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Q. The focal length of objective and eye lens of a astronomical telescope are respectively $2 \,m$ and $5\, cm .$ Final image is formed at (i) least distance of distinct vision (ii) infinity. The magnifying power in both cases will be

Ray Optics and Optical Instruments

Solution:

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
$m=\frac{-f_{o}}{f_{e}}=\frac{200}{5}=-40$