Question

The focal length of objective and eye piece of a telescope are ${{f}_{0}}$ and ${{f}_{e}}$ respectively. Then its magnification will be:

• A. ${{f}_{0}}+{{f}_{e}}$
• B. ${{f}_{0}}\times {{f}_{e}}$
• C. $\frac{{{f}_{0}}}{{{f}_{e}}}$
• D. $\frac{{{f}_{e}}}{{{f}_{0}}}$

Answer 1

Answer: (C)

$\frac{{{f}_{0}}}{{{f}_{e}}}$

Answer 2

Magnifying power of telescope for relaxed eye $M=-\frac{f_{o}}{f_{e}}=\frac{f_{o}}{f_{e}}$ (numerically)