# The magnifying power of a telescope with tube length $60cm$ is $5$. What is the focal length (in cm) of its eye piece?   Option: 1 30 Option: 2 40 Option: 3 10 Option: 4 20

$\begin{array}{l}{\text { Let the focal length of the objective as } f_{0}} \\ {\text { and focal length of the eyepiece as } f_{e}} \\ {\text { Magnifying power, } M=\frac{f_{0}}{f_{e}}} \\ {\text { Tube length, } L=f_{0}+f_{e}}\end{array}$

$\begin{array}{l}{\text { Given: Magnifying power }=5} \\ \\ {\therefore \frac{f_{0}}{f_{\mathrm{e}}}=5} \\ {\therefore {f}_{0}=5{f}_{e}} \\ and \ {\mathrm{L}=\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=60} \\ {5 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=60\quad\left[\because \mathrm{f}_{0}=5 \mathrm{f}_{\mathrm{e}}\right]} \\ {6\mathrm{f}_{\mathrm{e}}=60} \\ {\mathrm{f}_{\mathrm{e}}=\frac{60}{6}=10 \mathrm{cm}} \\ {\text { So the focal length of eye piece is } 10 \mathrm{cm} .}\end{array}$

So the correct answer is given in option 3.

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