Question

A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away from the objective. Calculate the magnitude of magnification produced.

Answer 1

Given $u=-200cm,f=50cm$
For image $I_1$ of object formed by objective lens,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
We have
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{50}+\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}$
$\Rightarrow v=+\frac{200}{3}cm$
Also, magnification produced by objective lens
$m_0=\frac{v}{u}=-\frac{200/3}{200}=\frac{1}{3}$
Image $I_1$ acts as an object for eye lens.
Here, $v=-25cm,f=5cm$
$\therefore \frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{u}=\frac{1}{v}-=\frac{1}{f}=-\frac{1}{25}-\frac{1}{5}=-\frac{1+5}{25}$
$\therefore u=-\frac{25}{6}cm$
And magnification produced by eye lens,
$m_e=\frac{v}{u}=\frac{-25}{(-25/6)}=6$
a. The separation between objective and eyepiece
$=|V|+|u|=\frac{200}{3}+\frac{25}{6}=\frac{425}{6}=70.73cm$
b. Magnification produced, $m=m_0\times m_e=-\frac{1}{3}\times 6=-2$
The negative sign shown that the final image is inverted.