Question

Magnification produced by astronomical telescope for normal adjustment is 10 and length of telescope is 1.1 m. The magnification when the 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

$\frac { f _ { \mathrm { o } } } { f _ { \mathrm { e } } }$ = 10

ƒ_o + ƒ_e = 1.1

\ ƒ_o = 100 cm and ƒ_e = 10 cm

Final magnification = ƒ_o

= 100 $\left[ \frac { 1 } { 25 } + \frac { 1 } { 10 } \right]$ = 14