Question

An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, find the powers of the objective and the eyepiece.

Answer 1

Given :

Magnifying power, m = 50

Length of the tube, L = 102 cm

Let the focal length of objective and eye piece be f0 and fe,

m=fo/fe, we get:

fo = 50/fe    …(1)

And,

L = fo + fe = 102 cm    …(2)

On substituting the value of fo from (1) in (2), we get:

50fe + fe = 102

⇒ 51fe = 102

⇒ fe = 2 cm = 0.02 m

And,

fo = 50 ×0.02 = 1 m

Power of the objective lens = 1/fo= 1 D

And,

Power of the eye piece lens = 1/fe=1/0.02= 50 D

Answer 2

Hey !

Refer the attachment !!

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