Question

If the focal length of the eye-piece of a telescope is doubled, it's magnifying power $\left(m\right)$ will be

• A. $2m$
• B. $3m$
• C. $\frac{m}{2}$
• D. $4m$

Answer 1

The correct option is C $\frac{m}{2}$

For a telescope, magnifying power, $m=\frac{-f_o}{f_e}$

where $f_o$ and $f_e$ are the focal lengths of the objective and eye-piece respectively.

Focal length of the eye-piece is doubled, i.e $f_e^{\prime }=2f_e$

New magnifying power, $m^{\prime }=\frac{-f_o}{f_e^{\prime }}=\frac{-f_o}{2f_e}=\frac{m}{2}$