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
2
n = 1.50
35-3. Compute the speed of light in (a) crown glass, (b) diamond, (c) water, and (d) ethyl
alcohol.
(a) vg 
(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
; 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
; v = 2.26x 108 m/s
(a) vw 
1.33
3 x 108 m/s
; va = 2.21 x 108 m/s
(b) va 
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  g  na sin  a ;
sin  g 
487
(1.0)sin 370
;
1.6
g = 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  a  nw sin  w ;
sin w 
(1)sin 600
sin w 
 0.651;
1.33
na sin  a
nw
air
nw = 1.5
600
water
w = 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?
 = 900 – 320 = 580;
sin w 
na sin  a  nw sin  w ;
(1)sin 580
 638;
1.33
sin w 
na sin  a
nw
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  x  na sin  a ;
nx 
na sin  a (1)(sin 600 )

;
sin 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? [ A = 900 – 350 = 550. ]
nA sin  A  nB sin  B ;
nB sin  A sin 550


;
nA sin  B sin 350
350
B
0
nB sin 55

 1.43;
nA sin 350
nr = 1.43
488

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?
(1)sin 450
nm 
;
sin 350
nA sin  A  nm sin  m ;
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
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 air = nw sin w = ng sin g = nair sin air
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:
e = 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 air = nw sin w = ng sin g = nair sin air;
e = i = 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.
g na
 ;
a ng
g 
a na
ng

(589 nm)(1)
;
1.47
489
g = 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 x is equal to ¾ of its air value:
x
 0.750;
air
nair

n
 x  0.750; nair  air ;
nx
air
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

air
n 
(1)(600 nm)
; g  air air 
;
g
ng
1.5
g = 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?
n 
nL r
(1)(620 nm)
 ; nL  air r 
;
nair b
b
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  A A
 ;
sin  B B
B 
A sin  B (400 nm)sin500

;
sin  A
sin300
B = 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).
sin  c 
n2 1.33

;
n1 1.54
490
c = 59.70
Physics, 6th Edition
Chapter 35. Refraction
35-19. The critical angle for a given medium relative to air is 400. What is the index of
refraction for the medium?
n2
;
n1
sin  c 
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  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  c 
n2
;
n1
sin c 
1.0
;
2.42
c = 24.40
Water:
sin  c 
n2
;
n1
sin c 
1.0
;
1.33
c = 48.80
Alcohol:
sin  c 
n2
;
n1
sin c 
1.0
;
1.36
c = 47.30
35-22. What is the critical angle for flint glass immersed in ethyl alcohol?
sin  c 
n2 1.36

;
n1 1.63
c = 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?
n
1.33
c < 450; sin  c  w 
;
np
np
np = 1.88
1.33
np 
;
sin 450
(Minimum for total internal reflection.)
491
np
nw = 1.33
Physics, 6th Edition
Chapter 35. Refraction
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  x  nw sin  w ;
nw sin  w (1.33)(sin 300 )
nx 

;
sin 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 air

;
vx  x
air 
(400 nm)(3 x 108 m/s)
;
(2.40 x 108 m/s)
x = 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?
ng
nA sin  A  ng sin  g ;
nA

sin  A sin 600

;
sin  g sin 300
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  2 n1
 ;
sin 1 n2
sin 1 
n2 sin  2 1.6sin 300

;
n1
1.4
1 = 34.80
35-28. In going from glass (n = 1.50) to water (n = 1.33), what is the critical angle for total
internal reflection?
sin  c 
n2 1.33

;
n1 1.50
492
c = 62.50
Physics, 6th Edition
Chapter 35. Refraction
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

g
;
a
air 
n = 1.76
(3 x 108 m/s)(650 nm)
; air = 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  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);
493
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  will the ray emerge from the other side?
sin 300 1.50

;
sin 1
1.0
sin 1 
sin 300
;
1.5
1  19.470
600
0
2 = 900 – 19.470 = 70.50;
3 = 49.470;

4 = 900 – 49.460;
0
sin 40.5 1.00

;
sin 
1.5
2
30
1
4 = 40.530
sin   1.5sin 40.50 ;
 = 77.10
n = 1.5
3

4
60
60
0
0
*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 4 ]
sin  4c 
nair
1

;  4c  41.80 ;
ng 1.5
Now find 1:
3 = 900 – 41.80 = 48.20
2 = 1800 – (48.20 + 600); 2 = 71.8 and 1 = 900 – 71.80 = 18.20
sin18.20 1.00

;
sin  min 1.50
sin  min  1.5sin18.20 ;
494
min = 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 1 is 400 and the glass
(n = 1.50) is 2 cm thick.
sin 400 1.5

;
sin  2 1.0
400
500
 2  25.40
3 = 900 – (25.40 + 500); 3 = 14.60
cos 25.40 
sin  3 
2 cm
;
R
2 3
R
2 cm
t
d
R = 2.21 cm ;
d
; d  R sin  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 1 and is refracted into the glass where it continues to the top
surface of the block. What is the minimum angle 1 at the side such that the light does
not go out of the glass at the top? (First find c )
sin  c 
1.33
 0.864;  c  59.7 0
1.54
2 = 90 – 59.7 = 30.3 ;
0
nw = 1.33
0
sin 1 n2
 ;
sin  2 n1
0
c
2
2 = 30.3
0
1
ng = 1.54
sin1
1.54

;
0
sin30.3 1.33
sin 1 
1.54sin 30.30
;
1.33
1 = 35.70
For angles smaller than 35.70, the light will leave the glass at the top surface.
495
Physics, 6th Edition
Chapter 35. Refraction
*35-37. Prove that the lateral displacement in Fig. 35-22 can be calculated from
 n cos1 
d  t sin 1 1  1

 n2 cos 2 
Use this relationship to verify the answer to Critical Thinking Problem 35-35.
cos 1 
tan 1 
d
;
p
p
sin 1 p  a

;
cos 1
t
a  t tan  2 ;
at
d
cos 1
tan  2 
a
;
t
1
sin  2
d
; p
cos  2
cos 1
sin  2
 d
t

sin 1 p  a
cos1
cos  2


cos1
t
t



t
1
2
d






a
sin  2 
p
1
1
Simplifying this expression and solving for d, we obtain: d  t sin 1 
Now, n2 sin2 = n1 sin1 or
t
2
d
t sin  2 cos 1
cos  2
n1
sin 1
n2
Substitution and simplifying, we finally obtain the expression below:
 n cos1 
d  t sin 1 1  1

 n2 cos 2 
Substitution of values from Problem 35-35, gives the following result for d:
d = 5.59 mm
496