shown in Figure 2Q4-1.



Chapter 2

2.8

3 February 2013

2.5 TE and TM Modes in dielectric slab waveguide Consider a planar dielectric guide with a core thickness 20 m, n1 = 1.455 n2 = 1.440, light wavelength of 1.30 m. Given the waveguide condition, and the expressions for phase changes  and  in TIR for the TE and TM modes respectively, 2   n2   2 sin  m       n1   tan  12 m    cos  m

1/ 2

and

2   n2   2 sin  m       n1   tan  12 m    2  n2    cos  m  n1 

1/ 2

using a graphical solution find the angle  for the fundamental TE and TM modes and compare their propagation constants along the guide. Solution

The waveguide condition is (2a)k1 cos  m  m  m we can now substitute for m which has different forms for TE and TM waves to find,

TE waves

TM waves

2   n2   2 sin  m      n1      tan ak1 cos  m  m    2 cos  m 

1/ 2

2   n2   2 sin  m      n1       tan ak1 cos  m  m   2 2   n2    cos  m  n1 

1/ 2

 f TE ( m )

 f TM ( m )

The above two equations can be solved graphically as in Example 2.1.1 to find m for each choice of m. Alternatively one can use a computer program for finding the roots of a function. The above equations are functions of m only for each m. Using a = 10 m,  = 1.3 m, n1 = 1.455 n2 = 1.440, the results are: TE Modes

m=0

m (degrees)

88.8361

m = k1sinm 

7,030,883 m-1

TM Modes

m=0

m (degrees)

88.8340

m= k1sinm

7,030,878m-1





Note that 5.24 m-1 and the -difference is only 7.510-5 %. The following intuitive calculation shows how the small difference between the TE and TM waves can lead to dispersion that is time spread in the arrival times of the TE and TM optical signals.



Chapter 2

2.9

3 February 2013

Suppose that  is the delay time between the TE and TM waves over a length L. Then,

   1 1  (5.24 m 1 )    TE  TM   L vTE vTM    (1.45  1015 rad/s) = 3.610-15 s m-1 = 0.0036 ps m-1. Over 1 km, the TE-TM wave dispersion is ~3.6 ps. One should be cautioned that we calculated dispersion using the phase velocity whereas we should have used the group velocity. 2.6 Group velocity We can calculate the group velocity of a given mode as a function of frequency  using a convenient math software package. It is assumed that the math-software package can carry out symbolic algebra such as partial differentiation (the author used Livemath, , though others can also be used). The propagation constant of a given mode is  = k1sin where  and  imply m and m. The objective is to express  and  in terms of . Since k1 = n1/c, the waveguide condition is



  sin 2   (n2 / n1 ) 2   tan a cos   m   2 cos   sin  so that









1/ 2



tan  arctan sec  sin 2   (n2 / n1 ) 2  m( / 2)  Fm ( ) a



(1)

where Fm() and a function of  at a given m. The frequency  is given by



c c Fm ( )  n1 sin  n1 sin 

(2)

Both  and  are now a function of  in Eqs (1) and (2). Then the group velocity is found by differentiating Eqs (1) and (2) with respect to  i.e.

i.e.

vg 

d  d   d  c  Fm ( ) cos    1         Fm ( )    2 d  d   d  n1  sin  sin    Fm ( ) 

vg 

Fm ( )  c  1  cot    n1 sin   Fm ( ) 

Group velocity, planar waveguide

(3)

where Fm = dFm/d is found by differentiating the second term of Eq. (1). For a given m value, Eqs (2) and (3) can be plotted parametrically, that is, for each  value we can calculate  and vg and plot vg vs. . Figure 2.11 shows an example for a guide with the characteristics in the figure caption. Using a convenient math-software package, or by other means, obtain the same vg vs. behavior, discuss intermodal dispersion, and whether the Equation (2.2.2) is appropriate. Solution [Revised 4 February 2013]

The results shown in Figure 2.11, and Figure 2Q6-1 were generated by the author using LiveMath based on Eqs (1) and (3). Obviously other math software packages can also be used. In the presence of say two



Chapter 2

2.10

3 February 2013

modes (TE0 and TE1) and near cutoff, the important conclusion from Figure 2.11 is that although the maximum group velocity is ~c/n2 (near cut-off) minimum group velocity is not c/n1 and can be much lower. For just 2 modes near cut-off, vgmax  c/n2 and vgmin  c/n1, that is, taking the group velocity as the phase velocity. Thus, it is only approximate. Further, in the presence of many modes, group velocity is well below c/n2 and below c/n1 as shown in Figure 2Q6-2 Notice that when there are many, many modes then the fastest is c/n1 (lowest mode) and the slowest is the highest mode at the operating frequency, in this example m = 65. The highest mode will have an incidence angle close to qc so that its group velocity will very roughly (c/n1)sinc = (c/n1)(n2/n1). Thus, in the presence of numerous modes vgmax  c/n1 and vgmin  (cn2/n12) so that the dispersion is L = 1/vgmin 1/vgmax  n1/cn12/(cn2)  (n1/n2)(n2n1)/c  (n2n1)/c which is the same expression as before since n1/n2 is close to unity. Equation (2.2.2) can be used in the presence of just a few modes near cut-off or in the presence of many modes.

Figure 2Q6-1 Group velocity vs. angular frequency  for three modes, TE0 (red), TE1 (blue) and TE4 (orange) in a planar dielectric waveguide. The horizontal black lines mark the phase velocity in the core (bottom line, c/n1) and in the cladding (top line, c/n1). (LiveMath used)



Chapter 2

2.11

3 February 2013

Figure 2Q6-2 Group velocity vs. angular frequency  for many modes

2.7 Dielectric slab waveguide Consider a dielectric slab waveguide that has a thin GaAs layer of thickness 0.2 m between two AlGaAs layers. The refractive index of GaAs is 3.66 and that of the AlGaAs layers is 3.40. What is the cut-off wavelength beyond which only a single mode can propagate in the waveguide, assuming that the refractive index does not vary greatly with the wavelength? If a radiation of wavelength 870 nm (corresponding to bandgap radiation) is propagating in the GaAs layer, what is the penetration of the evanescent wave into the AlGaAs layers? What is the mode field width (MFW) of this radiation? Solution

Given n1 = 3.66 (AlGaAs), n2 = 3.4 (AlGaAs), 2a = 210-7 m or a = 0.1 m, for only a single mode we need

V





2a



(n12  n22 )1 / 2 

2a(n12  n22 )1 / 2





 2 2 (0.1 μm)(3.66 2  3.40 2 )1 / 2

2

= 0.542 m.



2

The cut-off wavelength is 542 nm. When  = 870 nm,

V

2 (1 μm)(3.66 2  3.40 2 )1/ 2 = 0.979 < /2 (0.870 μm)



Therefore,  = 870 nm is a single mode operation. For a rectangular waveguide, the fundamental mode has a mode field width

2 wo  MFW  2a

0.979  1 V 1  (0.2 μm) = 0.404 m. 0.979 V

The decay constant  of the evanescent wave is given by,



V 0.979  =9.79 (m)-1 or 9.79106 m-1. a 0.1 μm

The penetration depth







 = 1/ = 1/ [9.79 (m)-1] = 0.102m.

The penetration depth is half the core thickness. The width between two e-1 points on the field decays in the cladding is Width = 2a + 2× = 0.2 m + 2(0.102) m = 0.404 m.



Chapter 2

2.12

3 February 2013

2.8 Dielectric slab waveguide Consider a slab dielectric waveguide that has a core thickness (2a) of 20 m, n1 = 3.00, n2 = 1.50. Solution of the waveguide condition in Eq. (2.1.9) (in Example 2.1.1) gives the mode angles 0 and 1for the TE0 and TE1 modes for selected wavelengths as summarized in Table 2.7. For each wavelength calculate  and m and then plot  vs. m. On the same plot show the lines with slopes c/n1 and c/n2. Compare your plot with the dispersion diagram in Figure 2.10 Table 2.7 The solution of the waveguide condition for a = 10 m, n1 = 3.00, n2 = 1.50 gives the incidence angles 0 and 1 for modes 0 and 1 at the wavelengths shown.

m

15

20

25

30

40

45

50

70

100

150

200

0

77.8

74.52

71.5

68.7

63.9

61.7

59.74

53.2

46.4

39.9

36.45

1

65.2

58.15

51.6

45.5

35.5

32.02

30.17

‐

‐

‐

‐

Solution

Consider the case example for = 25 m = 25×10-6 m. The free space propagation constant k = 2/ = 225×10-6 m = 2.513×105 m-1. The propagation constant within the core is k1 = n1k = (3.00)( 2.513×105 m-1) = 7.540×105 m-1. The angular frequency  = ck = (3×108 m s-1)( 2.513×105 m-1) = 7.54×1013 s-1. Which is listed in Table 2Q8-1 in the second row under = 25 m. The propagation constant along the guide, along z is given by Eq. (2.1.4) so that

m = k1sinm or

0 = k1sin0 = (7.540×105 m-1)sin(71.5) = 7.540×105 m-1 = 7.15×105 m-1.

which is the value listed in bold in Table 2Q8-1 for the m = 0 mode at  = 25 m. Similarly

1 = k1sin1 = (7.540×105 m-1)sin(51.6) = 7.540×105 m-1 = 5.91×105 m-1.

which is also listed in bold in Table 2Q8-1. We now have both 0 and  1 at  = 2.54×1013 s-1. We can plot this 1 point for the m =0 mode at 0 = 7.15×105 m-1 along the x-axis, taken as the -axis, and  = 2.54×1013 s-1 along the y-axis, taken as the -axis, as shown in Figure 2Q8-1. We can also plot the 1 point we have for the m = 1 mode. Propagation constants () at other wavelengths and hence frequencies () can be similarly calculated. The results are listed in Table 2Q8-1 and plotted in Figure 2Q8-1. This is the dispersion diagram. For comparison the dispersion  vs  for the core and the cladding are also shown. They are drawn so that the slope is c/n1 for the core and c/n2 for the cladding. Thus, the solutions of the waveguide condition as in Example 2.1.1 generates the data in Table 2Q8-1 for 2a = 10 m, n1 = 3; n2 = 1.5. Table2Q8-1 Planar dielectric waveguide with a core thickness (2a) of 20 m, n1 = 3.00, n2 = 1.50. m

15

20

25

30

40

45

50

70

100

150

200



Chapter 2

2.13

3 February 2013



12.6

9.43

7.54

6.283

4.71

4.19

3.77

2.69

1.89

1.26

0.94

0

77.8

74.52

71.5

68.7

63.9

61.7

59.74

53.2

46.4

39.9

36.45

1

65.2

58.15

51.6

45.5

35.5

32.02

30.17

‐

‐

‐

‐

0

12.3

9.08

7.15

5.85

4.23

3.69

3.26

2.16

1.37

0.81

0.56

11.4

8.01

5.91

4.48

2.74

2.22

1.89

‐

‐

‐

‐

1013 s‐1

105 1/m 1

105 1/m

Figure 2Q8-1 Dispersion diagram for a planar dielectric waveguide that has a core thickness (2a) of 20 m, n1 = 3.00, n2 = 1.50. Black, TE0 mode. Purple: TE1 mode. Blue: Propagation along the cladding. Red: Propagation along the core.

Author's Note: Remember that the slope at a particular frequency  is the group velocity at that frequency. As apparent, for the TE0 (m = 0) mode, this slope is initially (very long wavelengths) along the blue curve at low frequencies but then along the red curve at high frequencies (very short wavelengths). The group velocity changes from c/n2 to c/n1. 2.9 Dielectric slab waveguide Dielectric slab waveguide Consider a planar dielectric waveguide with a core thickness 10 m, n1 = 1.4446, n2 = 1.4440. Calculate the V-number, the mode angle m for m = 0 (use a graphical solution, if necessary), penetration depth, and mode field width, MFW = 2a + 2, for light wavelengths of 1.0 m and 1.5 m. What is your conclusion? Compare your MFW calculation with 2wo = 2a(V+1)/V. The mode angle 0, is given as 0 = 88.85 for  = 1 m and 0 = 88.72 for  = 1.5 m for the fundamental mode m = 0. Solution



Chapter 2

2.14

3 February 2013

 = 1 m, n1 = 1.4446, n2 = 1.4440, a = 5 m. Apply V to obtain

2a



n

2 1

 n22



1/ 2

V = 1.3079

Solve the waveguide condition 2   n2   2 sin  m      n1       tan ak1 cos  m  m   2 cos  m 

1/ 2

 f ( m )

graphically as in Example 2.1.1 to find: c = 88.35 and the mode angle (for m = 0) is o = 88.85. Then use  n  2  2n2  1  sin 2  m  1  n2   1  m 

m

1/ 2



to calculate the penetration depth:





= 1/= 5.33 m. MFW = 2a + 2 = 20.65 m



We can also calculate MFW from MFW = 2a(V+1)/V = 2(5 m)(1.3079+1)/(1.3079) = 17.6 m (Difference = 15%)

 = 1.5 m, V = 0.872, single mode. Solve waveguide condition graphically that the mode angle is o = 88.72.





= 1/= 9.08 m. MFW = 2a + 2 = 28.15 m.



Compare with MFW = 2a(V+1)/V = 2(5 m)(0.872+1)/(0.872) = 21.5 m (Difference = 24%) Notice that the MFW from 2a(V+1)/V gets worse as V decreases. The reason for using MFW = 2a(V+1)/V, is that this equation provides a single step calculation of MFW. The calculation of the penetration depth  requires the calculation of the incidence angle  and . Author's Note: Consider a more extreme case

 = 5 m, V = 0.262, single mode. Solve waveguide condition graphically to find that the mode angle is o = 88.40. 





= 1/= 77.22 m. MFW = 2a + 2 = 164.4 m.

Compare with MFW = 2a(V+1)/V = 2(5 m)(0.262 + 1)/(0.262) = 48.2 m (Very large difference.)



Chapter 2

2.15

3 February 2013

2.10 A multimode fiber Consider a multimode fiber with a core diameter of 100 m, core refractive index of 1.4750, and a cladding refractive index of 1.4550 both at 850 nm. Consider operating this fiber at  = 850 nm. (a) Calculate the V-number for the fiber and estimate the number of modes. (b) Calculate the wavelength beyond which the fiber becomes single mode. (c) Calculate the numerical aperture. (d) Calculate the maximum acceptance angle. (e) Calculate the modal dispersion  and hence the bit rate  distance product. Solution

Given n1 = 1.475, n2 = 1.455, 2a = 10010-6 m or a = 50 m and = 0.850 m. The V-number is,

V

2a



n

2 1



1/ 2

 n22

2π (50 μm)(1.4752  1.4552 )1/2 = 89.47 (0.850 μm)



Number of modes M,

V 2 89.47 2 M    4002 2 2 The fiber becomes monomode when,

V

2a



n

2 1

 n22 

1/ 2

2a n12  n22   2.405

 2.405

1/ 2

or



2 (50 μm)(1.4752  1.4552 )1 / 2 = 31.6 m 2.405

For wavelengths longer than 31.6 m, the fiber is a single mode waveguide. The numerical aperture NA is NA  (n12  n22 )1/ 2  (1.475 2  1.455 2 )1 / 2 = 0.242

If max is the maximum acceptance angle, then,  NA    arcsin(0.242 / 1) = 14  no 

 max  arcsin Modal dispersion is given by

 intermode n1  n2 1.475  1.455   L c 3  108 m s -1 = 66.7 ps m-1 or 67.6 ns per km Given that   0.29, maximum bit-rate is BL 

0.25 L

 total



0.25 L

 intermode



0.25 (0.29)(66.7 ns km -1 )



Chapter 2

2.16

3 February 2013

i.e.

BL = 13 Mb s-1 km (only an estimate)

We neglected material dispersion at this wavelength which would further decrease BL. Material dispersion and modal dispersion must be combined by 2 2  t2otal   intermode   material

For example, assuming an LED with a spectral rms deviation  of about 20 nm, and a Dm 200 ps km-1 nm-1 (at about 850 nm)we would find the material dispersion as

material = (200 ps km-1 nm-1)(20 nm)(1 km)  4000 ps km-1 or 4 ns km-1, which is substantially smaller than the intermode dispersion and can be neglected. 2.11 A water jet guiding light One of the early demonstrations of the way in which light can be guided along a higher refractive index medium by total internal reflection involved illuminating the starting point of a water jet as it comes out from a water tank. The refractive index of water is 1.330. Consider a water jet of diameter 3 mm that is illuminated by green light of wavelength 560 nm. What is the V-number, numerical aperture, total acceptance angle of the jet? How many modes are there? What is the cut-off wavelength? The diameter of the jet increases (slowly) as the jet flows away from the original spout. However, the light is still guided. Why?

Light guided along a thin water jet. A small hole is made in a plastic soda drink bottle full of water to generate a thin water jet. When the hole is illuminated with a laser beam (from a green laser pointer), the light is guided by total internal reflections along the jet to the tray. Water with air bubbles (produced by shaking the bottle) was used to increase the visibility of light. Air bubbles scatter light and make the guided light visible. First such demonstration has been attributed to JeanDaniel Colladon, a Swiss scientist, who demonstrated a water jet guiding light in 1841.

Solution

V-number V = (2a/)(n12n22)1/2 = (2×1.5×10-3/550×10-9)(1.33021.0002)1/2 = 15104 Numerical aperture NA= (n12n22)1/2 = (1.33021.0002)1/2 = 0.8814 Total acceptance angle, assuming that the laser light is launched within the water medium sinmax = NA/n0 = 0.113/1.33 or max = 41.4°. Total acceptance 2o = 82.8 Modes = M = V2/2 = (15104)2/2 = 1.14×108 modes (~100 thousand modes) The curoff wavelength corresponds to V = 2.405, that is V = (2a/)NA = 2.405

c = [2aNA]/2.405 = [(2)(4 m)(0.8814)]/2.405 = 3.5 mm



Chapter 2

2.17

3 February 2013

The large difference in refractive indices between the water and the air ensures that total internal reflection occurs even as the width of the jet increases, which changes the angle of incidence. 2.12 Single mode fiber Consider a fiber with a 86.5%SiO2-13.5%GeO2 core of diameter of 8 m and refractive index of 1.468 and a cladding refractive index of 1.464 both refractive indices at 1300 nm where the fiber is to be operated using a laser source with a half maximum width of 2 nm. (a) Calculate the V-number for the fiber. Is this a single mode fiber? (b) Calculate the wavelength below which the fiber becomes multimode. (c) Calculate the numerical aperture. (d) Calculate the maximum acceptance angle. (e) Obtain the material dispersion and waveguide dispersion and hence estimate the bit rate distance product (BL) of the fiber. Solution (a)

Given n1 = 1.475, n2 = 1.455, 2a = 810-6 m or a = 4 m and =1.3 m. The V-number is,

V (b)

2a



n

2 1



1/ 2

 n22

2 (4 μm)(1.4682  1.464 2 )1 / 2 = 2.094 (1.3 μm)

Since V < 2.405, this is a single mode fiber. The fiber becomes multimode when

V

2a



(n12  n22 )1 / 2  2.405

2a n12  n22   2.405

1/ 2

or



2 (4 μm)1.4682  1.464 2   2.405

1/ 2

=1.13 m

For wavelengths shorter than 1.13 m, the fiber is a multi-mode waveguide. (c)

The numerical aperture NA is



NA  n12  n22 (d)



1/ 2

 (1.4682  1.4642 )1/ 2 = 0.108

If max is the maximum acceptance angle, then,  NA    arcsin(0.108 / 1) = 6.2  no 

 max  arcsin

so that the total acceptance angle is 12.4. (e)

At  =1.3 m, from D vs. , Figure 2.22, Dm  7.5 ps km-1 nm-1, Dw  5 ps km-1 nm-1.

 1/ 2  Dm  Dw 1 / 2 L = |7.55 ps km-1 nm-1|(2 nm) = 15 ps km-1 + 10 ps km-1 = 0.025 ns km-1 Obviously material dispersion is 15 ps km-1 and waveguide dispersion is 10 ps km-1 The maximum bit-rate distance product is then



Chapter 2

2.18

3 February 2013

BL 

0.59 L 0.59  = 23.6 Gb s-1 km.  1 / 2 0.025 ns km -1

2.13 Single mode fiber Consider a step-index fiber with a core of diameter of 9 m and refractive index of 1.4510 at 1550 nm and a normalized refractive index difference of 0.25% where the fiber is to be operated using a laser source with a half-maximum width of 3 nm. At 1.55 m, the material and waveguide dispersion coefficients of this fiber are approximately given by Dm = 15 ps km-1 nm-1 and Dw = 5 ps km-1 nm-1. (a) Calculate the V-number for the fiber. Is this a single mode fiber? (b) Calculate the wavelength below which the fiber becomes multimode. (c) Calculate the numerical aperture. (d) Calculate the maximum total acceptance angle. (e) Calculate the material, waveguide and chromatic dispersion per kilometer of fiber. (f) Estimate the bit rate distance product (BL) of this fiber. (g) What is the maximum allowed diameter that maintains operation in single mode? (h) What is the mode field diameter? Solution (a)

The normalized refractive index difference  and n1 are given. Apply,  = (n1n2)/n1 = (1.451 n2)/1.451 = 0.0025, and solving for n2 we find n2 = 1.4474. The V-number is given by

V (b)



(n  n ) 2 1

2 1/ 2 2

2 (4.5 μm)(1.4510 2  1.4474 2 )1 / 2  = 1.87; single mode fiber. (1.55 μm)

For multimode operation we need

V  (c)

2a

2a



(n12  n22 )1 / 2 

2 (4.5 μm)(1.45102  1.4474 2 )1 / 2



 2.405

< 1.205 m.

The numerical aperture NA is NA  ( n12  n 22 )1 / 2  (1.4510 2  1.4474 2 )1 / 2 = 0.1025.

(d)

If max is the maximum acceptance angle, then,  NA   = arcsin(0.1025/1) = 5.89  no 

 max  arcsin

Total acceptance angle 2amax is 11.8 . (e)

Given, Dw = 5 ps km-1 nm-1 and Dm =  ps km-1 nm-1. Laser diode spectral width (FWHM) 1/2 = 3 nm Material dispersion 1/2/L = |Dm|1/2 = (15 ps km-1 nm-1)(3 nm) = 45 ps km-1 Waveguide dispersion 1/2/L = |Dw|1/2 = (5 ps km-1 nm-1)(3 nm)



Chapter 2

2.19

3 February 2013

= 5 ps km-1 Chromatic dispersion, 1/2/L = |Dch|1/2 = (5 ps km-1 nm-1 + 15 ps km-1 nm-1)(3 nm) = 30 ps km-1 (f)

Maximum bit-rate would be BL  i.e.

(g)

BL  20 Mb s-1 km (only an estimate)

To find the maximum diameter for SM operation solve,

V  (h)

0.59 L 0.59 0.59   = 20 Gb s-1 km 12 1  1/ 2 ( 1 / 2 / L) (30  10 s km )

2a



(n12  n22 )1/ 2 

2 (a μm)(1.4510 2  1.4474 2 )1/ 2  2.405 (1.55 μm)

2a = 11.5 m.

The mode filed diameter 2w is 2 w  2a(0.65  1.619V 3 / 2  2.879V 6 ) = 12.2 m

2.14 Normalized propagation constant b Consider a weakly guiding step index fiber in which (n1  n2) / n1 is very small. Show that

b

(  / k ) 2  n22 (  / k )  n2  n12  n22 n1  n2

Note: Since  is very small, n2/ n1  1 can be assumed were convenient. The first equation can be rearranged as

 / k  [n22  b(n12  n22 )]1/ 2  n22 (1  x)1/ 2 ; x  b(n12  n22 ) / n22 where x is small. Taylor's expansion in x to the first linear term would then provide a linear relationship between  and b. Solution

 k

1 2

 [n  b(n  n )]  n2 (1  x) 2 2

2 1

2 2

1 2

 n2  where x  b 12  1  n2 

Taylor expansion around x  0 and truncating the expression, keeping only the linear term yields,

 k

 n2 

 b  n2  nx b  x  n2 1    n2 1   12  1  n2  2 2  n  n22   2   2  n2  2 1  n2 



Chapter 2

2.20

3 February 2013

then using the assumption

 k

n1  1 we get n2

 n2  b(n1  n2 )

and b

(  / k )  n2 n1  n2

as required. 2.15 Group velocity of the fundamental mode Reconsider Example 2.3.4, which has a single mode fiber with core and cladding indices of 1.4480 and 1.4400, core radius of 3 m, operating at 1.5 m. Use the equation

b

(  / k )  n2 ;  = n2k[1 + b] n1  n2

to recalculate the propagation constant . Change the operating wavelength to  by a small amount, say 0.01%, and then recalculate the new propagation constant . Then determine the group velocity vg of the fundamental mode at 1.5 m, and the group delay g over 1 km of fiber. How do your results compare with the findings in Example 2.3.4? Solution

From example 2.3.4, we have b  0.3860859 , k  4188790 m 1 ,  

2c



 1.256637 1015 s 1

 

  n2 k[1  b]  (1.4400)(4188790m 1 ) 1  (0.3860859)

(1.4480  1.4400)   1.4480

  6044795 m 1    1.5 μm(1  1.001)  1.5015 μm , b  0.3854382 , k   4184606 m 1 ,   1.255382  1015 s1  

   n2 k [1  b]  (1.4400)(418406m 1 ) 1  (0.3854382)

(1.4480  1.4400)   1.4800

   6038736m 1 Group Velocity

    (1.255382  1.256637) 1015 s 1 g    2.0713 108 m s 1 6 1     (6.038736  6.044795) 10 m

 g  4.83μs over 1 km.



Chapter 2

2.21

3 February 2013

Comparing to Example 2.3.4

%diff 

2.0713  2.0706  100%  0.03% 2.0706

2.16 A single mode fiber design The Sellmeier dispersion equation provides n vs.  for pure SiO2 and SiO2-13.5 mol.%GeO2 in Table1.2 in Ch. 1. The refractive index increases linearly with the addition of GeO2 to SiO2 from 0 to 13.5 mol.%. A single mode step index fiber is required to have the following properties: NA = 0.10, core diameter of 9 m, and a cladding of pure silica, and operate at 1.3 m. What should the core composition be? Solution

The Sellmeier equation is n2  1 

A32 A12 A2 2   2  12 2  22 2  32

From Table1.2 in Ch.1. Sellmeier coefficients as as follows Sellmeier

0.890815

1 m 0.0690660

2 m 0.115662

3 m 9.900559

0.704048

0.0642700

0.129408

9.425478

A1

A2

A3

SiO2 (fused silica)

0.696749

0.408218

86.5%SiO2‐13.5%GeO2

0.711040

0.451885

Therefore, for  = 1.3 m pure silica has n(0) = 1.4473 and SiO2-13.5 mol.%GeO2 has n(13.5)= 1.4682. Confirming that for NA=0.10 we have a single mode fiber

n2 



Apply NA  n12  n22

2a





1/ 2

NA 

2 (4.5 μm) (0.1) = 2.175 (1.3 μm)



to obtain n1  NA2  n22



1/ 2

=(0.12+1.44732)1/2 = 1.4508

The refractive index n(x) of SiO2-x mol.%GeO2, assuming a linear relationship, can be written as x  x  n( x)  n(0)1    n(13.5) 13.5  13.5  Substituting n(x) = n1 = 1.4508 gives x = 2.26. 2.17 Material dispersion If Ng1 is the group refractive index of the core material of a step fiber, then the propagation time (group delay time) of the fundamental mode is

  L / v g  LN g1 / c Since Ng will depend on the wavelength, show that the material dispersion coefficient Dm is given approximately by



Chapter 2

2.22

3 February 2013

d  d 2n  Dm  Ld c d2 Using the Sellmeier equation and the constants in Table 1.2 in Ch. 1, evaluate the material dispersion at= 1.55 m for pure silica (SiO2) and SiO2-13.5%GeO2 glass. Solution

From Ch. 1 we know that

Ng  n  

dn d

Differentiate  with respect to wavelength  using the above relationship between Ng and n.



L LN g1  vg c



d L dN g1 L  dn d 2 n dn  L d 2n     2    2 d c d c  d d d  c d

Thus,

Dm 

d  d 2n  Ld c d2

(1)

From Ch. 1 we know that the Sellmeier equation is A32 A12 A2 2 n 1  2     12 2  22 2  32 2

(2)

The Sellmeier coefficients for SiO2-13.5%GeO2. The 1, 2, 3 are in m.

A1

A2

A3

1

2

3

SiO2‐13.5%GeO2

0.711040

0.451885

0.704048

0.0642700

0.129408

9.425478

We can use the Sellmeier coefficient in Table1.2 in Ch.1 to find n vs. , dn/d and d2n/d, and, from Eq. (1), Dm vs as in Figure 2Q17-1. At  = 1.55 m, Dm =14 ps km-1 nm-1



Chapter 2

2.23

3 February 2013

Figure 2Q17-1 Materials dispersion Dm vs. wavelength (LiveMath used). (Other math programs such as Matlab can also be used.) 2.18 Waveguide dispersion Waveguide dispersion arises as a result of the dependence of the propagation constant on the V-number, which depends on the wavelength. It is present even when the refractive index is constant; no material dispersion. Let us suppose that n1 and n2 are wavelength (or k) independent. Suppose that  is the propagation constant of mode lm and k = 2π/in which is the free space wavelength. Then the normalized propagation constant b and propagation constant are related by

 = n2k[1 + b]

(1)

The group velocity is defined and given by

vg 

d dk c d d

Show that the propagation time, or the group delay time,  of the mode is

L Ln2 Ln2  d (kb)   vg c c dk

(2)

V  ka[n12  n22 ]1/ 2  kan2 (2)1/ 2

(3)

 Given the definition of V,

and





d (Vb) d d  bkan2 (2)1/ 2  an2 (2)1/ 2 (bk ) dV dV dV

(4)

d Ln  d 2 (Vb)  2 V d c dV 2

(5)

Show that

and that the waveguide dispersion coefficient is



Chapter 2

2.24

3 February 2013

d n2  d 2 (Vb) Dw   V Ld c dV 2

(6)

Figure 2.53 shows the dependence of V[d2(Vb)/dV2] on the V-number. In the range 1.5 < V < 2.4,

V

d 2 (Vb) 1.984  dV 2 V2

Show that,

Dw  

n2  1.984 (n  n2 ) 1.984  1 2 c V c V2

(7)

Dw  

1.984  c( 2a ) 2 2n 2

(8)

which simplifies to

i.e.

Dw ( ps nm 1 km 1 )  

83.76  (μm ) [ a (μm )]2 n2

Waveguide dispersion coefficient

(9)

Consider a fiber with a core of diameter of 8 m and refractive index of 1.468 and a cladding refractive index of 1.464, both refractive indices at 1300 nm. Suppose that a 1.3 m laser diode with a spectral linewidth of 2 nm is used to provide the input light pulses. Estimate the waveguide dispersion per kilometer of fiber using Eqs. (6) and (8). 1.5

1 V[d2(Vb)/dV2] 0.5

0 0

1

2

3

V-number

Figure 2.53 d2(Vb)/dV2 vs V-number for a step index fiber. (Data extracted from W. A. Gambling et al. The Radio and Electronics Engineer, 51, 313, 1981.)

Solution

Waveguide dispersion arises as a result of the dependence of the propagation constant on the V-number which depends on the wavelength. It is present even when the refractive index is constant; no material dispersion. Let us suppose that n1 and n2 are wavelength (or k) independent. Suppose that  is the propagation constant of mode lm and k = 2/where is the free space wavelength. Then the normalized propagation constant b is defined as,



Chapter 2

2.25

3 February 2013

(  / k ) 2  n22 b n12  n22

(1)

Show that for small normalized index difference  = (n1  n2)/n1, Eq. (1) approximates to (  / k )  n2 n1  n2

(2)

 = n2k[1 + b]

(3)

b which gives  as,

The group velocity is defined and given by

vg 

d dk c d d

Thus, the propagation time  of the mode is



L L  d  Ln2 Ln2  d (kb)     vg c  dk  c c dk

(4)

where we assumed   constant (does not depend on the wavelength). Given the definition of V, V  ka[n12  n22 ]1/ 2  ka[(n1  n2 )(n1  n2 )]1/ 2 1/ 2

  n  n2     ka (n1  n2 )n1  1  n1    ka[2n2 n1]1/ 2  kan2 (2)1/ 2

(5)

From Eq. (5),





d (Vb) d d  bkan2 ( 2 )1 / 2  an2 (2 )1 / 2 (bk ) dV dV dV This means that  depends on V as,



Ln2 Ln2  d (Vb)  c c dV

(6)

Dispersion, that is, spread  in due to a spread  can be found by differentiating Eq. (6) to obtain, d Ln2  dV d d (Vb) Ln2   V  d 2 (Vb)     d c d dV dV c    dV 2 Ln2  d 2 (Vb)  V c dV 2

(7)

The waveguide dispersion coefficient is defined as

Dw 

n  d 2 (Vb) d  2 V Ld c dV 2

(8)



Chapter 2

2.26

3 February 2013

Figure 2.53 shows the dependence of V[d2(Vb)/dV2] on the V-number. In the range 2 < V < 2.4,

d 2 (Vb) 1.984 V  dV 2 V2 so that Eq. (8) becomes,

Dw  

n2  1.984 (n  n2 ) 1.984  1 2 c V c V2

(9)

We can simplify this further by using

 n  1.984 1.984n2    Dw   2   2 1/ 2  c V c  2an2 (2)  

Dw  

1/ 2

1.984  c ( 2 a ) 2 2 n 2

(10)

Equation (6) should really have Ng2 instead of n2 in which case Eq. (10) would be Dw  

1.984 N g 2 c(2a) 2 2n22



(11)

Consider a fiber with a core of diameter of 8 m and refractive index of 1.468 and a cladding refractive index of 1.464 both refractive indices at 1300 nm. Suppose that a1.3 m laser diode with a spectral linewidth of 2 nm is used to provide the input light pulses. Estimate the waveguide dispersion per kilometer of fiber using Eqs. (8) and (11).

V and

2a



n

2 1

 n22



1/ 2



2 (4 μm)(1.4682  1.464 2 )1 / 2 = 2.094 (1.3 μm)

 = (n1  n2)/n1 = (1.4681.464)/1.468 = 0.00273.

From the graph, Vd2(Vb)/dV2 = 0.45,

Dw   

n2  d 2 (Vb) (1.464)(2.73  10 3 )   (0.45) V c dV 2 (3  108 m s -1 )(1300  10 9 m)

Dw  4.610-6 s m-2 or 4.6 ps km-1 nm-1

Using Eq. (10) Dw   

1.984 1.984(1300  10 9 m)    c(2a) 2 2n2 (3  108 m s -1 )[2  4  10 6 m]2 2(1.464)]

Dw  4.610-6 s m-2 or 4.6 ps km-1 nm-1

For 1/2 = 2 nm we have, 1/2 = |Dw|L1/2 = (4.6 ps km-1 nm-1)(2 nm) = 9.2 ps/km



Chapter 2

2.27

3 February 2013

2.19 Profile dispersion Total dispersion in a single mode, step index fiber is primarily due to material dispersion and waveguide dispersion. However, there is an additional dispersion mechanism called profile dispersion that arises from the propagation constant of the fundamental mode also depending on the refractive index difference . Consider a light source with a range of wavelengths  coupled into a step index fiber. We can view this as a change in the input wavelength. Suppose that n1, n 2, hence depends on the wavelength . The propagation time, or the group delay time, g per unit length is

 g  1 / v g  d / d  (1 / c)(d / dk )

(1)

where k is the free space propagation constant (2/), and we used dcdk. Since  depends on n1,  and V, consider g as a function of n1,  (thus n2), and V. A change in will change each of these quantities. Using the partial differential chain rule,

 g  g n1  g V  g      n1  V   

(2)

The mathematics turns out to be complicated but the statement in Eq. (2) is equivalent to Total dispersion = Material dispersion (due to ∂n1/∂) + Waveguide dispersion (due to ∂V/∂) + Profile dispersion (due to ∂/∂) in which the last term is due to  depending on; although small, this is not zero. Even the statement in Eq. (2) above is over simplified but nonetheless provides an insight into the problem. The total intramode (chromatic) dispersion coefficient Dch is then given by Dch = Dm + Dw + Dp

(3)

in which Dm, Dw, Dp are material, waveguide, and profile dispersion coefficients respectively. The waveguide dispersion is given by Eq. (8) and (9) in Question 2.18, and the profile dispersion coefficient is (very) approximately 1 ,

Dp  

N g1  d 2 (Vb)  d  V   dV 2  d  c 

(4)

in which b is the normalized propagation constant and Vd2(Vb)/dV2 vs. V is shown in Figure 2.53,we can also use Vd2(Vb)/dV2  1.984/V2. Consider a fiber with a core of diameter of 8 m. The refractive and group indices of the core and cladding at  = 1.55 m are n1 = 1.4500, n 2 = 1.4444, Ng1 = 1.4680, Ng 2 = 1.4628, and d/d = 232 m-1. Estimate the waveguide and profile dispersion per km of fiber per nm of input light linewidth at this wavelength. (Note: The values given are approximate and for a fiber with silica cladding and 3.6% germania-doped core.) Solution 1

J. Gowar, Optical Communication Systems, 2nd Edition (Prentice Hall, 1993). Ch. 8 has the derivation of this equation..



Chapter 2

2.28

3 February 2013

Total dispersion in a single mode step index fiber is primarily due to material dispersion and waveguide dispersion. However, there is an additional dispersion mechanism called profile dispersion that arises from the propagation constant of the fundamental mode also depending on the refractive index difference . Consider a light source with a range of wavelengths  coupled into a step index fiber. We can view this as a change in the input wavelength. Suppose that n1, n 2, hence depends on the wavelength . The propagation time, or the group delay time, g per unit length is

g 

1 1  d     vg c  dk 

(1)

Since  depends on n1,  and V, let us consider g as a function of n1,  (thus n2) and V. A change in will change each of these quantities. Using the partial differential chain rule,

 g  g n1  g V  g     n1  V    

(2)

The mathematics turns out to be complicated but the statement in Eq. (2) is equivalent to Total dispersion = Materials dispersion (due to n1/) + Waveguide dispersion (due to V/) + Profile dispersion (due to /) where the last term is due  depending on; although small this is not zero. Even the above statement in Eq. (2) is over simplified but nonetheless provides an sight into the problem. The total intramode (chromatic) dispersion coefficient Dch is then given by Dch = Dm + Dw + Dp

(3)

where Dm, Dw, Dp are material, waveguide and profile dispersion coefficients respectively. The waveguide dispersion is given by Eq. (8) in Question 2.6 and the profile dispersion coefficient away is (very) approximately, Dp  

N g 1  d 2 (Vb)  d   V  c  dV 2  d 

(4)

where b is the normalized propagation constant and Vd2(Vb)/dV2 vs. V is shown in Figure 2.53. The term Vd2(Vb)/dV2  1.984/V2. Consider a fiber with a core of diameter of 8 m. The refractive and group indexes of the core and cladding at  = 1.55 m are n1 = 1.4504, n 2 = 1.4450, Ng1 = 1.4676, Ng 2 = 1.4625. d/d = 161 m-1.

V and

2a



n

2 1

 n22



1/ 2



2 (4 μm)(1.4504 2  1.4450 2 )1/ 2 = 2.03 (1.55 μm)

 = (n1  n2)/n1 = (1.4504-1.4450)/1.4504 = 0.00372

From the graph in Figure 2.53, when V = 2.03, Vd2(Vb)/dV2  0.50, Profile dispersion:



Chapter 2

2.29

3 February 2013

Dp  



N g1  d 2 (Vb)  d  1.4676  V 0.50161 m 1   2  c  dV  d  3  108 m s 1

Dp = 3.8 10-7 s m-1 m-1 or 0.38 ps km-1 nm-1

Waveguide dispersion: Dw  

1.984 1.984(1500  10 9 m)    c(2a ) 2 2n2 (3  108 m s -1 )[2  4  10 6 m]2 2(1.4450)]

Dw   5.6 ps km-1 nm-1



Profile dispersion is more than 10 times smaller than waveguide dispersion. 2.20 Dispersion at zero dispersion coefficient Since the spread in the group delay  along a fiber depends on the , the linewidth of the sourcewe can expand  as a Taylor series in . Consider the expansion at  = 0 where Dch = 0. The first term with  would have d /d as a coefficientthat is Dch, and at 0 this will be zero; but not the second term with ( that has a differential, d2/d or dDch/d. Thus, the dispersion at 0 would be controlled by the slope S0 of Dch vs.  curve at 0. Show that the chromatic dispersion at 0 is

 

L S 0 ( ) 2 2

A single mode fiber has a zero-dispersion at 0 = 1310 nm, dispersion slope S0 = 0.090 ps nm2 km. What is the dispersion for a laser with  = 1.5 nm? What would control the dispersion? Solution

Consider the Taylor expansion for , a function of wavelength, about its center around, say at 0, when we change the wavelength by For convenience we can the absolute value of  at 0 as zero since we are only interested in the spread . Then, Taylor's expansion gives,

  f ( )  







  0 

 

1 d 2  d ( )  ( ) 2   2 2! d d

1 d 2  1 d  d  1 d 2 Dch ( ) 2 ( ) 2  0   ( )  2 2! d 2! dt  d  2! dt





L 1 km 2 S 0 ( ) 2  0.090 ps nm -2 km -1 2 nm  = 1.01 ps 2 2

This can be further reduced by using a narrower laser line width since depends on ( 2.21 Polarization mode dispersion (PMD) A fiber manufacturer specifies a maximum value of 0.05 ps km-1/2 for the polarization mode dispersion (PMD) in its single mode fiber. What would be the dispersion, maximum bit rate and the optical bandwidth for this fiber over an optical link that is 200 km long if the only dispersion mechanism was PMD? Solution



Chapter 2

2.30

3 February 2013

Dispersion

  DPMD L1 / 2  0.05ps km1 / 2 ( 200km)1 / 2  0.707 ps

Bit rate

B

0.59 0.59   8.35 Gb s -1  0.707ps

Optical bandwidth f op  0.75B  (0.75)(8.35 Gb s 1 )  6.26 GHz

2.22 Polarization mode dispersion Consider a particular single mode fiber (ITU-T G.652 compliant) that has a chromatic dispersion of 15 ps nm-1 km-1. The chromatic dispersion is zero at 1315 nm, and the dispersion slope is 0.092 ps nm-2 km-1. The PMD coefficient is 0.05 ps km-1/2. Calculate the total dispersion over 100 km if the fiber is operated at 1315 nm and the source is a laser diode with a linewidth (FWHM)  = 1 nm. What should be the linewidth of the laser source so that over 100 km, the chromatic dispersion is the same as PMD? Solution Polarization mode dispersion for L = 100 km is  PMD  DPMD L1 / 2 = 0.05  100 ps = 0.5 ps We need the chromatic dispersion at 0, where the chromatic dispersion Dch = 0. For L = 100 km, the chromatic dispersion is L  ch  S0 (  )2 = 1000.092(1)2/2 = 4.60 ps 2 The rms dispersion is 2  rms   PMD   ch2 = 4.63 ps

The condition for  PMD   ch is  

2 DPMD = 0.33 nm S 0 L1 / 2

2.23 Dispersion compensation Calculate the total dispersion and the overall net dispersion coefficient when a 900 km transmission fiber with Dch = +15 ps nm-1 km-1 is spliced to a compensating fiber that is 100 km long and has Dch = 110 ps nm-1 km-1. What is the overall effective dispersion coefficient of this combined fiber system? Assume that the input light spectral width is 1 nm. Solution

Using Eq. (2.6.1) with  = 1 nm, we can find the total dispersion  = (D1L1 + D2L2) = [(+15 ps nm-1 km-1)(900 km) + (110 ps nm-1 km-1)(100 km)](1 nm) = 2,500 ps nm-1 for 1000 km. The net or effective dispersion coefficient can be found from  = DnetL,



Chapter 2

2.31

3 February 2013

Dnet =  /(L = (2,500 ps)/[(1000 km)(1 nm)] = 2.5 ps nm-1 km-1 2.24 Cladding diameter A comparison of two step index fibers, one SMF and the other MMF shows that the SMF has a core diameter of 9 m but a cladding diameter of 125 m, while the MMF has a core diameter of 100 m but a cladding diameter that is the same 125 m. Discuss why the manufacturer has chosen those values. Solution

For the single mode fiber, the small core diameter is to ensure that the V-number is below the cutoff value for singe mode operation for the commonly used wavelengths 1.1 m and 1.5 m. The larger total diameter is to ensure that there is enough cladding to limit the loss of light that penetrates into the cladding as an evanescent wave. For multimode fibers, the larger core size allows multiple modes to propagate in the fiber and therefore the spectral width is not critical. Further, the larger diameter results in a greater acceptance angle. Thus, LEDs, which are cheaper and easier to use than lasers, are highly suitable. The total diameter of the core and cladding is the same because in industry it is convenient to standardize equipment and the minor losses that might accumulate from light escaping from the cladding do not matter as much over shorter distances for multimode fibers – they are short haul fibers. 2.25 Graded index fiber Consider an optimal graded index fiber with a core diameter of 30 m and a refractive index of 1.4740 at the center of the core and a cladding refractive index of 1.4530. Find the number of modes at 1300 nm operation. What is its NA at the fiber axis, and its effective NA? Suppose that the fiber is coupled to a laser diode emitter at 1300 nm and a spectral linewidth (FWHM) of 3 nm. The material dispersion coefficient at this wavelength is about 5 ps km-1 nm-1. Calculate the total dispersion and estimate the bit rate  distance product of the fiber. How does this compare with the performance of a multimode fiber with same core radius, and n1 and n2? What would the total dispersion and maximum bit rate be if an LED source of spectral width (FWHM) 1/2  80 nm is used? Solution

The normalized refractive index difference  = (n1n2)/n1 = (1.47401.453)/1.474 = 0.01425 Modal dispersion for 1 km of graded index fiber is

 intermode 

Ln1 2 (1000)(1.474)   (0.01425) 2 = 2.910-11 s or 0.029 ns 8 20 3c 20 3 (3  10 )

The material dispersion (FWHM) is  m (1/ 2 )  LDm 1 / 2  (1000 m)(5 ps ns 1 km 1 )(3 nm) = 0.015 ns

Assuming a Gaussian output light pulse shape, rms material dispersion is,

m = 0.4251/2 = (0.425)(0.015 ns) = 0.00638 ns Total dispersion is 2  total   intermode   m2  0.029 2  0.006382 = 0.0295 ns.



Chapter 2

2.32

3 February 2013

so that

B = 0.25/total = 8.5 Gb

If this were a multimode step-index fiber with the same n1 and n2, then the rms dispersion would roughly be

 n1  n2 1.474  1.453   = 70 ps m-1 or 70 ns per km 8 -1 L c 3  10 m s Maximum bit-rate is BL  i.e.

0.25L

 intermode



0.25L 0.25  (0.28) (0.28)(70 ns km -1 )

BL = 12.8 Mb s-1 km (only an estimate!)

The corresponding B for 1 km would be around 13 Mb s-1 With LED excitation, again assuming a Gaussian output light pulse shape, rms material dispersion is

 m  (0.425) m (1/ 2 )  (0.425) LDm 1/ 2  (0.425)(1000 m)(5 ps ns 1 km 1 )(80 nm)

= 0.17 ns

Total dispersion is 2  total   intermode   m2  0.0292  0.17 2 = 0.172 ns

so that

B = 0.25/total = 1.45 Gb

The effect of material dispersion now dominates intermode dispersion. 2.26 Graded index fiber Consider a graded index fiber with a core diameter of 62.5 m and a refractive index of 1.474 at the center of the core and a cladding refractive index of 1.453. Suppose that we use a laser diode emitter with a spectral FWHM linewidth of 3 nm to transmit along this fiber at a wavelength of 1300 nm. Calculate, the total dispersion and estimate the bit-rate  distance product of the fiber. The material dispersion coefficient Dm at 1300 nm is 7.5 ps nm-1 km-1. How does this compare with the performance of a multimode fiber with the same core radius, and n1 and n2? Solution

The normalized refractive index difference  = (n1n2)/n1 = (1.4741.453)/1.474 = 0.01425 Modal dispersion for 1 km of graded index fiber is

 intermode 

Ln1 2 (1000)(1.474)   (0.01425) 2 = 2.910-11 s or 0.029 ns. 8 20 3c 20 3 (3  10 )

The material dispersion is  m (1/ 2 )  LDm 1/ 2  (1000 m)(7.5 ps ns 1 km 1 )(3 nm) = 0.0225 ns

Assuming a Gaussian output light pulse shaper,



Chapter 2

2.33

3 February 2013

intramode = 0.4251/2 = (0.425)(0.0225 ns) = 0.0096 ns Total dispersion is 2 2  rms   intermode   intramode  0.0292  0.00962 = 0.0305 ns.

B = 0.25/rms = 8.2 Gb for 1 km

so that

If this were a multimode step-index fiber with the same n1 and n2, then the rms dispersion would roughly be

 n1  n2 1.474  1.453   L c 3  108 m s-1 = 70 ps m-1 or 70 ns per km Maximum bit-rate would be BL 

0.25L

 intermode



0.25 (0.28)(70 ns km-1 )

BL = 12.7 Mb s-1 km (only an estimate!)

i.e.

The corresponding B for 1 km would be around 13 Mb s-1. 2.27 Graded index fiber A standard graded index fiber from a particular fiber manufacturer has a core diameter of 62.5 m, cladding diameter of 125 m, a NA of 0.275. The core refractive index n1 is 1.4555. The manufacturer quotes minimum optical bandwidth × distance values of 200 MHzkm at 850 nm and 500 MHzkm at 1300 nm. Assume that a laser is to be used with this fiber and the laser linewidth = 1.5 nm. What are the corresponding dispersion values? What type of dispersion do you think dominates? Is the graded index fiber assumed to have the ideal optimum index profile? (State your assumptions). What is the optical link distance for operation at 1 Gbs-1 at 850 and 1300 nm Solution

We are given the numerical aperture NA = 0.275. Assume that this is the maximum NA at the core n2  n12  NA 2 2  1.45552  0.2752 2  1.4293 1



1

n1  n2 1.4555  1.4293   0.018 n1 1.4555

We can now calculate intermodal dispersion

 intermodal

(1.4555)(0.018) 2 n1 2     45.43 ps km-1 1 5 20 3c 20 3(3 10 km s )

The total dispersion for   850 nm is

T 

0.19 0.19   0.95 ns km-1 6 1 f op 200  10 s km



Chapter 2

2.34

3 February 2013

So the intramodal dispersion is

 intramodal     2 T



1 2 2 intermodal



 0.95  10

  45.43 10  

1 12 2 2

9 2

 0.949 ns km -1

For = 1300 nm, the total dispersion is

T 

0.19 0.19   0.38 ns km-1 6 1 f op 500  10 s km

and

 intramodal     2 T



1 2 2 intermodal



 0.38  10

  45.43 10  

1 12 2 2

9 2

 0.377 ns km -1

For both 850 nm and 1300 nm intramodal dispersion dominates intermodal dispersion.

  2(1  )  2(1  0.018)  1.96 Gamma is close to 2 so this is close to the optimal profile index. 2.28 Graded index fiber and optimum dispersion The graded index fiber theory and equations tend to be quite complicated. If  is the profile index then the rms intermodal dispersion is given by 2 1/ 2

Ln       2    1    2c    1  3  2 



 2 4c1 c2 (  1) 16c22 2 (  1) 2   c1   2  1 (5  2)(3  2)  

1/ 2

(1)

where c1 and c2 are given by

c1 

 n     d   3  2  2   2  ; c2  ;   2 1    N    d   2 2(  2)  g1 

(2)

where  is a small unitless parameter that represents the change in  with . The optimum profile coefficient o is

o  2  

(4   )(3   ) (5  2 )

(3)

Consider a graded index fiber for use at 850 nm, with n1 = 1.475, Ng1 = 1.489,  = 0.015, d/d = 683 m-1.Plot  in ps km-1 vs from  = 1.8 to 2.4 and find the minimum. (Consider plotting  on a logarithmic axis.) Compare the minimum  and the optimum , with the relevant expressions in §2.8. Find the percentage change in  for a 10× increase in . What is your conclusion? Solution

2

R. Olshansky and D. Keck, Appl. Opt., 15, 483, 1976.



Chapter 2

2.35

3 February 2013

From the graph in Figure 2Q28-1 0 = 2.04.

Equation (3) gives the same value 0 = 2.040. From the graph in Figure 2Q28-1 intermode/L = 31.87 ps km-1 Consider Eq. (2.8.4) in §2.8,

 intermode L



n1 2 = 31.93 ps km-1. 20 3c

From the graph in Figure 2Q28-1, a 3.4% change of  leads to 10 times increase of dispersion. It is therefore important to control the refractive profile.

Figure 2Q28‐1

2.29 GRIN rod lenses Figure 2.32 shows graded index (GRIN) rod lenses. (a) How would you represent Figure 2.32(a) using two conventional converging lenses. What are O and O? (b) How would you represent Figure 2.32(b) using a conventional converging lens. What is O? (c) Sketch ray paths for a GRIN rod with a pitch between 0.25P and 0.5P starting from O at the face center. Where is O? (d) What use is 0.23P GRIN rod lens in Figure 2.32(c)?

Figure 2.32 Graded index (GRIN) rod lenses of different pitches. (a) Point O is on the rod face center and the lens focuses the rays onto O' on to the center of the opposite face. (b) The rays from O on the rod face center are collimated out. (c) O is slightly away from the rod face and the rays are collimated out.

Solution (a) and (b)



Chapter 2

2.36

3 February 2013

Figure 2Q29-1: (a) The beam bending from O to O using a GRIN rod can be achieved equivalently by using two converging lenses. O and O are the focal points of the lenses (approximately). (Schematic only). (b) The collimation of rays from a point source on the face of a GRIN rod can be equivalently achieved by a single converging lens whose focal length is 0.25P and O is the focal point. (Schematic only).

(c) Consider a GRIN rod with 0.4P

Figure 2Q29-2: Ray paths in a GRIN rod that has a pitch between 0.25P to 0.5P. (Schematic only.)

(d) Since the point O does not have to be right on the face of the GRIN rod, it can be used to collimate a point source O by bringing the rod sufficiently close to O; a fixed annular spacer can “fix” the required proximity of the rod to O. Since the source does not have to be in contact with the face of the rod, possible damage (such as scratches) to the face are avoided. 2.30 Optical Fibers Consider the manufacture of optical fibers and the materials used. (a) What factors would reduce dispersion? (b) What factors would reduce attenuation? Possible Answers

It is essential to control of refractive index profile, core radius, and minimize variations in the (a) refractive index due to variations in doping. (b) Minimize impurities. Reduce scattering by reducing density and hence refractive index n fluctuations (may not be readily possible). Use a glass material with a lower glass transition temperature so that the frozen n-variations are smaller. 2.31 Attenuation A laser emitter with a power 2 mW is used to send optical signals along a fiber optic link of length 170 km. Assume that all the light was launched into the fiber. The fiber is quoted as



Chapter 2

2.37

3 February 2013

having an attenuation of 0.5 dB/km. What is the output power from the optical link that a photodetector must be able to detect? Solution

Pout  Pin exp( L) where



 dB 4.34



0.5 dB km1  0.115 km1 4.34

so

Pout  2 mW exp(0.115 km1  170 km)  6.24 pW

2.32 Cut-back method of attenuation measurement Cut-back method is a destructive measurement technique for determining the attenuation  of a fiber. The first part of the experiment involves measuring the optical power Pfar coming out from the fiber at the far end as shown in Figure 2.54 Then, in the second part, keeping everything the same, the fiber is cut close to the launch or the source end. The output power Pnear is measured at the near end from the short cut fiber. The attenuation is then given by

 = (10/L)log(Pfar/Pnear) in which L is the separation of the measurement points, the length of the cut fiber, and  is in dB per unit length. The output Pnear from the short cut fiber in the second measurement is actually the input into the fiber under test in the first experiment. Usually a mode scrambler (mode stripper) is used for multimode fibers before the input. The power output from a particular fiber is measured to be 13 nW. Then, 10 km of fiber is cut-out and the power output is measured again and found to be 43 nW. What is the attenuation of the fiber?

Figure 2.54 Illustration of the cut-back method for measuring the fiber attenuation. S is an optical source and D is a photodetector



Chapter 2

2.38

3 February 2013

Solution

 = (10/L)log(Pfar / Pnear) = (10/10 km) log(10/43) = 0.63 dB km-1 2.33

Intrinsic losses

(a) Consider a standard single mode fiber with a NA of 0.14. What is its attenuation at 1625 and at 1490 nm? How do these compare with the attenuation quotes for the Corning SMF-28e+, 0.200.23 dB km-1 at 1625 nm and 0.21  0.24 dB km-1 at 1490 nm? (b) Consider a graded index fiber with a NA of 0.275. What would you expect for its attenuation at 850 nm and 1300 nm? How do your calculations compare with quoted maximum values of 2.9 dB km-1 at 850 nm and 0.6 dB km-1 at 1300 nm for 62.5 m graded index fibers? Actual values would be less. Solution (a)

When the wavelength is 1625 nm,

 FIR  A exp B /    B  48.5  11 -1   7.8  10 exp    0.085 dB km    1.625 

 FIR  A exp 

AR  0.63  2.06 NA  0.63  2.06  0.14  0.918 dB km-1 μm 4

0.918 dB km 1 μm 4  0.132 dB km 1 aR  4  4  1.625 μm AR

 total   FIR   R  0.085  0.132  0.217 dB km-1 When the wavelength is 1490 nm  B  48.5  11 -1   7.8 10 exp    0.0057 dB km 1 . 490     

 FIR  A exp 

AR  0.63  2.06 NA  0.63  2.06  0.14  0.918 dB km-1 μm 4

aR 

AR

4



0.918 dB km 1 μm 4  0.1863 dB km 1 4 1.490 μm

 total   FIR   R  0.0057  0.1863  0.192 dB km-1 (b)

Rayleigh scattering

AR = 0.63 + 1.75×NA = 0.63 + 1.75×0.275 = 1.111 dB km-1 m4 At 850 nm and 1300 nm the  FIR term is essentially zero and does not need to be included At 850 nm,



Chapter 2

2.39

3 February 2013

1.111 dB km1 μm4 aR  4   2.13 dB km1 4  0.850 μm AR

At 1300 nm,

aR  2.34

AR

4



1.111 dB km1 μm 4  0.390 dB km1 1.300 μm4

Scattering losses and fictive temperature

Rayleigh scattering process decreases with wavelength, and as mentioned in Ch. 1, it is inversely proportional to 4. The expression for the attenuation R in a single component glass such as silica due to Rayleigh scattering is approximately given by two sets of different equations in the literature 3 ,

R 

8 3 8 2 n p  T k BT f 34

and

R 

8 3 2 (n  1) 2  T k B T f 34

in which is the free space wavelength, n is the refractive index at the wavelength of interest,  is the isothermal compressibility (at Tf) of the glass, kT is the Boltzmann constant, and Tf is a quantity called the fictive temperature (or the glass transition temperature) at which the liquid structure during the cooling of the fiber is frozen to become the glass structure. Fiber is drawn at high temperatures and as the fiber cools eventually the temperature drops sufficiently for the atomic motions to be so sluggish that the structure becomes essentially "frozen-in" and remains like this even at room temperature. Thus, Tf marks the temperature below which the liquid structure is frozen and hence the density fluctuations are also frozen into the glass structure. Use these two equations and calculate the attenuation in dB/km due to Rayleigh scattering at around the = 1.55 m window given that pure silica (SiO2) has the following properties: Tf  1180°C; T  710-11 m2 N-1 (at high temperatures); n  1.45 at 1.55 m, p = 0.28. The lowest reported attenuation around this wavelength is about 0.14 dB/km. What is your conclusion? Solution

R 

8 3 8 2 n p T k BT f = 0.0308 km-1 or 4.34×0.0308 = 0.13 dB km-1 34

R 

8 3 2 (n  1) 2 T k BT f = 0.0245 km-1 or 4.34×0.0245 = 0.11 dB km-1 4 3

The first equation appears to be the closest to the experimental value. However, note that the reported attenuation also has a contribution from the fundamental IR absorption.

 FIR  A exp B /   , A = 7.81×1011 dB km-1; B= 48.5 m gives FIR = 0.02 dB km-1. Thus, adding FIR to R gives First equation + FIR attenuation = 0.13 + 0.02 = 0.015 dB km-1 Second equation + FIR attenuation = 0.11 + 0.02 = 0.013 dB km-1

3

For example, R. Olshansky, Rev. Mod. Phys, 51, 341, 1979.



Chapter 2

2.40

3 February 2013

The experimental value lies exactly in between.

2.35 Bending loss Bending losses always increase with the mode field diameter (MFD). Since the MFD increases for decreasing V, 2w  2×2.6a/V, smaller V fibers have higher bending losses. How does the bending loss vs. radius of curvature R behavior look like on a semilogarithmic plot (as in Figure 2.39(a) for two values of the V-number V1 and V2 if V2 > V1. It is found that for a single mode fiber with a cut-off wavelength c = 1180 nm, operating at 1300 nm, the microbending loss reaches 1 dB m-1 when the radius of curvature of the bend is roughly 6 mm for  = 0.00825, 12 mm for  = 0.00550, and 35 mm for  = 0.00275. Explain these findings. Solution

We expect the bending loss vs. R on a semilogarithmic plot to be as in Figure 2Q35-1 (schematic)

Figure 2Q35-1 Microbending loss  decreases sharply with the bend radius R. (Schematic only.)

From the figure, given  = 1, R increases from R1 to R2 when V decreases from V1 to V2. Expected







Equivalently at one R = R1

R with V

(1)

 with V

(2)

We can generalize by noting that the penetration depth into the cladding  1/V. Expected







Equivalently at one R = R1

R with 

(3)

 with 

(4)

Eqs. (3) and (4) correspond to the general statement that microbending loss  gets worse when penetration  into cladding increases; intuitively correct based on Figure 2.32. Experiments show that for a given  = 1, R increases with decreasing . Observation 





R with 

(5)



Consider the penetration depth  into a second medium (Example 2.1.3),



Chapter 2

2.41

3 February 2013



1





 1 1 1  2 2 1/ 2 2 ( n1  n2 ) 2n1 (2 )1 / 2

 with 



(6)

Thus,  increases with decreasing . Thus, from Eqs. (3) and (6), we expect Expected







R with  with  





(7)

Thus Eq, (7) agrees with the observation in Eq. (5). NOTE If we plot vs. R on a log-log plot, we would find the line in Figure 2Q35-2, that is,   Rx, x = 0.62. Very roughly, from theoretical considerations, we expect R R   exp 3 / 2     Rc  

  exp 

(8)

where Rc is a constant (“a critical radius type of constant”) that is proportional to . Thus, taking logs,

ln   3 / 2 R  constant

(9)

We are interested in the R behavior at a constant . We can lump the constant into ln and obtain,   R 2 / 3

(10)

As shown in Figure 2Q35-2, x = 0.62 is close to 2/3 given three points and the rough derivation above. 1

Normalized index different 

0.01

0.001

100

10 y = 0.0255x‐0.625 R² = 0.9993

R (mm)

Figure 2Q35-2 The relationship between  and the radius of curvature R for a given amount of bending loss.



Chapter 2

2.42

3 February 2013

2.36 Bend loss reduced fibers Consider the bend loss measurements listed in Table 2.8 for four difference types of fiber. The trench fibers have a trench placed in the cladding where the refractive index is lowered as shown in Figure 2.39 The nanoengineered fiber is shown in the Figure 2.55. There is a ring of region in the cladding in which there are gas-filled nanoscale voids. (They are introduced during fabrication.) A void in the ring has a circular cross section but has a length along the fiber that can be a few meters. These voids occupy a volume in the ring that is only 1 - 10%. Plot the bending loss semilogarithmically ( on a log scale and R on a linear scale) and fit the data to micobend = Aexp(R/Rc) and find A and Rc. What is your conclusion? Suppose that we set our maximum acceptable bending loss to 0.1 dB/turn in installation (the present goal is to bring the bending loss to below 0.1 dB/turn). What are the allowed radii of curvature for each turn? Table 2.8 Bend radius R in mm,  in dB/turn. Data over 1.55 - 1.65 m range. (Note, data used from a number of sources: (a) M.-J. Li et al. J. Light Wave Technol., 27, 376, 2009; (b) K. Himeno et al, J. Light Wave Technol., 23, 3494, 2005; (c) L.A. de Montmorillon, et al. "Bend-Optimized G.652D Compatible Trench-Assisted Single-Mode Fibers" Proceedings of the 55th IWCS/Focus, pp. 342-347, November, 2006.) Standard SMFa

Trench Fiber 1b

Trench Fiber 2c

Nanoengineered Fibera

1550 nm

1650 nm

1625 nm

1550 nm

R mm

 dB/turn

R mm


R mm


R mm


5.0

15.0

7.50

0.354

5.0

0.178

5.0

0.031

7.0

4.00

10.0

0.135

7.5

0.0619

7.5

0.0081

10.0

0.611

15.0

0.020

10.0

0.0162

10.0

0.0030

12.5

0.124

15.0

0.00092

15

0.00018

16.0

0.0105

17.5

0.0040

Solution



Chapter 2

2.43

3 February 2013

Bend Losses 100

Standard SMF

y = 418.95e-0.659x

Trench 1 Trench 2

10

Nanoengineered Expon. (Standard SMF)

1

Expon. (Trench 1)

dB/turn

y = 6.241e-0.383x

Expon. (Trench 2) Expon. (Nanoengineered)

0.1

0.01

0.001 y = 2.971e-0.533x

y = 0.4079e-0.51x 0.0001 0

10

20

30

R (mm)

Figure 2Q36-1 Attenuation per turn as a function of bend radius

For a bending loss of 0.1 dB/turn, the allowed radii of curvature are (very roughly) Standard SMF, 13 mm; trench 1, 10 mm; trench 2, 6 mm; nanoengineered, 3 mm. 2.37 Microbending loss Microbending loss B depends on the fiber characteristics and wavelength. We will calculate  approximately given various fiber parameters using the single mode fiber microbending loss equation (D. Marcuse, J. Op. Soc. Am., 66, 216, 1976)

B 

 1 / 2 2

2 3 / 2V 2 K1 (a)

2

R 1 / 2 exp(

2 3 R) 3 2

where R is the bend radius of curvature, a = fiber radius,  is the propagation constant, determined by b, normalized propagation constant, which is related to V,  = n2k[1 + b]; k = 2/ is the free-space wavevector;  = [2  n22k2];  = [ n12k2  ], and K1(x) is a first-order modified Bessel function, available in math software packages. The normalized propagation constant b can be found from b = (1.14280.996V-1)2. Consider a single mode fiber with n1 = 1.450, n2 = 1.446, 2a (diameter) = 3.9 m. Plot B vs. R for = 633 nm and 790 nm from R = 2 mm to 15 mm. Figure 2.56 shows the experimental results on a SMF that has the same properties as the fiber above. What is your conclusion? (You might wish to compare your calculations with the experiments of A.J. Harris and P.F. Castle, IEEE J. Light Wave Technol., LT4, 34, 1986).



Chapter 2

2.44

3 February 2013

Solution

Given: n1 = 1.450, n2 = 1.446, 2a (diameter) = 3.9 m;  = 790 nm, we can calculate the following: k = 2/= 7.953106 m-1; 

(n12  n22 )1 / 2 = 0.00275; 2n12

V

2a



(n12  n22 )1 / 2 

2 (3.9 μm)(1.4502  1.4462 )1 / 2 = 1.67; (0.790 μm)

2

0.996   b  1.1428   = 0.2977; V  

 = n2k[1 + b] = 1.1510107 m-1;

 = [2  n22k2] = 4.6544105 m-1;

 = [ n12k2  ] = 7.175105 m-1; Substitute these values into

B  to find

 1 / 2 2

2 3 / 2V 2 K1 (a)

2

R 1 / 2 exp(

 B  (1.03 103 ) R 1 / 2 exp(

2 3 R) 3 2

R ) 0.0020

which is plotted in Figure 2Q37-1 on the LHS . Given: n1 = 1.450, n2 = 1.446, 2a (diameter) = 3.9 m;  = 633 nm, we can calculate the following: k = 2/= 9.926106 m-1; 

(n12  n22 )1 / 2 = 0.00275; 2n12

V

2a



(n12  n22 )1 / 2 

2 (3.9 μm)(1.4502  1.4462 )1 / 2 = 2.08; (0.633 μm)



Chapter 2

2.45

3 February 2013 2

0.996   b  1.1428   = 0.44127; V  

 = n2k[1 + b] = 1.437107 m-1;

 = [2  n22k2] = 7.073105 m-1;

 = [ n12k2  ] = 7.99105 m-1; Substitute these values into

B  to find

 1 / 2 2

2 3 / 2V 2 K1 (a)

2

R 1 / 2 exp(

 B  (2.08 103 ) R 1 / 2 exp(

2 3 R) 3 2

R ) 0.00089

which is plotted on the RHS of Figure 2Q37-1.

Figure 2Q37-1 Bending loss B vs. bend radius R (LiveMath used.)

Results compare reasonably with the experiments in Figure 2.56 given the approximate nature of the theory. Note that the calculated attenuation is per meter (for 1 meter) whereas the attenuation in Figure 2.56 is for a 10 cm fiber, so that for a 1 m of fiber, the observed attenuation will be 10 times higher. 2.38 Fiber Bragg grating A silica fiber based FBG is required to operate at 850 nm. What should be the periodicity of the grating ? If the amplitude of the index variation n is 2×10-5 and total length of the FBG is 5 mm, what are the maximum reflectance at the Bragg wavelength and the bandwidth of the FBG? Assume that the effective refractive index n is 1.460. What are the reflectance and the bandwidth if n is 2×10-4? Solution

Using equation for Bragg wavelength B  2n  one can get  

B 2n

= 291.1 nm. The results of further

calculations for n = 2×10-5 and 2×10-4 are collected in Table. n

FBG #1 FBG #2 2×10‐5

1×10‐4



Chapter 2

2.46

3 February 2013

(1/m)

73.92

739.2

L

0.37

3.7

Grating is

weak

strong

R  tanh 2 (L)

0.125

0.998

NA

0.47

0.099

N/A

strong 

42B , nm n

weak 

2B nL

, nm

The parameter L for FBG#1 with n = 2×10-5 is equal to 0.37 which is a weak grating. The parameter L for FBG#2 with n = 2×10-4 is equal to 3.7 which is a strong grating. 2.39 Fiber Bragg grating sensor array Consider a FBG strain sensor array embedded in a silica fiber that is used to measure strain at various locations on an object. Two neighboring sensors have grating periodicities of 1 = 534.5 nm and 2 = 539.7 nm. The effective refractive index is 1.450 and the photoelastic coefficient is 0.22. What is the maximum strain that can be measured assuming that (a) only one of the sensors is strained; (b) when the sensors are strained in opposite directions ? What would be the main problem with this sensor array? What is the strain at fracture if the fiber fractures roughly at an applied stress of 700 MPa and the elastic modulus is 70 GPa? What is your conclusion? Solution

Initially the Bragg wavelengths of two sensors are B1  2n1 =1550.05 nm and B 2  2n 2 = 1565.13 nm, respectively. When the second sensor is stretched its effective refractive index changes due to photoelastic effect and there is also a change in the period, both of which leads to

B 2  B 2 (1  12 n 2 pe ) and B 2 shifts towards B1 . (a) The separation between the Bragg wavelengths is B = B 2  B1 = 1565.13 – 1550.05 = 15.08 nm The shift due to strain is (only B is strained)

B 2 =  B 2 1  12 n 2 pe   2n 2 1  12 n 2 pe 

Maximum strain occurs when

B = B 2  B1  2n 2  2n1  2n 2 1  12 n 2 pe 





 2  1 = 0.012 or 1.2%  2 (1  12 n 2 pe )

The strain at fracture is given by strain = stress / elastic modulus = 700×106 / 70×109 = 0.01 or 1%. The fiber is likely to fracture before it reaches the maximum strain.



Chapter 2

2.47

3 February 2013

(b) Consider the sensors strained in opposite directions. The separation between the Bragg wavelengths is still B 2  B1 =1565.13 – 1550.05 = 15.08 nm. Note that B 2  B1 = 2n(  2   2 ) Shift due to strain is now

B =  B 2 1  12 n 2 pe   B1 1  12 n 2 pe   2n(1   2 )1  12 n 2 pe 

Which must be B 2  B1 so that





2n( 2   2 )  2n(1   2 ) 1  12 n 2 pe  



2  2 = 0.0063 or 0.63% (about half the above value). (1   2 ) 1  12 n 2 pe





The main problem is precise compensation of temperature.


Solutions Manual to Optoelectronics and Photonics: Principles and Practices, Second Edition Safa Kasap

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