Question

An optical instrument used for angular magnification has a 25 D objective and 20 D eyepiece. The tube length is 25 cm when the eye is least strained. (a) Whether it is a microscope or a telescope? (b) What is the angular magnification produced?

Answer 1

Given:
Power of the objective lens = 25 D
The focal length of the objective lens is given by

`f_0 =1/(25D)` = 0.04 cm

Power of the eyepiece = 5 D
The focal length of the eyepiece is given by

`f_e =1/(20D)` =0.05 m = 5 cm

(a) As  f₀ <  f_e , he instrument must be a microscope. 

(b) Tube length,l = 25 cm

Image distance for the eye lens, v_e = 25 cm 

We know :

l = v_o +f_e

⇒ v₀ = l - f_e =25 -5 = 20 cm

On applying the lens formula for the objective lens , we get :

`1/v_0 -1/u_0 = 1/f_0`

`=> -1/u_0 =1/4 - 1/20`

`=> -1/u_0 = (5-1)/20 =4/20`

`=> u_0 = 20/4 = 5  text/"cm"`

The angular magnification of microscope (m) is given by

`m = v_0/u_0 (D/f_e)`

= `20/5 (25/5)` = 20 cm

So, the required angular magnification of the microscope is 20.