Unformatted text preview: en viewed from air, the ′
′
logo’s actual depth appears to be d w + d g . Substituting the expression for d g into the ′
expression for d w , we obtain n
′
′
d w = (d w + d g ) air nw nair = dw nw n n w
air + dg ng nw nair = dw nw n air + dg ng 1.000 1.000 = (1.50 cm ) + ( 3.20 cm ) = 3.23 cm 1.333 1.52 ______________________________________________________________________________
25. REASONING AND SOLUTION The light rays coming from the bottom of the beaker are
refracted at two interfaces, the wateroil interface and the oilair interface. When the rays
enter the oil from the water, they appear to have originated from an apparent depth d ′ below
the wateroil interface. This apparent depth is given by Equation 26.3 as n 1.48 = 16.7 cm
d ′ = d oil = (15.0 cm) 1.33 n water When the rays reach the top of the oil, a distance of 15.0 cm above the water, they can be
regarded as having originated from a depth of 15.0 cm + 16.7 cm = 31.7 cm below the
oilair interface. When the rays enter the air, they are refracted again and appear to have Chapter 26 Problems 1365 come from an apparent depth d′′ below the oilair interface. This apparent depth is given by
Equation 26.3 as
n 1.00 = 21.4 cm
d ′′ = (31.7 cm) air = (31.7 cm) 1.48 n oil ______________________________________________________________________________
26. REASONING The light ray traveling in the oil can only penetrate into the water if it does
not undergo total internal reflection at the boundary between the oil and the water. Total
internal reflection will occur if the angle of incidence θ = 71.4° is greater than the critical
angle θc for these two media. The critical angle is found from sin θc = n2
n1 (26.4) where n2 = 1.333 is the index of refraction of water (see Table 26.1), and n1 = 1.47 is the
index of refraction of the oil.
SOLUTION Solving Equation 26.4 for θc, we obtain n2 −1 1.333 o = sin = 65.1 n1 1.47 θc = sin −1 Comparing this result to θ = 71.4°, we see that the angle of incidence is greater than the
critical angle (θ > θc). Therefore, the ray of light will not enter the water ; it will instead
undergo total internal reflection within the oil.
______________________________________________________________________________
27. SSM REASONING AND SOLUTION According to Equation 26.4, the critical angle is
related to the refractive indices n1 and n2 by sin θ c = n2 / n1 , where n1 > n2. Solving for n1,
we find
n
1.000
n1 = 2 =
= 1.54
sin θ c sin 40.5°
______________________________________________________________________________ 28. REASONING AND SOLUTION Only the light which has an angle of incidence less than
or equal θc can escape. This light leaves the source in a cone whose apex angle is 2θc. The
radius of this cone at the surface of the water (n = 1.333, see Table 26.1) is R = d tan θc.
Now 1.000 θ c = sin −1 = 48.6° 1.333 so
R = (2.2 m) tan 48.6° = 2.5 m 1366 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS ______________________________________________________________________________
29. REASONING The refractive index nLiquid of the liquid can be less than the refractive index
of the glass nGlass. However, we must consider the phenomenon of total internal reflection.
Some of the light will enter the liquid as long as the angle of incidence is less than or equal to
the critical angle. At incident angles greater than the critical angle, total internal reflection
occurs, and no light enters the liquid. Since the angle of incidence is 75.0º, the critical angle
cannot be allowed to fall below 75.0º. The critical angle θc is determined according to
Equation 26.4:
nLiquid
sin θ c =
nGlass As nLiquid decreases, the critical angle decreases. Therefore, nLiquid cannot be less than the
value calculated from this equation, in which θc = 75.0º and nGlass = 1.56.
SOLUTION Using Equation 26.4, we find that
sin θ c = nLiquid
nGlass or nLiquid = nGlass sin θ c = (1.56 ) sin 75.0° = 1.51 30. REASONING Total internal reflection occurs only when light goes from a higher index
material toward a lower index material (see Section 26.3). Since total internal reflection
occurs at both the ab and ac interfaces, the index of refraction of material a is larger than
that of either material b or c: na > nb and na > nc . We now need to determine which
index of refraction, nb or nc, is larger. The critical angle is given by Equation 26.4 as
sin θc = n2/n1, where n2 is the smaller index of refraction. Therefore, the larger the value of
n2, the larger the critical angle. It is evident from the drawing that the critical angle for the
ac interface is larger than the critical angle for the ab interface. Therefore nc must be
larger than nb. The ranking of the indices of refraction, largest to smallest, is: na, nc, nb. SOLUTION For the ab interface, the critical angle is given by Equation 26.4 as
sin θc = nb/na. Therefore, the index of refraction for material b is
nb = na sin θ c = (1.80 ) sin 40.0° = 1.16 For the ac interface, we note that the angle of incide...
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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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