A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

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.