Physics Solution Manual for 1100 and 2101

2 determine the critical angle c for total internal n

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Unformatted text preview: n 1) x L θ2 θ2 t φ Glass Glass ( n2) ( n2 ) Air ( n1) θ1 SOLUTION If we apply Snell's law (see Equation 26.2) to the bottom interface we obtain n1 sin θ1 = n2 sin θ 2 . Similarly, if we apply Snell's law at the top interface where the ray emerges, we have n2 sin θ 2 = n3 sin θ 3 = n1 sin θ 3 . Comparing this with Snell's law at the bottom face, we see that n1 sin θ1 = n1 sin θ 3 , from which we can conclude that θ3 = θ1. Therefore, the emerging ray is parallel to the incident ray. From the geometry of the ray and thickness of the pane, we see that L cos θ 2 = t , from which it follows that L = t /cos θ 2 . Furthermore, we see that x = L sin φ = L sin (θ1 – θ 2 ) . Substituting for L, we find 1362 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS x = L sin(θ1 – θ 2 ) = t sin(θ1 – θ 2 ) cos θ 2 Before we can use this expression to determine a numerical value for x, we must find the value of θ2. Solving the expression for Snell's law at the bottom interface for θ2, we have sin θ 2 = n1 sin θ1 (1.000) (sin 30.0°) = = 0.329 n2 1.52 θ 2 = sin –1 0.329 = 19.2° or Therefore, the amount by which the emergent ray is displaced relative to the incident ray is t sin (θ1 – θ 2 ) (6.00 mm) sin (30.0° –19.2°) = = 1.19 mm cos θ 2 cos 19.2° ______________________________________________________________________________ x= 22. REASONING The fish appears to be a distance less than 1 L from the front wall of the 2 aquarium, where L is the distance between the back and front walls. The phenomenon of apparent depth is at play here. According to Equation 26.3, the apparent depth or distance d ′ n is related to the actual depth 1 L by d ′ = 1 L air , where nair and nwater are the refractive 2 2 n water indices of air and water. Referring to Table 26.1, we see that nair < nwater, so that d ′ < 1 L . 2 Since the fish is a distance 1 2 1 2 L in front of the plane mirror, the image of the fish is a distance L behind the plane mirror. Thus, the image is a distance 3 2 L from the front wall. The image of the fish appears to be at a distance less than 3 L from the front wall, because of 2 the phenomenon of apparent depth. The explanation given above applies here also, except that the actual depth or distance is 3 L instead of 1 L . 2 2 SOLUTION a. Using Equation 26.3, we find that the apparent distance d ′ between the fish and the front wall is n 1.000 d ′ = 1 L air = 1 ( 40.0 cm ) = 15.0 cm 2 n 2 1.333 water b. Again using Equation 26.3, we find that the apparent distance between the image of the fish and the front wall is n d ′ = 3 L air 2 n water = 3 2 ( 40.0 cm ) 1 = 45.0 cm 1.333 Chapter 26 Problems 23. SSM REASONING Following the discussion in Conceptual Example 4, we have the drawing at the right to use as a guide. In this drawing the symbol d refers to depths in the water, while the symbol h refers to heights in the air above the water. Moreover, symbols with a prime denote apparent distances, and unprimed symbols denote actual distances. We will use Equation 26.3 to relate apparent distances to actual distances. In so doing, we will use the fact that the refractive index of air is essentially nair = 1 and denote the refractive index of water by nw = 1.333 (see Table 26.1). 1363 h′ h d′ Air Water d SOLUTION To the fish, the man appears to be a distance above the air-water interface that is given by Equation 26.3 as h′ = h ( nw /1) . Thus, measured above the eyes of the fish, the man appears to be located at a distance of nw h ′ + d = h +d 1 (1) To the man, the fish appears to be a distance below the air-water interface that is given by Equation 26.3 as d ′ = d (1/ nw ) . Thus, measured below the man’s eyes, the fish appears to be located at a distance of 1 h + d ′ = h + d (2) nw Dividing Equation (1) by Equation (2) and using the fact that h = d, we find h′+ d = h +d′ nw h +d 1 1 h + d nw = nw + 1 = nw 1 1+ nw (3) In Equation (3), h ′ + d is the distance we seek, and h + d ′ is given as 2.0 m. Thus, we find h′ + d = nw ( h + d ′ ) = (1.333) ( 2.0 m ) = 2.7 m ______________________________________________________________________________ 1364 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 24. REASONING The drawing at the right shows the situation. As discussed in the text, when the observer is directly above, the apparent depth d ′ of the object is related to the actual depth by Equation 26.3: dw = 1.50 cm water (nw = 1.333) n2 d′ = d n1 dg = 3.20 cm glass (ng = 1.52) In this case, we must apply logo Equation 26.3 twice; once for the rays in the glass, and once again for the rays in the water. SOLUTION We refer to the drawing for our notation and begin at the logo. To an observer ( ) depth is, ′ in the water directly above the logo, the apparent depth of the logo is d g = d g nw / ng . When ( viewed ′ ′ dw = dw + d g directly from above in air, the logo’s apparent ) ( nair / nw ) , where we have used the fact that wh...
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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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