# Q 9.29  (b) 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?

Given,

focal length of the objectove lens = $f_{objective}$= 140cm

focal length of the eyepiece lens = $f_{eyepiece}$ = 5 cm

Height of tower $h_{tower}$ = 100m

Distance of object which is acting like a object $u$ = 3km = 3000m.

The angle subtended by the tower at the telescope

$tan\theta=\frac{h_{tower}}{u}=\frac{100}{3000}=\frac{1}{30}$

Now, let the height of the image of the tower by the objective lens is  $h_{image}$.

angle made by the image by the objective lens  :

$tan\theta'=\frac{h_{image}}{f_{objective}}=\frac{h_{image}}{140}$

Since both, the angles are the same we have,

$tan\theta=tan\theta'$

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

$h_{image}= \frac{140}{30}=4.7cm$

Hence the height of the image of the tower formed by the objective lens is 4.7 cm.

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