p35_035 - 1 f = n − 1 µ 1 r 1 − 1 r 2 where r 1 and...

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35. (a) When the eye is relaxed, its lens focuses far-away objects on the retina, a distance i behind the lens. We set p = intheth inlensequat iontoobta in1 /i =1 /f ,where f is the focal length of the relaxed eFective lens. Thus, i = f =2 . 50 cm. When the eye focuses on closer objects, the image distance i remains the same but the object distance and focal length change. If p is the new object distance and f 0 is the new focal length, then 1 p + 1 i = 1 f 0 . We substitute i = f and solve for f 0 : f 0 = pf f + p = (40 . 0 cm)(2 . 50 cm) 40 . 0cm+2 . 50 cm =2 .
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Unformatted text preview: 1 f = ( n − 1) µ 1 r 1 − 1 r 2 ¶ where r 1 and r 2 are the radii of curvature of the two surfaces of the lens and n is the index of refraction of the lens material. ±or the lens pictured in ±ig. 35-34, r 1 and r 2 have about the same magnitude, r 1 is positive, and r 2 is negative. Since the focal length decreases, the combination (1 /r 1 ) − (1 /r 2 ) must increase. This can be accomplished by decreasing the magnitudes of both radii....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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