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The focal length of objective and eye lens of a astronomical telescope are respectively $2 \mathrm{~m}$ and $5 \mathrm{~cm}$. Final image is formed at (i) least distance of distinct vision (ii) infinity. The magnifying power in both cases will be
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$-48,-40$
When the final image is at the least distance of distinct vision, then
$m=-\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=\frac{200}{5}\left(1+\frac{5}{25}\right)=\frac{200 \times 6}{5 \times 5}=-48$
When the final image is at infinity, then
$m=\frac{-f_o}{f_e}=\frac{200}{5}=-40$
$m=-\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=\frac{200}{5}\left(1+\frac{5}{25}\right)=\frac{200 \times 6}{5 \times 5}=-48$
When the final image is at infinity, then
$m=\frac{-f_o}{f_e}=\frac{200}{5}=-40$
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