Physics Solution Manual for 1100 and 2101

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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 water-oil interface and the oil-air interface. When the rays enter the oil from the water, they appear to have originated from an apparent depth d ′ below the water-oil 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 oil-air 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 oil-air 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 a-b and a-c 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 a-c interface is larger than the critical angle for the a-b 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 a-b 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 a-c 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.

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