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  4. When a telescope is in normal adjustment, the distance of the objective from the eyepiece is found to be 100 cm. If the magnifying power of the telescope, at normal adjustment, is 24, the focal lengths of the lenses are

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Q. When a telescope is in normal adjustment, the distance of the objective from the eyepiece is found to be $100\, cm$. If the magnifying power of the telescope, at normal adjustment, is $24$, the focal lengths of the lenses are

Ray Optics and Optical Instruments

Solution:

In normal adjustment of telescope, magnifying power is
$M = \frac{f_0}{f_e}$
where $f_0$ and $f_e$ be the focal lengths of the objective lens and eye-piece respectively.
In normal adjustment of telescope, distance between objective lens and eye piece is $L = f_0 + f_e$
According to the problem
$24 = \frac{f_0}{f_e} \quad ...(i)$
$100 = f_0 + f_e\quad ...(ii)$
Solving $(i)$ and $(ii)$, we get
$f_0 = 96\,cm, f_e = 4\,cm$