JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 15)
In normal adujstment, for a refracting telescope, the distance between objective and eye piece is $$30 \mathrm{~cm}$$. The focal length of the objective, when the angular magnification of the telescope is 2 , will be :
20 cm
30 cm
10 cm
15 cm
Explanation
$$\because$$ $$m = {{{f_o}} \over {{f_e}}}$$
$$ \Rightarrow 2 = {{{f_o}} \over {{f_e}}}$$ ...... (i)
and, $$l = {f_o} + {f_e}$$
$$ \Rightarrow 30 = {f_o} + {f_e}$$ ..... (ii)
$$ \Rightarrow 30 = {f_o} + {{{f_o}} \over 2}$$
$$ \Rightarrow 30 \times {2 \over 3} = {f_o}$$
$$ \Rightarrow {f_o} = 20$$ cm
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