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+1 vote
1.7k views
in Optics by (63.5k points)

A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

1 Answer

+2 votes
by (64.8k points)
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Best answer

f0 = 144 cm, fe = 6cm.

magnifying power of the telescope, m = -f0/fe

⇒ m = -144/6 = - 24

Seperation between the objective and the eyepiece

L = f0 + fe = 144 + 6 = 150cm

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