The focal length of the objective of an astronomical telescope is 1 m and it is in normal adjustment. Initially the telescope is focused to a heavenly body. If the same telescope is to be focused to an object at a distance of 21 m from the objective, then identify the correct choice

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eye piece should be displaced by 5 cm away from the objective

eye piece should be displaced by 5 cm towards from the objective

eye piece should be displaced by 2 cm towards the objective

eye piece should be displaced by 2 cm away from the objective

answer is D.

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Detailed Solution

$u_{1} = \infty; v_{1} = f_{0} = 1 m$

$u_{2} = 21 m; \frac{1}{f} = \frac{1}{v_{2}} - \frac{1}{u_{2}} \Rightarrow \frac{1}{1} = \frac{1}{v_{2}} + \frac{1}{21} \frac{1}{v_{2}} = \frac{1}{1} - \frac{1}{21} = \frac{20}{21} v_{2} = \frac{21}{20} s h i f t \frac{21}{20} - 1 = \frac{1}{20} m = 5 c m$