(a) For the telescope described in Exercise 9.34 (a),

Solution:

Focal length of the objective lens, fo = 140 cm

Focal length of the eyepiece, fe = 5 cm

(a) In normal adjustment, the separation between the objective lens and the eyepiece $=f_{\mathrm{o}}+f_{\mathrm{e}}=140+5=145 \mathrm{~cm}$

(b) Height of the tower, h1 = 100 m

Distance of the tower (object) from the telescope, u = 3 km = 3000 m

 

The angle subtended by the tower at the telescope is given as:

$\theta=\frac{h_{1}}{u}$

$=\frac{100}{3000}=\frac{1}{30} \mathrm{rad}$

The angle subtended by the image produced by the objective lens is given as:

$\theta=\frac{h_{2}}{f_{\mathrm{o}}}=\frac{h_{2}}{140} \mathrm{rad}$

Where,

h2 = Height of the image of the tower formed by the objective lens

$\frac{1}{30}=\frac{h_{2}}{140}$

$\therefore h_{2}=\frac{140}{30}=4.7 \mathrm{~cm}$

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

(c) Image is formed at a distance, d = 25 cm

The magnification of the eyepiece is given by the relation:

$m=1+\frac{d}{f_{\mathrm{e}}}$

$=1+\frac{25}{5}=1+5=6$

Height of the final image $=m h_{2}=6 \times 4.7=28.2 \mathrm{~cm}$

Hence, the height of the final image of the tower is 28.2 cm.