Question

Answer the following questions:

(a) For the telescope described in Exercise (a)[A small telescope has an objective lens of focal length $140cm$ and an eyepiece of focal length $5.0cm$. What is the magnifying power of the telescope for viewing distant objects when

(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?],

What is the separation between the objective lens and the eyepiece?

(b) If this telescope is used to view a $100m$ tall tower $3km$ away, what is the height of the image of the tower formed by the objective lens?

(c) What is the height of the final image of the tower if it is formed at $25cm$?

Answer 1

(a) Focal length of objective lens is $f_o=140cm$

Focal length of an eyepiece is $f_e=5cm$

At normal adjustment, the separation between the objective lens and the eyepiece is $f_e+f_o=145cm$

(b) Distance of tower is $3000m;h_1=100m$

angle subtended by tower at telescope is ${\theta }_1=h_1/u=1/30rad$

angle subtended by image, ${\theta }_2=h_2/f_o=h_2/140rad$

as two angles are equal, $h_2=4.7cm$

(c) Image formed at $d=25cm$.

$m=1+d/f=6$

$f_e$ is focal length of eyepiece.

Height of the image is $m\times h_2=6\times 4.7=28.2cm$