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

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

Tardigrade Admin · Accepted Answer

When the final image is at the least distance of distinct vision, then m=(fo/fe)(1+(fe/D))=(200/5)(1+(5/25))=(200 × 6/5 × 5)=-48 When the final image is at infinity, then m=(-fo/fe)=(200/5)=-40

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

- 48, - 40

- 40, - 48

- 40, 48

- 48, 40

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

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