Question

The magnifying power of an astronomical telescope is $0.5$. The power of its eyepiece is $10D$. What is the power of its objective

• A. $+1D$
• B. $+20D$
• C. $-2D$
• D. $+4D$

Answer 1

Answer: (B)

$+20D$

The magnifying power of an astronomical

$\text{telescope}=\frac{f_0}{f_e}=\frac{P_e}{P_0}(:.P_0=\frac{1}{f_0})$

where $f_0=$ focal length of objective.

$P_0=$ power of objective

$P_c=$ Power of eye lens.

$f_e=$ focal length of eye lens.

given

magnifying power $=0.5$

$\begin{array}{c}{P}_e=10\\ 0.5=\frac{+10}{P_0}\\ P_0=+20D\end{array}$

So option $B$ is correct