Solucionario Capitulo 37 - Paul E. Tippens

Physics, 6th Edition

Chapter 37. Interference, Diffraction, and Polarization.

Chapter 37. Interference, Diffraction, and Polarization Young’s Experiment; Interference 37-1. Light from a laser has a wavelength of 632 nm. Two rays from this source follow paths that differ in length. What is the minimum path difference require to cause (a) constructive interference, (b) destructive interference? (a) Constructive: ∆p = nλ = (1)(632 nm); ∆p = 632 nm. (b) Destructive: ∆p = nλ/2 = ½(632 nm);

∆p = 316 nm

37-2. Find the difference in path length required in Problem 37-1 to provide the very next instances of constructive and destructive interference. (a) 2nd Constructive: ∆p = nλ = (2)(632 nm); ∆p = 1264 nm. (b) 2nd Destructive: ∆p = nλ/2 =

3λ 3(632 nm) = ; ∆p = 948 nm 2 2

37-3. Monochromatic light illuminates two parallel slits 0.2 mm apart. On a screen 1.0 m from the slits, the first bright fringe is separated from the central fringe by 2.50 mm. What is the wavelength of the light? yd = nλ ; x

λ=

yd (2.5 mm)(0.2 mm) = ; nx (1)(1000 mm)

λ = 500 nm

37-4. Monochromatic light from a sodium flame illuminates two slits separated by 1.0 mm. A viewing screen is 1.0 m from the slits, and the distance from the central bright fringe to the bright fringe nearest it is 0.589 mm. What is the frequency of the light? yd c = (1)λ = ; x f

cx (3 x 108 m/s)(1 m) f = = ; yd (5.89 x 10-4 m)(0.001 mm) 254

f = 5.09 x 1014 Hz



37-5. Two slits 0.05 mm apart are illuminated by green (λ = 520 nm). A diffraction pattern is formed on a viewing screen 2.0 m away. What is the distance from the center of the screen to the first bright fringe? What is the distance to the third dark fringe? yd = nλ ( n = 1, 2, 3, . . .) x

The first bright fringe occurs when n = 1:

y=

(1)λ x (5.2 x 10-4 mm)(2000 mm) = ; d (0.05 mm)

The third dark fringe occurs when n = 5:

y=

y = 20.8 mm

yd λ =n (n = 1, 3, 5, . . .) x 2

5λ x 5(5.2 x 10-4 mm)(2000 mm) = ; 2d (2)(0.05 mm)

y = 52.0 mm

*37-6. For the situation described in Problem 37-5, what is the separation of the two first-order bright fringes located on each side of the central band? y=

(1)λ x = 20.8 mm; d

∆y = 2(20.8 mm) ;

∆y = 41.6 mm

*37-7. Young’s experiment is performed using monochromatic light of wavelength 500 nm. The slit separation is 1.20 mm and the screen is 5.00 m away. How far apart are the bright fringes? The separation of each fringe is same as for n = 1. (1)λ x (5.00 x 10-7 m)(5.00 m) y= = ; d (1.2 x 10-3m)

y = 2.08 mm

*37-8. In Young’s experiment it is noted that the second dark fringe appears at a distance of 2.5 cm from the central bright fringe. Assume that the slit separation is 60 µm and that the screen is 2.0 m away. What is the wavelength of the incident light? (Note that n = 3) yd λ = (3) ; x 2

λ=

2 yd 2(0.025 m)(60 x 10-6 m) = ; 3x (3)(2.0 m)

255

λ = 500 nm



The Diffraction Grating 37-9. A diffraction grating having 300 lines/mm is illuminated by light of wavelength 589 nm. What are the angles at which the first and second-order bright fringes are formed? d=

1 = 3.33 x 10-3 mm or d = 3.33 x 10-6 m 300 lines/mm

d sin θ = nλ ;

sin θ =

nλ (1)(589 x 10-9 m) ; = d 3.33 x 10-6 m

θ = 10.20

d sin θ = nλ ;

sin θ =

nλ (2)(589 x 10-9 m) ; = d 3.33 x 10-6 m

θ = 20.70

37-10. A diffraction grating has 250,000 lines/m. What is the wavelength of incident light if the second order bright fringe occurs at 12.60. d=

1 = 4.00 x 10-6 m; 5 2.50 x 10 lines/m

λ=

(4.00 x 10-6 m) sin12.60 ; 2

d sin θ = nλ ;

λ=

d sin θ n

λ = 436 nm

37-11. A small sodium lamp emits light of wavelength 589 nm, which illuminates a grating marked with 6000 lines/cm. Calculate the angular deviation of the first and second order bright fringes? d=

1 = 1.67 x 10-4 cm or d = 1.67 x 10-6 m 6000 lines/cm

d sin θ = nλ ;

sin θ =

nλ (1)(589 x 10-9 m) ; = d 1.67 x 10-6 m

θ = 20.70

d sin θ = nλ ;

sin θ =

nλ (2)(589 x 10-9 m) ; = d 1.67 x 10-6 m

θ = 45.00

256



37-12. A parallel beam of light illuminates a diffraction grating with 6000 lines/cm. The secondorder bright fringe is located 32.0 cm from the central image on a screen 50 cm from the grating. Calculate the wavelength of the light? d=

1 = 1.67 x 10-6 m; 6000 lines/cm

32 cm tan θ = ; θ = 32.60 ; 50 cm

d sin θ = nλ

(n = 1, 2, 3, . . .)

d sin θ (1.67 x 10-6 m)sin32.60 λ= = ; 2 2

λ = 449 nm

*37-13. The visible light spectrum ranges in wavelength from 400 to 700 nm. find the angular width of the first-order spectrum produced by passing white light through a grating marked with 20,000 lines/in. ( Angular width = θ700 - θ400 ) d=

1 = 5.00 x 10-5 in.; 20, 000 lines/in.

 1m  -6 d = (5 x 10-5in.)   = 1.27 x 10 m  39.37 in. 

d sin θ = nλ ;

nλ4 (1)(400 x 10-9 m) ; sin θ 4 = = d 1.27 x 10-6 m

θ4 = 18.40

d sin θ = nλ ;

nλ7 (1)(700 x 10-9 m) ; sin θ 7 = = d 1.27 x 10-6 m

θ7 = 33.40

Angular width = 33.40 – 18.40;

∆θ = 15.00

*37-14. An infrared spectrophotometer uses gratings to disperse infrared light. One grating is ruled with 240 lines/mm. What is the maximum wavelength that can be studied with this grating? (The maximum dispersion occurs when θ = 900 and sin θ = 1 ) d=

1 = 4.17 x 10-3 mm or d = 4.17 x 10-6 m 240 lines/mm

λ=

d sin θ (4.17 x 10-6 m)(1) = ; 1 1

257

λmax = 4.17 µm



Resolving Power of Instruments 37-15. Light of wavelength 600 nm falls on a circular opening of diameter 0.32 mm. A diffraction pattern forms on a screen 80 cm away. What is the distance from the center of the pattern to the first dark fringe? R 1.22λ θ0 = = ; f D

1.22λ f 1.22(600 x 10-9 m)(0.80 m) R= = ; D 0.32 x 10-3 m

R = 1.83 mm

37-16. The limiting angle of resolution for an objective lens in an optical instrument is 3 x 10-4 rad for a light 650 nm light source. What is the diameter of the circular opening?

θ0 =

1.22λ 1.22(650 x 10-9 m) ; ; D= D 3 x 10-4 rad

D = 2.64 mm

37-17. A certain radio telescope has a parabolic reflector that is 70 m in diameter. Radio waves from outer space have a wavelength of 21 cm. Calculate the theoretical limit of resolution for this telescope?

θ0 =

1.22λ 1.22(0.21 m) = ; D 70 m

θo = 3.66 x 10-3 rad

37-18. Using a telescope whose objective lens has a diameter of 60 m, how far apart can two objects be resolved if they are located a distance in space equal to that from the earth to the sun (93 million miles)? Assume that light is at the central wavelength of 500 nm

θ0 =

1.22λ s0 = ; D p

s0 =

1.22λ p 1.22(500 x 10-9 m)(9.3 x 107 mi) = ; D 60 m

so = 0.946 mi

258

or

4,990 ft



37-19. What is the angular limit of resolution of a person’s eye when the diameter of the opening is 3 mm? Assume that the wavelength of the light is 500 nm. 1.22λ 1.22(500 x 10-9 m) θ0 = = ; D 3 x 10-3m

θ o = 2.03 x 10-4 rad

37-20. At what distance could the eyes of Problem 37-19 resolve wires in a door screen that are separated by 2.5 mm?

θ0 =

s0 ; p

p=

s0 2.5 x 10-3 m = ; θ 0 2.03 x 10-4 rad

p = 12.3 m

Challenge Problems 37-21. In Young’s experiment 600-nm light illuminates a slit located 2.0 m from a screen. The second bright fringe formed on the screen is 5 mm from the central maximum. What is the slit width? yd nλ x 2(600 x 10-9 m)(2.0 m) = nλ ; d = = ; x y 5 x 10-3m

d = 480 µm

37-22. A transmission grating is ruled with 5000 lines/cm. For light of wavelength 550 nm, what is the angular deviation of the third order bright fringe? d=


d sin θ = nλ ;

sin θ =

nλ (3)(550 x 10-9 m) ; = d 2.00 x 10-6 m

259

θ = 55.60



37-23. Monochromatic light passes through two slits separated by 0.24 mm. In the pattern formed on a screen 50 cm away, the distance between the first bright fringe of the left of the central maximum and the first fringe on the right is 2.04 mm. What is the wavelength of the light? (The deviation of the first bright fringe is half of 2.04 mm or 1.02 mm.) yd yd (1.02 x 10-3 m)(0.24 x 10-3 m) = nλ ; λ = = ; x nx (1)(0.50 m)

λ = 490 nm

37-24. A transmission grating ruled with 6000 lines/cm forms a second-order bright fringe at an angle of 530 from the central fringe. What is the wavelength of the incident light? d=


d sin θ = nλ ;

λ=

d sin θ (1.67 x 10-6 m) sin 530 = ; n 2

λ = 667 nm

*37-25. If the separation of the two slits in Young’s experiment is 0.10 mm and the distance to the screen is 50 cm, find the distance between the first dark fringe and the third bright fringe when the slits are illuminated with light of wavelength 600 nm. Bright :

yd = nλ x

(n = 1, 2, 3, . . .)

∆y = yB 3 − yD1 =

Dark :

yd λ =n x 2

(n = 1, 3, 5, . . .)

(3)λ x (1)λ x  λx  − = (2.5)  ; d 2d  d 

2.5(600 x 10-9 m)(0.5 m) ∆y = ; 0.10 x 10-3 m

260

∆ y = 7.50 mm



*37-26. Light from a mercury-arc lamp is incident on a diffraction grating ruled with 7000 lines per inch. The spectrum consists of a yellow line (579 nm) and a blue line (436 nm). Compute the angular separation (in radians) of these lines in the third-order spectrum. d=

 1m  -6 d = (1.43 x 10-4in.)   = 3.63 x 10 m  39.37 in. 

1 = 1.43 x 10-4 in.; 7000 lines/in. d sin θ = nλ ;

sin θ y =

d sin θ = nλ ;

sin θ b =

nλ y

(3)(579 x 10-9 m) ; 3.63 x 10-6 m

θ4 = 28.60

nλb (3)(436 x 10-9 m) ; = d 3.63 x 10-6 m

θ 7 = 21.10

d

=

∆θ = 28.60 – 21.10;

∆θ = 7.480

*37-27. A telescope will be used to resolve two points on a mountain 160 km away. If the separation of the points is 2.0 m, what is the minimum diameter for the objective lens? Assume that the light has an average wavelength of 500 nm.

θ0 =

1.22λ s0 1.22λ p 1.22(500 x 10-9 m)(1.6 x 105 m) = ; D= = ; D = 4.86 cm D p s0 2m

261



Critical Thinking Problems 37-28. A Michelson interferometer, as shown in Fig. 37-18, can be used to measure small distance. The beam splitter partially reflects and partially transmits monochromatic light of wavelength λ from the source S. One mirror M1 is fixed and another M2 is movable. The light rays reaching the eye from each mirror differ, causing constructive and destructive interference patterns to move across the scope as the mirror M2 is moved a distance x. Show that his distance is given by x=m

λ 2

where m is the number of dark fringes that cross an indicator line on the scope as the mirror M2 moves a distance x. x The difference in path lengths for each ray of light will be the distance x and back, or 2x,

Source.S M1

Thus, there is constructive interference when: 2x = mλ Thus,

Fixed mirror

m = 0, 1, 2, . . .

x=m

λ 2

m = 0, 1, 2, . . .

Observer

37-29. A Michelson interferometer (see previous problem) is used to measure the advance of a small screw. How far has the screw advanced if krypton-86 light (λ = 606 nm) is used and 4000 fringes move across the field of view as the screw advances? x=

mλ (4000)(606 x 10-9 m) = ; 2 2

262

x = 1.21 mm



*37-30. A diffraction grating has 500 lines/mm ruled on its glass surface. White light passes through the grating and forms several spectra on a screen 1.0 m away. Is the deviation of colors with a grating different than those experienced for prisms? On the screen, what is the distance between the first order blue line (400 nm) and the first order red line (680 nm). How many complete spectra (400-700 nm) are possible for these conditions? (a) When white light passes through a prism, the blue light bends more than the red light since the angle of refraction is less for longer wavelengths. (See Fig. 35-8 in text.) The reverse is true for gratings. The longer wavelengths are deviated more. (b) d =

1 = 2.00 x 10-3 mm; 500 lines/mm

d = 2.00 x 10-6 m

d sin θ = nλ ;

sin θ r =

nλr (1)(680 x 10-9 m) ; = d 2.00 x 10-6 m

θ r = 19.90

d sin θ = nλ ;

sin θ b =

nλb (1)(400 x 10-9 m) ; = d 2.00 x 10-6 m

θ 7 = 11.50

tan θ =

y ; y = x tan θ ; x

θ

1m

Red Blue

∆y = yr − yb = x tan θ r − x tan θ b

∆y = (1 m) tan 19.90– (1 m) tan 11.50;

∆y = 15.9 cm

(c) The maximum number of complete spectra is found by setting θ = 900 for red line. d sin θ = nλ ;

n=

d sin 900 2.00 x 10-6 m = ; n = 2.86 λ 700 x 10-9 m

Therefore, only two complete spectra are possible for this grating.

263



*37-31. The tail lights of an automobile are separated by 1.25 m. Assume that the pupil of a person’s eye has a diameter of 5 mm and that the light has an average wavelength of 640 nm. At night, on a long straight highway, how far away can the two tail lights be resolved? Suppose you squint your eyes forming a slit where the limiting angle changes from θ0 = 1.22 λ/D to θ0 = λ/d. What is the new distance for resolution of the images?

θ0 =

1.22λ s0 = ; D p

p=

s0 D (1.25 m)(0.005 m) = ; 1.22λ 1.22(640 x 10-9 m)

λ s0 = ; d p

p=

s0 D (1.25 m)(0.005 m) = ; λ (640 x 10-9 m)

θ0 =

p = 8000 m

p = 9765 m

*37-32. The intensity of unpolarized light is reduced by one-half when it passes through a polarizer. In the case of the plane-polarized light reaching the analyzer, the intensity I of the transmitted beam is given by I = I 0 cos2 θ where I0 is the maximum intensity transmitted and θ is the angle through which the analyzer has been rotated. Consider three Polaroid plates stacked so that the axis of each is turned 300 with respect to the spreading plate. By what percentage will the incident light be reduced in intensity when it passes through all three plates? The first plate transmits ½ of the incident light intensity. Each succeeding plate transmits a vector resolution at angle 300 and transmits a fraction, cos 300 = 0.866, of the amplitude, or (cos 300)2 of the intensity. Since cos 300 = 0.866, (cos 300)2 = 0.75. Thus the intensity of the stack is: (0.5)(0.75)(0.75 = 0.281 or 28.1%

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Solucionario Capitulo 37 - Paul E. Tippens

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