question_answer

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                               

A)  \[\frac{L}{l}+1\]              

B)  \[\frac{L}{l}-1\]

C)  \[\frac{L+1}{L-1}\]         

D)  \[\frac{L}{l}\]

Correct Answer: D

Solution :

We know, magnification of telescope, we have \[M=\frac{{{f}_{o}}}{{{f}_{e}}},\] Here      \[\frac{{{f}_{e}}}{{{f}_{e}}+u}=\frac{-I}{L}\] Þ        \[\frac{{{f}_{e}}}{{{f}_{e}}-({{f}_{o}}+{{f}_{e}})}\,=\frac{-I}{L}\] Þ      \[\frac{{{f}_{e}}}{{{f}_{o}}}=\frac{I}{L}i.e.M=\frac{L}{I}\]