Question

A small telescope has an objective lens of focal length 150 cm and an eye-piece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment?
If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

Answer 1

If the telescope is in normal adjustment, i.e., the final image is at infinity then its magnification is given by,

$M=\frac{f_o}{f_e}$

Since $f_o=150cm,f_e=5cm$

$M=\frac{150}{5}=30$

If tall tower is at distance 3 km from the objective lens of focal length 150 cm. It will form its image at distance $v_o$. So, using lens formula we get,

$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$

$\frac{1}{150cm}=\frac{1}{v_o}-\frac{1}{(-3km)}$

$\frac{1}{v_o}=\frac{1}{1.5m}-\frac{1}{3000m}$

$v_o=\frac{3000\times 1.5}{3000-1.5}=\frac{4500}{2998.5}=1.5m$

Magnification, $m_o=\frac{I}{O}=\frac{h_i}{h_o}=\frac{v_o}{u_o}$

$\frac{h_i}{100m}=\frac{1.5m}{3km}=\frac{1.5}{3000}$

$h_i=\frac{1.5\times 100}{3000}=\frac{1}{20}m$

$h_i=0.05m$
So the height of the image of the tower is 5 cm.