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

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-48,-40

-40,-48

- 40, 48

- 48, 40

answer is A.

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Detailed Solution

When the final image is at the least distance of distinct vision, then

$m = \frac{f_{o}}{f_{e}} (1 + \frac{f_{e}}{D}) = \frac{200}{5} (1 + \frac{5}{25}) = \frac{200 \times 6}{5 \times 5} = - 48$

When the final image is at infinity, then

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

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