Focal length of the objective lens, f₀ = 1.25 cm

Focal length of the eyepiece, f_e = 5 cm

Least distance of distinct vision, d = 25 cm

Angular magnification of the compound microscope = 30X

Total magnifying power of the compound microscope, m = 30

The angular magnification of the eyepiece is given by the relation:

m_e  = ( 1 + d/f_e) = (1 + 25/5 ) = 6
The angular magnification of the objective lens (m₀) is related to m_e as: m₀m_e = m

m₀= m/m_e  = 30/6 = 5

We also have the relation:

m₀ = Image distance for the objective lens (v₀)/Object distance for the objective lens (-u₀)

5 = v₀/(-u₀)

Therefore , v₀= -5u₀————–Eq-1
Applying the lens formula for the objective lens:

1/f₀ =1/v₀ -1/u ₀

1/1.25 = 1/(-5u₀)  –  1/u₀) = -6/5u₀

Therefore u₀ = -6/5 x 1.25 = -1.5 cm

and v₀ = -5u₀ 

= -5 x (-1.5) = 7.5cm

The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.

Applying the lens formula for the eyepiece:

1/v_e -1/u_e = 1/f_e

Where, v_e = Image distance for the eyepiece = – d = – 25 cm

u_e = Object distance for the eyepiece

1/u_e = 1/v_e -1/f_e

= -1/25 -1/5 = -6/25

Therefore , u_e =-4.17 cm

Separation between the objective lens and the eyepiece = |u_e| + |v₀| = 4.17 + 7.5 = 11.67 cm Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.