This chapter discusses refraction of light, including: - The index of refraction, which is the ratio of the speed of light in a vacuum to its speed in a medium. - Snell's law, which relates the angles of incidence and refraction based on the indices of refraction of the media. - Total internal reflection, which occurs when light travels from an optically dense medium to a less dense one at an angle greater than the critical angle. - Wavelength changes when light moves between media due to the index of refraction. Several example problems are worked through applying these concepts to compute angles, indices of refraction, wavelengths and speeds of light in various materials.

1. Chapter 35. Refraction Physics, 6th Edition
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?
c 3 x 108 m/s
n= = ; n = 1.88
v 1.6 x 108 m/s
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:
c c 3
vx = ( 2 3 )c; n= = ; n= and n = 1.50
vx ( 3
2 )c 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.)
3 x 108 m/s 3 x 108 m/s
(a) vg = ; vg = 1.97 x 108 m/s (b) vd = ; vd = 1.24 x 108 m/s
1.50 2.42
3 x 108 m/s 3 x 108 m/s
(a) vw = ; v = 2.26x 108 m/s (b) va = ; va = 2.21 x 108 m/s
1.33 1.36
35-4 If light travels at 2.1 x 108 m/s in a transparent medium, what is the index of refraction?
(3 x 108 m/s)
n= ; n = 1.43
2.1 x 108 m/s
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?
(1.0) sin 37 0
ng sin θ g = na sin θ a ; sin θ g = ; θg = 22.10
1.6
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2. Chapter 35. Refraction Physics, 6th Edition
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 = 1.5
na sin θ a = nw sin θ w ; sinθ w = 600
nw air
water
(1)sin 600
sinθ w = = 0.651; θw = 40.6 0
1.33
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
θ = 900 – 320 = 580; na sin θ a = nw sin θ w ; sinθ w =
nw
(1)sin 580
sinθ w = = 638; θw = 39.60
1.33
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?
na sin θ a (1)(sin 600 )
nx sin θ x = na sin θ a ; nx = = ; n = 1.35
sinθ x sin 400
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. ]
nB sin θ A sin 550 θ A 350
nA sin θ A = nB sin θ B ; = = ; 350
nA sin θ B sin 350
B
nB sin 550
= = 1.43; nr = 1.43
nA sin 350
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3. Chapter 35. Refraction Physics, 6th Edition
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?
(1)sin 450
nA sin θ A = nm sin θ m ; nm = ; nm = 1.23
sin 350
*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
into air again. Compute the angle of emergence. θair nair = 1
The angle of refraction into one medium becomes the nw = 1.33
angle of incidence for the next, and so on . . .
ng = 1.55
nair sin θair = nw sin θw = ng sin θg = 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 = θi = 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 θair = nw sin θw = ng sin θg = nair sin θair; θe = 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 na (589 nm)(1)
= ; λg = = ; λg = 401 nm
λa ng ng 1.47
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4. Chapter 35. Refraction Physics, 6th Edition
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 nair λ n
= 0.750; = x = 0.750; nair = air ; nx = 1.33
λair nx λair 0.750
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 λair n λ (1)(600 nm)
= ; λg = air air = ; λg = 400 nm
nair λg ng 1.5
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 = ; nL = 1.30
nair λb λb 478 nm
*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 λ A sin θ B (400 nm)sin500
= ; λB = = ; λB = 613 nm
sin θ B λB sin θ A sin300
Total Internal Reflection
35-18. What is the critical angle for light moving from quartz (n = 1.54) to water (n = 1.33).
n2 1.33
sin θ c = = ; θc = 59.70
n1 1.54
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5. Chapter 35. Refraction Physics, 6th Edition
35-19. The critical angle for a given medium relative to air is 400. What is the index of
refraction for the medium?
n2 nair 1
sin θ c = ; nx = 0
= ; nx = 1.56
n1 sin 40 sin 400
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?
n2 nair 1
sin θ c = ; nx = 0
= ; nx = 1.39
n1 sin 46 sin 460
35-21. What is the critical angle relative to air for (a) diamond, (b) water, and (c) ethyl alcohol.
n2 1.0
Diamond: sin θ c = ; sinθ c = ; θc = 24.40
n1 2.42
n2 1.0
Water: sin θ c = ; sinθ c = ; θc = 48.80
n1 1.33
n2 1.0
Alcohol: sin θ c = ; sinθ c = ; θc = 47.30
n1 1.36
35-22. What is the critical angle for flint glass immersed in ethyl alcohol?
n2 1.36
sin θ c = = ; θc = 56.50
n1 1.63
*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
nw 1.33 1.33
θc < 450; sin θ c = = ; np = ;
np np sin 450
np = 1.88 (Minimum for total internal reflection.)
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6. Chapter 35. Refraction Physics, 6th Edition
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?
nw sin θ w (1.33)(sin 300 )
nx sin θ x = nw sin θ w ; nx = = ; n = 1.50, glass
sinθ x sin 26.30
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 (400 nm)(3 x 108 m/s)
= ; λair = ; λx = 500 nm
vx λx (2.40 x 108 m/s)
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?
ng sin θ A sin 600
nA sin θ A = ng sin θ g ; = = ; nr = 1.73
nA sin θ g sin 300
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 n2 sin θ 2 1.6sin 300
= ; sin θ1 = = ; θ1 = 34.80
sin θ1 n2 n1 1.4
35-28. In going from glass (n = 1.50) to water (n = 1.33), what is the critical angle for total
internal reflection?
n2 1.33
sin θ c = = ; θc = 62.50
n1 1.50
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7. Chapter 35. Refraction Physics, 6th Edition
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= ; n = 1.76
1.7 x 108 m/s
vg λg (3 x 108 m/s)(650 nm)
= ; λair = ; λair = 1146 nm
vair λa 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?
n2 1.0
sin θ c = ; n1 = ; n1 = 1.62
n1 sin 380
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 2m
= = ; q= ; q = 1.50 m
p nw 1.33 1.33
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 = 1.50q = (1.50)(3 cm); p = 4.50 cm
p ng 1.50
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8. Chapter 35. Refraction Physics, 6th Edition
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 300
= ; sin θ1 = ; θ1 = 19.47 0 600
sin θ1 1.0 1.5 n = 1.5
0
30 θ2
θ2 = 90 – 19.47 = 70.5 ; θ3 = 180 − (60 + 70.5 )
0 0 0 0 0 0
θ1
θ3
θ4 θ
θ3 = 49.47 ;0
θ4 = 90 – 49.46 ;
0 0
θ4 = 40.53 0
600 600
sin 40.50 1.00
= ; sin θ = 1.5sin 40.50 ; θ = 77.10
sin θ 1.5
*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 ]
nair 1
sin θ 4 c = = ; θ 4c = 41.80 ; Now find θ1: θ3 = 900 – 41.80 = 48.20
ng 1.5
θ2 = 1800 – (48.20 + 600); θ2 = 71.8 and θ1 = 900 – 71.80 = 18.20
sin18.20 1.00
= ; sin θ min = 1.5sin18.20 ; θmin = 27.90
sin θ min 1.50
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9. Chapter 35. Refraction Physics, 6th Edition
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 θ1 is 400 and the glass
(n = 1.50) is 2 cm thick.
400
sin 400 1.5 500
= ; θ 2 = 25.40
sin θ 2 1.0
θ2 θ3 t
2 cm
θ3 = 900 – (25.40 + 500); θ3 = 14.60 R
d
2 cm
cos 25.40 = ; R = 2.21 cm ;
R
d
sin θ 3 = ; d = R sin θ 3 = (2.21 cm) sin14.60 ; d = 5.59 mm
R
*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 the glass at the top? (First find θc ) nw = 1.33
1.33 θc
sin θ c = = 0.864; θ c = 59.70
1.54
θ2
θ2 = 90 – 59.7 = 30.3 ;
0 0 0
θ2 = 30.3 0
θ1
ng = 1.54
sin θ1 n2 sinθ1 1.54
= ; = ;
sin θ 2 n1 0
sin30.3 1.33
1.54sin 30.30
sin θ1 = ; θ1 = 35.70
1.33
For angles smaller than 35.70, the light will leave the glass at the top surface.
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10. Chapter 35. Refraction Physics, 6th Edition
*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.
d d
cosθ1 = ; p=
p cos θ1
sin θ1 p + a a
tan θ1 = = ; tan θ 2 = ; θ1
cosθ1 t t
sin θ 2 d θ2 θ1
a = t tan θ 2 ; a=t ; p= t θ2 t
cos θ 2 cosθ1
d
p
 d sin θ 2  a θ1
sin θ1 p + a  cos θ + t cos θ  θ1 d
= = 1 2

cosθ1 t  t 
 
 
t sin θ 2 cos θ1
Simplifying this expression and solving for d, we obtain: d = t sin θ1 −
cos θ 2
n1
Now, n2 sinθ2 = n1 sinθ1 or sin θ 2 = 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:
d = 5.59 mm
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