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Q.
The optical length of an astronomical telescope with magnifying power of $10$. For normal vision is $44 \,cm$, what is focal length of the objective?
AFMCAFMC 2001
Solution:
To see with relaxed eye final image should be formed at infinity.
The distance between the objective and eyepiece is adjusted so that image $A' B'$ formed by objective is at focus $F_{e}'$ of the eyepiece. This adjustment of telescope is called normal adjustment. In this position the length of the telescope is
$f_{o}+f_{e}\left(f_{o}\right.$ is focal length of objective, $f_{e}$ of eyepiece)
$\therefore 44=f_{0}+f_{e}$ ... (i)
Also, magnifying power $m=-\frac{f_{ o }}{f_{ e }}$
$-10 =-\frac{f_{ e }}{f_{e}} $
$\Rightarrow f_{o} =10 f_{e}$ ... (ii)
Using Eqs. (i) and (ii), we get
$10 f_{e}+f_{e} =44 $
$f_{e} =4\, cm$
Also, focal length of objective is
$f_{0}=10 f_{e}$
$=10 \times 4=40\, cm$
Therefore, focal length of objective is $40 \,cm$.
