Question

The magnification produced by an astronomical telescope for normal adjustment is $10$ and the length of the telescope is $1.1m$. The magnification when the image is formed at half distance of the distinct vision ($D=25cm$) is:

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

Answer 1

The correct option is C $18$

Focal length of objective $=f_o$

Focal length of eyepiece $=f_e$

Given: $f_o+f_e=1.1m$

and $M=\frac{f_o}{f_e}=10$

$\Rightarrow f_e=0.1m$

and $f_o=1.1-0.1=1m$

New magnification:

$m^{\prime }=m\times (\frac{f_e}{x}+1)$, [where $x=D/2=0.125m$]

$=10\times (\frac{0.1}{0.125}+1)=18$