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Q. The magnifying power of a telescope is 10 and length of telescope is $1.1 m$ for normal adjustment. The magnification when image is formed at least distance of distinct vision is -

Ray Optics and Optical Instruments

Solution:

$10=\frac{f_{0}}{f_{e}} \& f_{0}+f_e$
$\therefore f_{0}=1 m\, \& \,f _{e}=0.1 m$
When final image is at D; then
$mp = f _{0}\left(\frac{1}{ D }+\frac{1}{ f _{ e }}\right)$
$=100\left(\frac{1}{25}+\frac{1}{10}\right)=14$