Question

The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is $20cm$. The focal length of objective and eyepiece are respectively

• A. $10cm$, $10cm$
• B. $18cm$, $2cm$
• C. $15cm$, $5cm$
• D. $11cm$, $9cm$

Answer 1

Answer: (B)

For final image at infinity, magnifying power of a telescope is given by
$m=\frac{f_0}{f_e}=9$
where, $m=$ magnification,
$f_0=$ focal length of objective
and $f_e=$ focal length of eyepiece
$\Rightarrow f_o=9f_e$...(i)
Also, distance between objective and eyepiece
$=f_0+f_e=20$ (given)
$\Rightarrow 9f_e+f_e=20\Rightarrow f_e=2cm$
$f_0=9f_e=18cm$