Given:

Focal length of the objective lens, = 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:

…(1)

By putting the values in equation 1, we get

⇒ m = (1 + 25/5)

⇒ m = 6

The angular magnification of the objective lens (m₀) is related to m_e as:

m_o × m_e = m

⇒ m_o = m/ me

⇒ m₀ = 30/6

⇒ m₀ = 5

We also have the relation:

From the available data we find that,

5 = v₀/(-u₀)

Therefore, v_o = -5u_o ………. (2)

Applying the lens formula for the objective lens:

…(3)

Where, f₀ = focal length of the objective lens

v₀ = Distance of image formation

u₀ = Distance of object from objective

Plugging the values in equation (3)

⇒ u₀ = (-6/5) × 1.25

⇒ u₀ = -1.5 cm

By putting the value of u₀ in equation (2) we can find v₀,

v₀ = -5 × -1.5

v₀ = 7.5 cm

1fo = 1vo–1uo 11.25 = 1−5uo−1uo

= −65uo

Therefore, u_o = −65 × 1.25

= -1.5 cm

And v_o = -5u_o

= -5 x (-1.5)

= 7.5 cm

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

Applying the lens formula for the eyepiece:

Applying the lens formula for the objective lens:

…(3)

Where, f_e = focal length of the objective lens

v_e = Image distance for eyepiece = -25cm

u₀ = Distance of object from objective

Plugging the values in equation (3)

⇒ u_e = (-6/25)

⇒ u_e = -4.17 cm

Separation between the objective lens and the eyepiece d = |u_e| + |v_o|

d = 4.17 + 7.5

d = 11.67 cm

Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.

NOTE: Separation of lenses or length of telescope in case of refractive telescopes is always equal to the sum of focal lengths of the two lenses because one lens converges the rays from infinity at its focus and the eyepiece magnifies the image formed( virtual object) at its focus.