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A small telescope has an objective lens of focal length $144 \mathrm{~cm}$ and an eyepiece of focal length $6.0 \mathrm{~cm}$. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
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Given, $\mathrm{f}_0=144 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=6 \mathrm{~cm}$.
Magnifying power of the telescope,
$$
\mathrm{m}=\frac{-\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}} \Rightarrow \mathrm{m}=\frac{-144}{6}=-24
$$
Seperation between the objective and the eyepiece $\mathrm{L}=\mathrm{f}_0+\mathrm{f}_{\mathrm{c}}=144+6=150 \mathrm{~cm}$
Magnifying power of the telescope,
$$
\mathrm{m}=\frac{-\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}} \Rightarrow \mathrm{m}=\frac{-144}{6}=-24
$$
Seperation between the objective and the eyepiece $\mathrm{L}=\mathrm{f}_0+\mathrm{f}_{\mathrm{c}}=144+6=150 \mathrm{~cm}$
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