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Unformatted text preview: nce is 90.0° – 40.0° = 50.0°. The
index of refraction for material c is
nc = na sin θ c = (1.80 ) sin 50.0° = 1.38 Chapter 26 Problems 1367 As expected, the ranking of the indices of refraction, highesttolowest, is na = 1.80,
nc = 1.38, nb = 1.16.
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31. REASONING AND SOLUTION
a. Using Equation 26.4 and the refractive index for crown glass given in Table 26.1, we find
that the critical angle for a crown glassair interface is 1.00 = 41.0° 1.523 θ c = sin −1 The light will be totally reflected at point A since the incident angle of 60.0° is greater than
θc. The incident angle at point B, however, is 30.0° and smaller than θc. Thus, the light will
exit first at point B .
b. The critical angle for a crown glasswater interface is 1.333 = 61.1° 1.523 θ c = sin −1 The incident angle at point A is less than this, so the light will first exit at point A .
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32. REASONING Total internal reflection can occur only when light is traveling from a higher
index material toward a lower index material. Thus, total internal reflection is possible
when the material above or below a layer has a smaller index of refraction than the layer
itself. With this criterion in mind, the following table indicates the various possibilities:
Is total internal reflection possible?
Layer Top surface of layer Bottom surface of layer a Yes No b Yes Yes c No Yes 1368 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS SOLUTION The critical angle for each interface at which total internal reflection is
possible is obtained from Equation 26.4: nair −1 1.00 = sin = 50.3° na 1.30 θ c = sin −1 Layer a, top surface ∆θ = 75.0° − 50.3° = 24.7° na −1 1.30 = sin = 60.1° 1.50 nb Layer b, top surface θ c = sin −1 ∆θ = 75.0° − 60.1° = 14.9° nc −1 1.40 = sin = 69.0° nb 1.50 Layer b, bottom surface θ c = sin −1 ∆θ = 75.0° − 69.0° = 6.0° nair −1 1.00 = sin = 45.6° nc 1.40 Layer c, bottom surface θ c = sin −1 ∆θ = 75.0° − 45.6° = 29.4°
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33. REASONING The time it takes for the light to travel from A
to B is equal to the distance divided by the speed of light in the
substance. The distance is known, and the speed of light v in
the substance is equal to the speed of light c in a vacuum
divided by the index of refraction n1 (Equation 26.1). The
index of refraction can be obtained by noting that the light is
incident at the critical angle θc (which is known). According to
Equation 26.4, the index of refraction n1 is related to the n2 = 1.63
n1
d θc B A critical angle and the index of refraction n2 by n1 = n2 / sin θc . SOLUTION The time t it takes for the light to travel from A to B is
t= Distance
d
=
Speed of light
v
in the substance (1) Chapter 26 Problems 1369 The speed of light v in the substance is related to the speed of light c in a vacuum and the
index of refraction n1 of the substance by v = c/n1 (Equation 26.1). Substituting this
expression into Equation (1) gives
t= dn
d
d
=
=1
v c
c
n 1 (2) Since the light is incident at the critical angle θc, we know that n1 sin θ c = n2 (Equation
26.4). Solving this expression for n1 and substituting the result into Equation (2) yields
n d 2 ( 4.60 m ) 1.63 sin θc dn sin 48.1° = 3.36 ×10−8 s
t= 1= =
c
c
3.00 × 108 m/s
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34. REASONING In the ratio nB/nC each refractive index can be related to a critical angle for
total internal reflection according to Equation 26.4. By applying this expression to the AB
interface and again to the AC interface, we will obtain expressions for nB and nC in terms of
the given critical angles. By substituting these expressions into the ratio, we will be able to
obtain a result from which the ratio can be calculated SOLUTION Applying Equation 26.4 to the AB interface, we obtain sin θ c, AB = nB
nA or nB = nA sin θ c, AB Applying Equation 26.4 to the AC interface gives sin θ c, AC = nC
nA or nC = nA sin θ c, AC With these two results, the desired ratio can now be calculated:
nB nA sin θ c, AB sin 36.5°
=
=
= 0.813
nC nA sin θ c, AC sin 47.0° ________________________________________________________________________ 1370 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 35. SSM REASONING
Total internal
reflection will occur at point P provided
that the angle α in the drawing at the right
exceeds the critical angle. This angle is
determined by the angle θ2 at which the
light rays enter the quartz slab. We can
determine θ2 by using Snell’s law of
refraction and the incident angle, which is
given as θ1 = 34°. P α
θ2 θ1 SOLUTION Using n for the refractive index of the fluid that surrounds the crystalline
quartz slab and nq for the refractive index of quartz and applying Snell’s law give
n sin θ1 = nq sin θ 2 sin θ 2 = or n
sin θ1
nq (1) But when α eq...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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