Question

The magnifying power of an astronomical telescope for normal adjustment is 10 and the length of the telescope is 110 cm. The magnifying power of the telescope when the image is formed at the least distance of distinct vision for normal eye is

• A. $-12$
• B. $-14$
• C. $-16$
• D. $-18$

Answer 1

The correct option is B $-14$

Magnifying power for normal adjustment $M_1=10=\frac{f_o}{f_e}$ ...(1)

Also $f_o+f_e=110cm$ ..(2)

By solving (1) and (2) we get,

$f_o=100cm$ and $f_e=10cm$

Magnifying power when image is formed at least distance of distinct vision

$M_2=-f_o[\frac{1}{D}+\frac{1}{f_e}]$

Putting the values we get,

$M_2=-100[\frac{1}{25}+\frac{1}{10}]$

$M_2=-14$ is the magnifying power of the telescope when image is formed at the least distance vision for normal eye.