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  4. In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I . The magnification of the telescope is

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Q. In an astronomical telescope in normal adjustment a straight black line of length $L$ is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is $I$ .
The magnification of the telescope is

AIPMTAIPMT 2015Ray Optics and Optical Instruments

Solution:

We know that magnification of telescope:
$M =\frac{ f _{0}}{ f _{ e }}$
The magnification of the eyepiece lens is given by:
$M_{ e }=\frac{ f _{ e }}{ f _{ e }+ u }=\frac{ L _{ I }}{ L _{ O }}$
Here, $L_{ I }$ is the length of the image and $L _{0}$ is the length of the object.
Now for the eypiece lens, the distance of the object is $-\left(f_{0}+f_{e}\right)$
$\frac{f_{e}}{f_{e}-\left(f_{0}+f_{e}\right)}=\frac{-1}{L}$
$\Rightarrow \frac{ f _{ e }}{ f _{0}}=\frac{1}{ L }$
i.e $M =\frac{ f _{0}}{ f _{ e }}=\frac{ L }{ l }$
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