Question

The magnification produced by a astronomical telescope for normal adjustment is $10$ and the length of the telescope is $1.1m$. The magnification, when the image is formed atleast distance of distinct vision is

• A. $6$
• B. $18$
• C. $16$
• D. $14$

Answer 1

The correct option is D $14$

Given, $m=10$, length of telescope $=1.1m$
We know that,
Magnification, $m=\frac{f_o}{f_e}$
$10=\frac{f_o}{f_e}$
$f_o=10f_e$
$\because f_e+f_o=1.1m$
$\Rightarrow f_e+10f_e=1.1m$ $[\because f_o=10f_e]$
$\Rightarrow f_e(1+10)=1.1m$
$\Rightarrow f_e=0.1m$ or $10cm$
Magnification least distance of distinct vision,
$M_D=\frac{f_o}{f_e}(1+\frac{f_e}{D})$
$=10(1+\frac{10}{25})$ $[\because D=25cm]$
$=10\times \frac{35}{25}$
$=14$