HW2Sol - Solutions to WebAssign HW #2 - Giancoli Ch. 32- 4,...

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Solutions to WebAssign HW #2 - Giancoli Ch. 32- 4, 13, 25, 41, 45, 53, 58, 59, 60, 64; (due 2/4) 4. The angle of incidence is the angle of reflection. See the diagram for the appropriate lengths. () tan 1.64m 0.38m (0.38m) 0.69m 2.30m Hh h x x x θ == =→ = l 13. The ball is a convex mirror with a focal length ( ) 11 22 4.6cm 2.3cm. fr = Use Eq. 32- 3 to locate the image. ( )( ) o i oi o 25.0cm 2.3cm 111 2.106cm 2.1cm 25.0cm 2.3cm df d dd f += = = = −− The image is 2.1 cm behind the surface of the ball, virtual, and upright. Note that the magnification is ( ) i o 2.106cm 0.084. 25.0cm d m d =− = =+ ( a ) To produce a smaller image located behind the surface of the mirror requires a ( b ) Find the image distance from the magnification. ( )( ) ii o i i oo o 26cm 3.5cm 20.2cm 20cm 4.5cm hd d h md h = = = (2 sig. fig.) As expected, i 0. d < The image is located ( c ) Find the focal length from Eq. 32.3. ( )( ) ( ) o i 26cm 20.2cm 1 90.55cm 91cm 20.2cm f ddf d d →= = = ++ ( d ) The radius of curvature is twice the focal length.
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HW2Sol - Solutions to WebAssign HW #2 - Giancoli Ch. 32- 4,...

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