Question

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

• A. – 48, – 40
• B. – 40, – 48
• C. – 40, 48
• D. – 48, 40

Answer 1

Answer: (A)

– 48, – 40

Answer 2

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

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

When the final image is at infinity, then

$m = \frac{- f_{o}}{f_{e}} = \frac{200}{5} = - \mspace{6mu} 40$