hw05_sol - 1. The image is 10 cm behind the mirror and you...

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1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm.
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2. The bird is a distance d 2 in front of the mirror; the plane of its image is that same distance d 2 behind the mirror. The lateral distance between you and the bird is d 3 = 5.00 m. We denote the distance from the camera to the mirror as d 1 , and we construct a right triangle out of d 3 and the distance between the camera and the image plane ( d 1 + d 2 ). Thus, the focus distance is () ( ) 22 2 2 12 3 4.30m+3.30m 5.00m 9.10m. dd =++ = + =
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56. We recall that for a converging (C) lens, the focal length value should be positive ( f = + 35 cm). (a) Eq. 34-9 gives i = pf/ ( p f ) = –88 cm. (b) Eq. 34-7 give m = / ip = +3.5. (c) The fact that the image distance is a negative value means the image is virtual (V). (d) A positive value of magnification means the image is not inverted (NI). (e) The image is on the same side as the object (see Fig. 34-16(b)).
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58. (a) A convex (converging) lens, since a real image is formed. (b) Since i = d – p and i / p = 1/2, p d == = 2 3 2400 3 26 7 . . cm cm. bg (c) The focal length is () 1 1 2 40.0 cm 11 1 1 2 8.89 cm .
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This note was uploaded on 04/05/2011 for the course PHYSICS 133 taught by Professor Steigman during the Fall '09 term at Ohio State.

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hw05_sol - 1. The image is 10 cm behind the mirror and you...

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