Question

The lens of a telescope, after focussing for infinity is taken out and a slit of length L, is placed in its position. A sharp image of the slit is formed by the eyepiece at a certain distance from it on other side. The length of the image is $l$, so the magnification of the telescope is

• A. $\frac{2L}{1}$
• B. $\frac{L}{21}$
• C. $\frac{L}{l}$
• D. $\frac{1}{l}$

Answer 1

The correct option is C $\frac{L}{l}$

If the line is the object for eye piece, then we have
$u=-(f_o+f_e)$ and $f=f_e$

$\frac{1}{v}-\frac{1}{-(f_o+f_e)}=\frac{1}{f_e}$

$⟹v=\frac{(f_o+f_e)f_e}{f_o}$

The magnification due to the eye piece is given by
$m=\frac{v}{u}=\frac{f_e}{f_o}=\frac{\text{image size}}{\text{object size}}=\frac{l}{L}$

At normal adjustment magnification of telescope

$M=\frac{f_o}{f_e}=\frac{L}{l}$