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Q.
If tube length of astronomical telescope is $105\, cm$ and magnifying power is $20$ for normal setting. The focal length of objective is
AMUAMU 2003
Solution:
To see with relaxed eye, final image should be formed at infinity. The length of telescope is given by
$L=f_{o}+f_{e}$ ... (i)
and magnifying power
$M=\frac{f_{o}}{f_{e}}$ ... (ii)
Given, $L=105, M=20$
$\therefore f_{o}+f_{e} =105 $ ... (iii)
$\frac{f_{o}}{f_{e}} =20$ ... (iv)
From Eqs. (iii) and (iv), we get
$ 20 f_{e}+f_{e}=105 $
$\Rightarrow 21 f_{e}=105 $
$\Rightarrow f_{e}=\frac{105}{21}=5 \,cm $
$\Rightarrow f_{o}=105-5=100 \,cm$
