Given:

Focal length of the objective lens, f₀ = 8 mm (0.8 cm)

Focal length of the eyepiece, f_e = 2.5 cm

Object distance for the objective lens, u₀ = -9.0 mm ( -0.9 cm)

Least distance of distant vision, d = 25 cm

Image distance for the eyepiece, v_e = -d = -25 cm

Let object distance for the eyepiece = u_e.

By using lens maker formula,

…(1)

From equation 1 we have,

…(2)

Where, f_e = focal length of the eye piece lens

v_e = Distance of image formation

u_e = Distance of object from eyepiece

Plugging the values in equation (2)

⇒

⇒ u₂ = -2.27 cm

We can find out the distance of image for objective lens using the equation (1)

For objective lens, we have

From equation 1 we have,

…(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)

⇒ v₀ = 7.2 cm

We know that the distance between the objective lens and eyepiece, d is given by,

d = |u_e| + v_o

d = 2.72 + 7.2

d = 9.47 cm

We know that magnifying power of a compound lens is given by,

m = ) …(4)

Where,

m = magnification

v₀ = image distance from objective lens.

U₀ = object distance from objective lens

d = minimum distance of distinct vision (25 cm)

f_e = Focal length of eyepiece

Now putting the values in equation (4)

m = )

m = 88.

Hence the magnification power of the microscope is 88 and it is a dimensionless quantity as obvious from calculation.