Question

The focal lengths of the lens of an astronomical telescope are $50\text{cm}$ and $5\text{cm}$. The length of the telescope when the image is formed at the least distance of distance vision is ( in $\text{cm}$ )

• A. $\frac{250}{6}$
• B. $\frac{275}{6}$
• C. $\frac{325}{6}$
• D. $\frac{350}{6}$

Answer 1

The correct option is C $\frac{325}{6}$

focal length of objective $f_o=50\text{cm}$
focal length of eye piece $f_e=5\text{cm}$
Length of telescope is $L=f_0+|u_e|$
for least distance of distinct vision $D=|v_e|=25\text{cm}$

Lens formula at eye piece
$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$
$\frac{1}{5}=-\frac{1}{25}-\frac{1}{u_e}$
$\frac{1}{u_e}=-[\frac{1}{25}+\frac{1}{5}]=-\frac{6}{25}$
$u_e=-\frac{25}{6}\text{cm}$
$\therefore L=f_0+u_e=50+\frac{25}{6}=\frac{325}{6}\text{cm}$