Question

Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is 25 cm.

Answer 1

For the compound microscope, we have:
Power of the objective lens = 25 D
The focal length of the objective lens is given by

`f_0 = 1/(25 D) = 0.04  m = 4  cm`

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

`f_e = 1/(5D) =0.2  m =20  cm`

Least distance of clear vision, D = 25 cm
Separation between the objective and the eyepiece, L = 30 cm
Magnifying power is maximum when the image is formed by the eyepiece at the least distance of clear vision, i.e., D = 25 cm.
For the eyepiece, we have:

v_e = -25 cm and f_e = 20 cm

The lens formula is given by

`1/v_e = 1/u_e + 1/f_e`

`=> 1/u_e = 1/v_e -1/f_e`

`=>1/-25 -1/20= -((4+5))/100`

⇒ u_e = `(-100)/9` = 11.11 cm

Let u_o and v_o be the object and image distance for the objective lens.
So, for the objective lens, the image distance will be

V₀ = L - u₀ 

v₀ = 30 -11.11 

⇒ v₀ = 18.89 cm   ....(1)

As the image produced is real , v₀ =+18.89 cm 

`1/u_0 = 1/v_0 -1/f_0` 

=`1/18.89 - 1/4 = -0.197`

`u_0 =-5.07 cm`  .....(2)

Maximum magnifying power of the compound microscope:

m=`v_0/u_0(1+D/f_e)`

=` -(18.89/-5.07) (1+25/20)`

=3.7225 × 2.25 =8.376

So, the maximum magnifying power of the compound microscope is 8.376.