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34_InstructorSolutions

# 34_InstructorSolutions - GEOMETRIC OPTICS 34 y = 4.85 cm...

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34-1 G EOMETRIC O PTICS 34.1. IDENTIFY and SET UP: Plane mirror: s s = − (Eq.34.1) and // 1 myy ss = =− =+ (Eq.34.2). We are given s and y and are asked to find and . s y ′′ EXECUTE: The object and image are shown in Figure 34.1. 39.2 cm ss ′ =−=− (1 ) ( 4 . 8 5 cm ) ym y == + 4.85 cm y ′ = Figure 34.1 The image is 39.2 cm to the right of the mirror and is 4.85 cm tall. EVALUATE: For a plane mirror the image is always the same distance behind the mirror as the object is in front of the mirror. The image always has the same height as the object. 34.2. IDENTIFY: Similar triangles say tree tree mirror mirror hd = . SET UP: mirror 0.350 m, d = mirror 0.0400 m h = and tree 28.0 m 0.350 m. d EXECUTE: tree tree mirror mirror 28.0 m 0.350 m 0.040 m 3.24 m. 0.350 m d hh d + = EVALUATE: The image of the tree formed by the mirror is 28.0 m behind the mirror and is 3.24 m tall. 34.3. IDENTIFY: Apply the law of reflection. SET UP: If up is the + y -direction and right is the + x- direction, then the object is at 00 (, ) x y −− and 2 P is at x y . EXECUTE: Mirror 1 flips the y -values, so the image is at x y which is 3 . P EVALUATE: Mirror 2 uses 1 P as an object and forms an image at 3 P . 34.4. IDENTIFY: /2 f R = SET UP: For a concave mirror 0. R > EXECUTE: (a) 34.0 cm 17.0 cm 22 R f = EVALUATE: (b) The image formation by the mirror is determined by the law of reflection and that is unaffected by the medium in which the light is traveling. The focal length remains 17.0 cm. 34.5. IDENTIFY and SET UP: Use Eq.(34.6) to calculate s and use Eq.(34.7) to calculate . y The image is real if s is positive and is erect if 0. m > Concave means R and f are positive, 22.0 cm; 11.0 cm. Rf R = += = + EXECUTE: (a) Three principal rays, numbered as in Sect. 34.2, are shown in Figure 34.5. The principal ray diagram shows that the image is real, inverted, and enlarged. Figure 34.5 34

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34-2 Chapter 34 (b) 11 1 s sf += 1 1 1 (16.5 cm)(11.0 cm) so 33.0 cm 16.5 cm 11.0 cm s f s sfs s f s f =−= = = = + −− 0 s ′ > so real image, 33.0 cm to left of mirror vertex 33.0 cm 2.00 16.5 cm s m s =− =− =− ( m < 0 means inverted image) 2.00(0.600 cm) 1.20 cm ym y ′ = == EVALUATE: The image is 33.0 cm to the left of the mirror vertex. It is real, inverted, and is 1.20 cm tall (enlarged). The calculation agrees with the image characterization from the principal ray diagram. A concave mirror used alone always forms a real, inverted image if s > f and the image is enlarged if f < s < 2 f . 34.6. IDENTIFY: Apply s and s m s = − . SET UP: For a convex mirror, 0 R < . 22.0 cm R = − and 11.0 cm 2 R f . EXECUTE: (a) The principal-ray diagram is sketched in Figure 34.6. (b) s . (16.5 cm)( 11.0 cm) 6.6 cm 16.5 cm ( 11.0 cm) sf s = . 6.6 cm 0.400 16.5 cm s m s =+ . (0.400)(0.600 cm) 0.240 cm y = . The image is 6.6 cm to the right of the mirror. It is 0.240 cm tall. 0 s ′ < , so the image is virtual. 0 m > , so the image is erect. EVALUATE: The calculated image properties agree with the image characterization from the principal-ray diagram.
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34_InstructorSolutions - GEOMETRIC OPTICS 34 y = 4.85 cm...

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