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+2 votes
19.2k views
in Optics by (64.8k points)

A small telescope has an objective lens of focal length 140 cm and an eye piece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (ie. when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25 cm)?

1 Answer

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Best answer

f0 = 140 cm, fe = 5 cm. 

(a) When the final image is formed at infinty.

m = -f0/fe = 140/-5 = -28

(b) When the final image isformed at the least distance of distinct vision.

m = f0/fe(1 + fe/d) = 140/5(1 + 5/25)

= 140/5((25 + 5)/25) = 140/5 x 30/25 = 33.6

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