Question

In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of the image is I. The magnification of the telescope is :

Answer 1

=> M = L/I

Hope this helps you

Answer 2

Given that,

Height of object = L

Height of image = I

We know that,

The magnification of the telescope is

$m'=\dfrac{f_{o}}{f_{e}}$....(I)

Object distance $u=-(f_{o}+f_{e})$

If given height of object and height of image then,

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{h'}{h}=\dfrac{f_{e}}{f_{e}+u}$

Where, h'= height of image

h = height of object

Put the value into the formula

$\dfrac{-I}{L}=\dfrac{f_{e}}{f_{e}+u}$

Put the value of u

$\dfrac{-I}{L}=\dfrac{f_{e}}{f_{e}-(f_{o}+f_{e})}$

$\dfrac{f_{e}}{f_{o}}=\dfrac{I}{L}$

We need to calculate the magnification of the telescope

Using equation (I)

$m'=\dfrac{f_{o}}{f_{e}}$

Put the value into the formula

$m'=\dfrac{L}{I}$

Hence, The magnification of the telescope is $\dfrac{L}{I}$