Question

The magnification produced by an astronomical telescope for normal adjustment is 10 and the length of telescope is 1.1m. The magnification when image is formed at least distance of distinct vision (D = 25 cm) is –

• A. 14
• B. 6
• C. 16
• D. 18

Answer 1

Answer: (A)

14

Answer 2

10 = = f₀ + f_e

1.1 f₀ = 1m = 100 cm & f_e = 0.1 m = 10 cm

Now; When final image is at D

MP = f₀ = 100 $\left( \frac { 1 } { 25 } + \frac { 1 } { 10 } \right)$ = 14