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CHAPTER 33
THE NATURE AND PROPAGATION OF LIGHT (14TH ED.)
33
33.1. IDENTIFY: For reflection,  r   a .
SET UP: The desired path of the ray is sketched in Figure 33.1.
140 cm
EXECUTE: tan  
, so   506. r  90    394 and r  a  394.
115 cm
EVALUATE: The angle of incidence is measured from the normal to the surface.
Figure 33.1
33.9. IDENTIFY and SET UP: Use Snell’s law to find the index of refraction of the plastic and then use
c
n  to calculate the speed v of light in the plastic.
v
EXECUTE: na sina  nb sinb .
 sin  a 
 sin 627 
nb  na 
  100 
  1194.
sin

 sin 481 
b

c
c
n  so v   (300 108 m/s)/1194  251108 m/s.
v
n
EVALUATE: Light is slower in plastic than in air. When the light goes from air into the plastic it is bent
toward the normal.
33.11. IDENTIFY: The figure shows the angle of incidence and angle of refraction for light going from the water
into material X. Snell’s law applies at the air-water and water-X boundaries.
SET UP: Snell’s law says na sin a  nb sin b . Apply Snell’s law to the refraction from material X into the
water and then from the water into the air.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
33-1
33-2
Chapter 33
EXECUTE: (a) Material X to water: na  nX , nb  nw  1333. a  25 and b  48.
 sin b 
 sin 48 
na  nb 
  (1333) 
  234.
sin

 sin 25 
a

(b) Water to air: As Figure 33.11 shows,  a  48. na  1333 and nb  100.
n 
sin b   a  sin  a  (1333)sin 48  82.
 nb 
EVALUATE: n  1 for material X, as it must be.
Figure 33.11
33.15. IDENTIFY: The critical angle for total internal reflection is  a that gives b  90 in Snell’s law.
SET UP: In Figure 33.15 the angle of incidence  a is related to angle  by a    90.
EXECUTE: (a) Calculate  a that gives b  90. na  160, nb  100 so na sina  nb sinb gives
100
and  a  387.   90  a  513.
160
1333
(b) na  160, nb  1333. (160)sina  (1333)sin90. sin a 
and  a  564.
160
  90  a  336.
(160)sina  (100)sin90. sina =
EVALUATE: The critical angle increases when the ratio
na
decreases.
nb
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The Nature and Propagation of Light
33-3
Figure 33.15
33.19. IDENTIFY and SET UP: For glass  water, crit  487. Apply Snell’s law with a  crit to calculate
the index of refraction na of the glass.
EXECUTE: na sincrit  nb sin90, so na 
nb
1333

 177
sincrit sin 487
EVALUATE: For total internal reflection to occur the light must be incident in the material of larger
refractive index. Our results give nglass  nwater , in agreement with this.
33.23. IDENTIFY: The index of refraction depends on the wavelength of light, so the light from the red and violet
ends of the spectrum will be bent through different angles as it passes into the glass. Snell’s law applies at
the surface.
SET UP: na sina  nb sinb . From the graph in Figure 33.17 in the textbook, for   400 nm (the violet
end of the visible spectrum), n  167 and for   700 nm (the red end of the visible spectrum), n  162.
The path of a ray with a single wavelength is sketched in Figure 33.23.
Figure 33.23
EXECUTE: For   400 nm, sin b 
na
100
sin a 
sin350, so b  201. For   700 nm,
nb
167
100
sin350, so b  207.  is about 0.6.
162
EVALUATE: This angle is small, but the separation of the beams could be fairly large if the light travels
through a fairly large slab.
sinb 
33.29. IDENTIFY and SET UP: Reflected beam completely linearly polarized implies that the angle of incidence
n
equals the polarizing angle, so  p  545. Use Brewster’s law, tanp  b , to calculate the refractive
na
index of the glass. Then use Snell’s law to calculate the angle of refraction. See Figure 33.29.
n
EXECUTE: (a) tanp  b gives nglass  nair tanp  (100) tan545  140.
na
(b) na sin a  nb sin b .
sinb 
na sin a (100)sin545

 05815 and b  355.
nb
140
EVALUATE:
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
33-4
Chapter 33
Note:   1800  r  b and r  a .
Thus   1800  545  355  900;
the reflected ray and the refracted ray are
perpendicular to each other. This agrees
with Figure 33.28 in the textbook.
Figure 33.29
33.42. IDENTIFY: Because the prism is a right-angle prism, the normals at point A and at surface BC are
perpendicular to each other (see Figure 33.42). Therefore the angle of incidence at A is 50.0°, and this is
the critical angle at that surface. Apply Snell’s law at A and at surface BC. For light incident at the critical
angle, the angle of refraction is 90°.
Figure 33.42
SET UP: Apply Snell’s law: na sin a  nb sin b . Use n = 1.00 for air, and let n be the index of refraction
of the glass.
EXECUTE: Apply Snell’s law at point A.
n sin(50.0°) = (1.00) sin(90°) = 1.00.
n = 1.305.
Now apply Snell’s law at surface BC.
(1.00) sin  = (1.305) sin(40.0°).
 = 57.0°.
EVALUATE: The critical angle at A would not be 50.0° if the prism were not a right-angle prism.
33.45. IDENTIFY: Use Snell’s law to determine the effect of the liquid on the direction of travel of the light as it
enters the liquid.
SET UP: Use geometry to find the angles of incidence and refraction. Before the liquid is poured in, the
ray along your line of sight has the path shown in Figure 33.45a (next page).
80 cm
 0500.
160 cm
a  2657.
tana 
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The Nature and Propagation of Light
33-5
Figure 33.45a
After the liquid is poured in,  a is the same and the refracted ray passes through the center of the bottom
of the glass, as shown in Figure 33.45b.
40 cm
 0250.
160 cm
b  1404.
tanb 
Figure 33.45b
EXECUTE: Use Snell’s law to find nb , the refractive index of the liquid:
na sin a  nb sin b .
nb 
na sin a (100)(sin 2657)

 184.
sin b
sin1404
EVALUATE: When the light goes from air to liquid (larger refractive index) it is bent toward the normal.
33.50. IDENTIFY: The ray shown in the figure that accompanies the problem is to be incident at the critical angle.
SET UP: b  90. The incident angle for the ray in the figure is 60.
 n sin  a   156 sin 60 
EXECUTE: na sina  nb sinb gives nb   a

  135.
 sin b   sin 90 
EVALUATE: Total internal reflection occurs only when the light is incident in the material of the greater
refractive index.
© Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.