A) \[L\text{/}l\]
B) \[\frac{L}{l}+1\]
C) \[\frac{L}{l}-1\]
D) \[\frac{L+1}{L-1}\]
Correct Answer: A
Solution :
Treat the line on the objective lens as object and eyepiece as the lens. \[u=-\,({{f}_{o}}+{{f}_{e}}),\,f={{f}_{e}}\] \[\frac{1}{V}-\frac{1}{-\,({{f}_{o}}+{{f}_{e}})}=\frac{1}{{{f}_{e}}}\] \[v=\frac{({{f}_{o}}+{{f}_{e}}){{f}_{e}}}{{{f}_{o}}}\] \[m=\left| \frac{v}{u} \right|=\frac{{{f}_{e}}}{{{f}_{o}}}=\frac{l}{L}\] Magnification of telescope\[=\frac{{{f}_{o}}}{{{f}_{e}}}=\frac{L}{l}\]
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