996_PartUniversity Physics Solution

# 996_PartUniversity Physics Solution - 34-34 Chapter 34 f d...

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34-34 Chapter 34 Thus 1 00 1 , fd rr f ⎛⎞ ′ = ⎜⎟ ⎝⎠ as was to be shown. (b) SET UP: The image at the focal point of the first lens, a distance 1 f to the right of the first lens, serves as the object for the second lens. The image is a distance 1 f d to the right of the second lens, so 21 1 () . s fd df =− = − EXECUTE: 22 1 2 2 12 s f f s s ff == −− 2 0 f < so f f and 2 , f df s f ′ = −+ as was to be shown. (c) SET UP: The effect of the diverging lens on the ray bundle is sketched in Figure 34.111b. EXECUTE: From similar triangles in the sketch, 2 f s = Thus 0 02 rf rs = Figure 34.111b From the results of part (a), 01 . d = Combining the two results gives 1 f f f ds = 1 1 2 1 2 1 1 2 1 , f f f f f fs f df f d f d f f d = + + as was to be shown. (d) SET UP: Put the numerical values into the expression derived in part (c). EXECUTE: f f = 2 216 cm 12.0 cm, 18.0 cm, so 6.0 cm f d = + 0 d = gives 36.0 cm; f = maximum f 4.0 cm d = gives 21.6 cm; f = minimum f 30.0 cm f = says 2 216 cm 30.0 cm 6.0 cm d = + 6.0 cm 7.2 cm d += and 1.2 cm d = EVALUATE: Changing d produces a range of effective focal lengths. The effective focal length can be both smaller and larger than . f f + 34.112. IDENTIFY: M θ = . ,and yy f s θθ ′′ . This gives . yf M s y = . SET UP: Since the image formed by the objective is used as the object for the eyepiece, ′ = . EXECUTE: 1 2 2 2 .. yf yf sf f M s ss === = Therefore, 1 2 48.0 cm 1.33 cm, 36 f s M = and this is just outside the eyepiece focal point. Now the distance from the mirror vertex to the lens is 49.3 cm, and so 1 2 2 111 1 1 12.3 cm. 1.20 cm 1.33 cm s ss f +=⇒= = Thus we have a final image which is real and 12.3 cm from the eyepiece. (Take care to carry plenty of figures in the calculation because two close numbers are subtracted.) EVALUATE: Eq.(34.25) gives 40 M = , somewhat larger than M for this telescope. 34.113. IDENTIFY and SET UP: The image formed by the objective is the object for the eyepiece. The total lateral magnification is tot 1 2 1 . 8.00 mm mm m f (objective); 2 7.50 cm f = (eyepiece)

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Geometric Optics 34-35 (a) The locations of the object, lenses and screen are shown in Figure 34.113.
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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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996_PartUniversity Physics Solution - 34-34 Chapter 34 f d...

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