981_PartUniversity Physics Solution

981_PartUniversity Physics Solution - Geometric Optics...

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Geometric Optics 34-19 (c) The principal-ray diagram is given in Figure 34.69. Figure 34.69 EVALUATE: For a real object, only virtual images are formed by a convex mirror. The virtual object considered in this problem must have been produced by some other optical element, by another lens or mirror in addition to the convex one we considered. 34.70. IDENTIFY: Apply abba nnnn s sR += , with R →∞ since the surfaces are flat. SET UP: The image formed by the first interface serves as the object for the second interface. EXECUTE: For the water-benzene interface to get the apparent water depth: 1.33 1.50 0 0 7.33 cm. 6.50 cm ab nn s ss s +=⇒ + =⇒= ′′ For the benzene-air interface, to get the total apparent distance to the bottom: 1.50 1 00 6 . 6 2 c m . (7.33 cm 2.60 cm) s s +=⇒= + EVALUATE: At the water-benzene interface the light refracts into material of greater refractive index and the overall effect is that the apparent depth is greater than the actual depth. 34.71. IDENTIFY: The focal length is given by 12 11 1 (1 ) n f RR ⎛⎞ =− ⎜⎟ ⎝⎠ . SET UP: 1 4.0 cm R or 8.0 cm ± . 2 8.0 cm R or 4.0 cm ± . The signs are determined by the location of the center of curvature for each surface. EXECUTE: 1 (0.60) 4.00 cm 8.00 cm f ±± , so 4.44 cm, 13.3 cm. f = The possible lens shapes are sketched in Figure 34.71. 123 456 78 13.3 cm; 4.44 cm; 4.44 cm; 13.3 cm; 13.3 cm; 13.3 cm; 4.44 cm; 4.44 cm. fff ff =+ = EVALUATE: f is the same whether the light travels through the lens from right to left or left to right, so for the pairs (1,6), (4,5) and (7,8) the focal lengths are the same. Figure 34.71 34.72. IDENTIFY: Apply 11 1 s sf and the concept of principal rays. SET UP: 10.0 cm s = . If extended backwards the ray comes from a point on the optic axis 18.0 cm from the lens and the ray is parallel to the optic axis after it passes through the lens. EXECUTE: (a) The ray is bent toward the optic axis by the lens so the lens is converging. (b) The ray is parallel to the optic axis after it passes through the lens so it comes from the focal point; 18.0 cm f = .
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34-20 Chapter 34 (c) The principal ray diagram is drawn in Figure 34.72. The diagram shows that the image is 22.5 cm to the left of the lens. (d) 11 1 s sf += gives (10.0 cm)(18.0 cm) 22.5 cm 10.0 cm 18.0 cm sf s == = −− . The calculated image position agrees with the principal ray diagram. EVALUATE: The image is virtual. A converging lens produces a virtual image when the object is inside the focal point. Figure 34.72 34.73. IDENTIFY: Since the truck is moving toward the mirror, its image will also be moving toward the mirror. SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is s , where f = R /2. EXECUTE: Since the mirror is convex, f = R /2 = (&1.50 m)/2 = &0.75 m. Applying the equation for a spherical mirror gives fs s s s f +=⇒= .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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981_PartUniversity Physics Solution - Geometric Optics...

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