Q. A compound microscope consists of an objective lens of focal length 2.0cm and an eye piece of focal length 6.25cm separated by a distance of 15cm.How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm),and(b)at infinity? What is the magnifying power of the microscope in each case?

Answer

A Focal length of the objective lens f₁ = 2.0cm

A Focal length of the eye piece f₂ = 6.25cm

Distance among the objective lens and the eye pieced = 15cm

(a)Least distance of distinct vision d' = 25cm

∴ Image distance for the eye piece v₂ = -25cm

An Object distance for the eye piece = u₂

As-per to the lens formula we have the relation

Image distance for the objective lens v₁ = d + u₂ = 15 - 5 = 10 cm

An Object distance for the objective lens = u₁
As-per to the lens formula we have the relation

Magnitude of the object distance u₁ =2.5cm

A magnifying power of a compound microscope is given by the relation:

Therefore the magnifying power of the microscope is 20.

(b) The absolute image is formed at infinity.

∴Image distance for the eye piece v₂ = ∞

An Object distance for the eye piece = u₂

As-per to the lens formula we have the relation:

A Image distance for the objective lens v₁=d + u₂= 15-6.25 = 8.75cm 

An Object distance for the objective lens=u₁

As-per to the lens formula, we have the relation

A Magnitude of the object distance |u₁|=2.59cm

A magnifying power of a compound microscope is given by the relation

Therefore the magnifying power of the microscope is 13.51