Physics, 6th Edition
Chapter 35. Refraction
Chapter 35. Refraction The Index of Refraction (Refer to Table 35-1 for values of n.) 35-1. The speed of light through a certain medium is 1.6 x 108 m/s in a transparent medium, what is the index of refraction in that medium? n=
c 3 x 108 m/s = ; v 1.6 x 108 m/s
n = 1.88
35-2. If the speed of light is to be reduced by one-third, what must be the index of refraction for the medium through which the light travels? The speed c is reduced by a third, so that: vx = ( 2 3 )c;
n=
c c = ; 2 v x ( 3 )c
n=
3 and n = 1.50 2
35-3. Compute the speed of light in (a) crown glass, (b) diamond, (c) water, and (d) ethyl alcohol. (Since n = c/v, we find that v = c/n for each of these media.) (a) vg =
3 x 108 m/s ; vg = 1.97 x 108 m/s 1.50
(b) vd =
3 x 108 m/s ; vd = 1.24 x 108 m/s 2.42
(a) vw =
3 x 108 m/s ; v = 2.26x 108 m/s 1.33
(b) va =
3 x 108 m/s ; va = 2.21 x 108 m/s 1.36
35-4 If light travels at 2.1 x 108 m/s in a transparent medium, what is the index of refraction? n=
(3 x 108 m/s) ; 2.1 x 108 m/s
n = 1.43
The Laws of Refraction 35-5. Light is incident at an angle of 370 from air to flint glass (n = 1.6). What is the angle of refraction into the glass? ng sin θ g = na sin θ a ;
sin θ g =
231
(1.0) sin 370 ; 1.6
θg = 22.10
Physics, 6th Edition
Chapter 35. Refraction
35-6. A beam of light makes an angle of 600 with the surface of water. What is the angle of refraction into the water? na sin θ a = nw sin θ w ;
sinθ w =
(1)sin 600 sinθ w = = 0.651; 1.33
na sin θ a nw
air
nw = 1.5
600
water
θw = 40.6
0
35-7. Light passes from water (n = 1.33) to air. The beam emerges into air at an angle of 320 with the horizontal water surface? What is the angle of incidence inside the water? na sin θ a = nw sin θ w ;
θ = 900 – 320 = 580;
sinθ w =
(1)sin 580 = 638; 1.33
sinθ w =
na sin θ a nw
θ w = 39.60
35-8. Light in air is incident at 600 and is refracted into an unknown medium at an angle of 400. What is the index of refraction for the unknown medium? nx sin θ x = na sin θ a ;
nx =
na sin θ a (1)(sin 600 ) = ; sinθ x sin 400
n = 1.35
35-9. Light strikes from medium A into medium B at an angle of 350 with the horizontal boundary. If the angle of refraction is also 350, what is the relative index of refraction between the two media? [ θA = 900 – 350 = 550. ] nA sin θ A = nB sin θ B ;
nB sin θ A sin 550 = = ; nA sin θ B sin 350
nB sin 550 = = 1.43; nA sin 350
nr = 1.43
232
350 B
θ
A
350
Physics, 6th Edition
Chapter 35. Refraction
35-10. Light incident from air at 450 is refracted into a transparent medium at an angle of 340. What is the index of refraction for the material? nA sin θ A = nm sin θ m ;
nm =
(1)sin 450 ; sin 350
nm = 1.23
*35-11. A ray of light originating in air (Fig. 35-20) is incident on water (n = 1.33) at an angle of 600. It then passes through the water entering glass (n = 1.50) and finally emerging back θ air
into air again. Compute the angle of emergence. The angle of refraction into one medium becomes the
nair = 1 nw = 1.33
angle of incidence for the next, and so on . . .
ng = 1.55
nair sin θair = nw sin θw = ng sin θ g = nair sin θair
nair = 1
Thus it is seen that a ray emerging into the same medium as that from which it originally entered has the same angle:
θe = θ i = 600
*35-12. Prove that, no matter how many parallel layers of different media are traversed by light, the entrance angle and the final emergent angle will be equal as long as the initial and final media are the same.
The prove is the same as shown for Problem 35-11:
nair sin θ air = nw sin θ w = ng sin θg = nair sin θ air;
θ e = θi = 600
Wavelength and Refraction 35-13. The wavelength of sodium light is 589 nm in air. Find its wavelength in glycerine. From Table 28-1, the index for glycerin is: n = 1.47.
λg na = ; λa ng
λg =
λa na (589 nm)(1) = ; ng 1.47 233
λg = 401 nm
Physics, 6th Edition
Chapter 35. Refraction
35-14. The wavelength decreases by 25 percent as it goes from air to an unknown medium. What is the index of refraction for that medium? A decrease of 25% means λx is equal to ¾ of its air value:
λx = 0.750; λair
nair λ n = x = 0.750; nair = air ; nx λair 0.750
nx = 1.33
35-15. A beam of light has a wavelength of 600 nm in air. What is the wavelength of this light as it passes into glass (n = 1.50)? ng nair
=
λair n λ (1)(600 nm) ; λg = air air = ; λg ng 1.5
λg = 400 nm
35-16. Red light (620 nm) changes to blue light (478 nm) when it passes into a liquid. What is the index of refraction for the liquid? What is the velocity of the light in the liquid? nL λr n λ (1)(620 nm) = ; nL = air r = ; nair λb λb 478 nm
nL = 1.30
*35-17. A ray of monochromatic light of wavelength 400 nm in medium A is incident at 300 at the boundary of another medium B. If the ray is refracted at an angle of 500, what is its wavelength in medium B? sin θ A λ A = ; sin θ B λB
λB =
λ A sin θ B (400 nm)sin500 = ; sin θ A sin300
λB = 613 nm
Total Internal Reflection 35-18. What is the critical angle for light moving from quartz (n = 1.54) to water (n = 1.33).
234
Physics, 6th Edition
Chapter 35. Refraction sin θ c =
n2 1.33 = ; n1 1.54
θc = 59.70
35-19. The critical angle for a given medium relative to air is 400. What is the index of refraction for the medium? sin θ c =
n2 ; n1
nx =
nair 1 = ; 0 sin 40 sin 400
nx = 1.56
35-20. If the critical angle of incidence for a liquid to air surface is 460, what is the index of refraction for the liquid? sin θ c =
n2 ; n1
nx =
nair 1 = ; 0 sin 46 sin 460
nx = 1.39
35-21. What is the critical angle relative to air for (a) diamond, (b) water, and (c) ethyl alcohol. Diamond:
sin θ c =
n2 ; n1
sinθ c =
1.0 ; 2.42
θ c = 24.40
Water:
sin θ c =
n2 ; n1
sinθ c =
1.0 ; 1.33
θ c = 48.80
Alcohol:
sin θ c =
n2 ; n1
sinθ c =
1.0 ; 1.36
θ c = 47.30
35-22. What is the critical angle for flint glass immersed in ethyl alcohol? sin θ c =
n2 1.36 = ; n1 1.63
θc = 56.50
*35-23. A right-angle prism like the one shown in Fig. 35-10a is submerged in water. What is the minimum index of refraction for the material to achieve total internal reflection? np nw = 1.33
235
Physics, 6th Edition
Chapter 35. Refraction nw 1.33 θc < 450; sin θ c = n = n ; p p
np =
1.33 ; sin 450
np = 1.88 (Minimum for total internal reflection.)
Challenge Problems 35-24. The angle of incidence is 300 and the angle of refraction is 26.30. If the incident medium is water, what might the refractive medium be? nx sin θ x = nw sin θ w ;
nx =
nw sin θ w (1.33)(sin 300 ) = ; sinθ x sin 26.30
n = 1.50, glass
35-25. The speed of light in an unknown medium is 2.40 x 108 m/s. If the wavelength of light in this unknown medium is 400 nm, what is the wavelength in air? c λair = ; vx λx
λair
(400 nm)(3 x 108 m/s) = ; (2.40 x 108 m/s)
λx = 500 nm
35-26. A ray of light strikes a pane of glass at an angle of 300 with the glass surface. If the angle of refraction is also 300, what is the index of refraction for the glass? nA sin θ A = ng sin θ g ;
sin θ A sin 600 = = ; nA sin θ g sin 300 ng
nr = 1.73
35-27. A beam of light is incident on a plane surface separating two media of indexes 1.6 and 1.4. The angle of incidence is 300 in the medium of higher index. What is the angle of refraction? sin θ 2 n1 = ; sin θ1 n2
n2 sin θ 2 1.6sin 300 sin θ1 = = ; n1 1.4
236
θ1 = 34.80
Physics, 6th Edition
Chapter 35. Refraction
35-28. In going from glass (n = 1.50) to water (n = 1.33), what is the critical angle for total internal reflection? sin θ c =
n2 1.33 = ; n1 1.50
θ c = 62.50
35-29. Light of wavelength 650 nm in a particular glass has a speed of 1.7 x 108 m/s. What is the index of refraction for this glass? What is the wavelength of this light in air? 3 x 108 m/s n= ; 1.7 x 108 m/s vg vair
=
λg ; λa
λair =
n = 1.76
(3 x 108 m/s)(650 nm) ; λair = 1146 nm 1.7 x 108 m/s
35-30. The critical angle for a certain substance is 380 when it is surrounded by air. What is the index of refraction of the substance? sin θ c =
n2 ; n1
n1 =
1.0 ; sin 380
n1 = 1.62
35-31. The water in a swimming pool is 2 m deep. How deep does it appear to a person looking vertically down? q nair 1.00 = = ; p nw 1.33
q=
2m ; 1.33
q = 1.50 m
35-32. A plate of glass (n = 1.50) is placed over a coin on a table. The coin appears to be 3 cm below the top of the glass plate. What is the thickness of the glass plate? q nair 1.00 = = ; p ng 1.50
p = 1.50q = (1.50)(3 cm);
237
p = 4.50 cm
Physics, 6th Edition
Chapter 35. Refraction
Critical Thinking Problems *35-33. Consider a horizontal ray of light striking one edge of an equilateral prism of glass (n = 1.50) as shown in Fig. 35-21. At what angle θ will the ray emerge from the other side? sin 300 1.50 = ; sin θ1 1.0
sin θ1 =
sin 300 ; 1.5
600
θ1 = 19.47 0 0
30 0
0
θ2
0
θ 2 = 90 – 19.47 = 70.5 ; θ3 = 1 80 − (60 + 70.5 ) 0
0
θ 3 = 49.470;
0
θ4 = 900 – 49.460;
sin 40.50 1.00 = ; sin θ 1.5
θ4 = 40.530
sin θ = 1.5sin 40.50 ;
θ1
600
n = 1.5 θ3
θ
θ4
600
θ = 77.10
*35-34. What is the minimum angle of incidence at the first face of the prism in Fig. 35-21 such that the beam is refracted into air at the second face? (Larger angles do not produce total internal reflection at the second face.) [ First find critical angle for θ4 ] sin θ 4 c =
nair 1 = ; θ 4c = 41.80 ; ng 1.5
Now find θ1: θ3 = 900 – 41.80 = 48.20
θ2 = 1800 – (48.20 + 600); θ 2 = 71.8 and θ1 = 900 – 71.80 = 18.20 sin18.20 1.00 = ; sin θ min 1.50
sin θ min = 1.5sin18.20 ;
238
θmin = 27.90
Physics, 6th Edition
Chapter 35. Refraction
35-35. Light passing through a plate of transparent material of thickness t suffers a lateral displacement d, as shown in Fig. 35-22. Compute the lateral displacement if the light passes through glass surrounded by air. The angle of incidence θ 1 is 400 and the glass = 1.50) is 2 cm thick. sin 400 1.5 = ; sin θ 2 1.0
400 500
θ 2 = 25.40
θ 3 = 900 – (25.40 + 500); θ3 = 14.60 cos 25.40 = sin θ 3 =
(n
2 cm ; R
θ2 θ3
2 cm
t
R
d
R = 2.21 cm ;
d ; d = R sin θ 3 = (2.21 cm) sin14.60 ; R
d = 5.59 mm
*35-36. A rectangular shaped block of glass (n = 1.54) is submerged completely in water (n = 1.33). A beam of light traveling in the water strikes a vertical side of the glass block at an angle of incidence θ1 and is refracted into the glass where it continues to the top surface of the block. What is the minimum angle θ1 at the side such that the light does not go out of nw = 1.33
the glass at the top? (First find θ c ) sin θ c =
θc
1.33 = 0.864; θ c = 59.70 1.54
θ2 = 900 – 59.70 = 30.30;
θ2 = 30.30
θ1
θ2 ng = 1.54
239
Physics, 6th Edition
Chapter 35. Refraction sin θ1 n2 = ; sin θ 2 n1
sinθ1 1.54 = ; 0 sin30.3 1.33 1.54sin 30.30 sin θ1 = ; 1.33
θ1 = 35.70
For angles smaller than 35.70, the light will leave the glass at the top surface. *35-37. Prove that the lateral displacement in Fig. 35-22 can be calculated from n cos θ1 d = t sin θ1 1 − 1 n2 cos θ 2 Use this relationship to verify the answer to Critical Thinking Problem 35-35. cosθ1 =
tan θ1 =
d ; p
p=
sin θ1 p + a = ; cosθ1 t
a = t tan θ 2 ;
d cos θ1 tan θ 2 =
a ; t
θ1
sin θ 2 d a =t ; p= cosθ 2 cosθ1
sin θ 2 d +t sin θ1 p + a cos θ1 cosθ 2 = = cosθ1 t t
t
θ2
θ1
θ2
d
a
sin θ 2 =
p
θ1
Simplifying this expression and solving for d, we obtain: d = t sin θ1 −
Now, n2 sinθ2 = n1 sinθ1 or
t θ1
t sin θ 2 cos θ1 cos θ 2
n1 sin θ1 n2
Substitution and simplifying, we finally obtain the expression below: n cos θ1 d = t sin θ1 1 − 1 n2 cos θ 2 Substitution of values from Problem 35-35, gives the following result for d: 240
d
Physics, 6th Edition
Chapter 35. Refraction d = 5.59 mm
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