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  4. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be 20 cm. The focal lengths of the lenses are

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Q. The magnifying power of a telescope is $9$. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be $20 \,cm$. The focal lengths of the lenses are

Ray Optics and Optical Instruments

Solution:

As, $M=\frac{f_{ o }}{f_{ e }}$ $9=\frac{f_{o}}{f_{e}}$
or $f_{o}=9 f_{ e }$
Also, $L=f_{o}+f_{e}$
or $20=f_{o}+f_{e}$
or $20=9 f_{e}+f_{e}$
or $20=10 f_{e}$
or $f_{ e }=2 \,cm$
$\therefore $ $f_{o}=9 \times 2\, cm =18 \,cm$