Question

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

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

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
$\Rightarrow f_{o}=9 f_{e}$...(i)
Also, distance between objective and eyepiece
$=f_{0}+f_{e}=20$ (given)
$\Rightarrow 9 f_{e}+f_{e}=20 \Rightarrow f_{e}=2 cm$
$f_{0}=9 f_{e}=18 cm$