Question

Magnifying power of a telescope is 5 in case of relaxed eyes and length of telescope is 72 $cm$. Find magnification in case when final image is formed at least distance of clear vision. (Assume least distance$D$ to be $24cm$)

• A. 6
• B. 7
• C. 7.5
• D. 8.5

Answer 1

The correct option is C 7.5

$m=\frac{f_o}{f_e}\frac{f_o}{f_e}=5$ (for relaxed eyes)
$L=f_o+f_e{f}_o+f_e=72$
$5f_e+f_e=72$
$f_e=12cm$
$f_o=60cm$
Magnification in case of final image at least distance of clear vision
$m^{\prime }=\frac{f_o}{f_e}(1+\frac{f_e}{D})=5(1+\frac{12}{24})$
= 5(1.5) = 7.5