Question

In an astronomical telescope in normal adjustment a straight black line of length $L$ is draw 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

• A. $\frac{L}{I}$
• B. $\frac{L}{l}+1$
• C. $\frac{L}{I}-I$
• D. $\frac{L+I}{L-I}$

Answer 1

The correct option is A $\frac{L}{I}$


Let $f_o$ and $f_e$ be the focal length of the objective and eye piece.
For normal adjustment, the distance from the objective to the eye piece $=f_o+f_e$
Treating the line on the objective as object and the eye piece as lens,
$u=-(f_o+f_e)$ and $f=f_e$
using , $1/v-1/u=1/f$ we get $\frac{1}{v}+\frac{1}{f_o+f_e}=\frac{1}{f_e}$
or $\frac{1}{v}=\frac{1}{f_e}-\frac{1}{f_o+f_e}=\frac{f_o}{(f_o+f_e)f_e}$
or $v=\frac{(f_o+f_e)f_e}{f_o}$
Therefore, magnification for eye piece$=\frac{v}{u}=\frac{f_e}{f_o}=\frac{I}{L}$
Thus, magnification of telescope in normal adjustment $=\frac{f_o}{f_e}=\frac{L}{I}$