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A small telescope has an objective lens of focal length $140 \mathrm{~cm}$ and an eye piece of focal length $5.0 \mathrm{~cm}$. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (ie. when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision $(25 \mathrm{~cm})$ ?
(a) the telescope is in normal adjustment (ie. when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision $(25 \mathrm{~cm})$ ?
Solution:
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Verified Answer
$$
\mathrm{f}_0=140 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=5 \mathrm{~cm} \text {. }
$$
(a) When the final image is formed at infinty.
$$
\mathrm{m}=\frac{-\mathrm{f}_0}{\mathrm{f}_{\mathrm{c}}}=\frac{140}{-5}=-28
$$
(b) When the final image is formed at the least distance of distinct vision.
$$
\begin{aligned}
\mathrm{m} &=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\left(1+\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{d}}\right)=\frac{140}{5}\left(1+\frac{5}{25}\right)=\frac{140}{5}\left(\frac{25+5}{25}\right) \\
&=\frac{140}{5} \times \frac{30}{25}=33.6
\end{aligned}
$$
\mathrm{f}_0=140 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=5 \mathrm{~cm} \text {. }
$$
(a) When the final image is formed at infinty.
$$
\mathrm{m}=\frac{-\mathrm{f}_0}{\mathrm{f}_{\mathrm{c}}}=\frac{140}{-5}=-28
$$
(b) When the final image is formed at the least distance of distinct vision.
$$
\begin{aligned}
\mathrm{m} &=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\left(1+\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{d}}\right)=\frac{140}{5}\left(1+\frac{5}{25}\right)=\frac{140}{5}\left(\frac{25+5}{25}\right) \\
&=\frac{140}{5} \times \frac{30}{25}=33.6
\end{aligned}
$$
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