A simple camera telephoto lens consists of two lenses. The objective lens has a

Question

A simple camera telephoto lens consists of two lenses. The objective lens has a focal length fi -39.2 cm. Precisely 35.7 cm behind this lens is a concave lens with a focal length f210.0 cm. The object to be photographed is 3.95 m in front of the objective lens Part A How far behind the concave lens should the film be placed? Express your answer using two significant figures. d- cm Submit Request Answer Part B What is the linear magnification of this lens combination? Express your answer using two significant figures. Submit Request Answer

Explanation / Answer

So 1/s'1 + 1/s1 = 1/f1...So 1/s'1 = 1/f1 - 1/s1 = 1/0.392 - 1/3.95

s'1 = 0.435m = 43.5cm

This makes the first image appears 43.5 - 35.7= 7.8 cm behind the second lens ...so s2 = -7.8

Now 1/s'2 + 1/s 2= 1/f2 (second lens)

So 1/s' 2= 1/f 2- 1/s2 = 1/(-10.0) - 1/(-7.8) = 0.0461...So s'2 = 35.45cm (which fortunately corresponds to the location of the film)

So m = m1*m2 = s'1/s*s'2/s2 =43.5/395*35.45/7.8= 0.521

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