An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102cm, find the powers of the objective and the eyepiece.
A
10;100
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B
1;50
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C
10;500
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D
2;100
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Solution
The correct option is B1;50
Magnifying power of an astronomical telescope is given by
m=−f0fe=−50 f0=50fe --- (1)
Length of the tube L=f0+fe=102cm --- (2)
Substituting (1) in (2), 51fe=102
i.e fe=2cm
and fo=100cm
Power of objective Po=1fo=1D
Power of eyepiece Pe=1fe=50D