Question

The magnification produced by an astronomical telescope for normal adjustment is 10 and the length of the telescope is 11 m. The magnification, when the image is formed atleast distance of distinct vision is

• A. 6
• B. 14
• C. 16
• D. 18

Answer 1

Answer: (B)

14

Magnification $\frac{f_0}{f_e}=10$
or $f_0=10f-e$
$f_0+f_e=1.1m$
$f_e+10f_e=1.1\times 100cm$
$f_e=10cm$
Magnification least distance of distant vission
$M_b=\frac{f_0}{f_e}(1+\frac{f_0}{f_s})$
$=10\frac{f_0}{f_e}(1+\frac{10}{25})=10\left(\frac{35}{25}\right)=14$