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The magnifying power of a telescope is $9 .$ When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be $20 \mathrm{~cm}$. The focal length of lenses are
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$18 \mathrm{~cm}, 2 \mathrm{~cm}$
$\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=9, \quad \therefore \mathrm{f}_{0}=9 \mathrm{f}_{\mathrm{e}}$
Also $\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=20(\because$ final image is at infinity $)$ $9 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=20, \mathrm{f}_{\mathrm{e}}=2 \mathrm{~cm}, \quad \therefore \mathrm{f}_{0}=18 \mathrm{~cm}$
Also $\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=20(\because$ final image is at infinity $)$ $9 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=20, \mathrm{f}_{\mathrm{e}}=2 \mathrm{~cm}, \quad \therefore \mathrm{f}_{0}=18 \mathrm{~cm}$
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