A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15 cm. 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 (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answer:

According to the question, we have –

Focal length of the given objective lens is f₁=2.0 cm

Focal length of the given eyepiece is f₂=6.25 cm

Distance between the eyepiece and the objective lens is d =15 cm

(a) Least distance of distinct vision is denoted by d’ = 25 cm

Image distance for the eyepiece,

$v_{2}$= -25cm

Object distance for the eyepiece= is represented by u₂

Using the lens formula, we have the relation:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$

$\frac{1}{u_{2}}=\frac{1}{v_{2}}-\frac{1}{f_{2}}$

$\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$

$u_{2}=-5cm$

Now, the image distance for the objective lens becomes –

 $v_{1}=d+u_{2}=15 – 5=10 cm$

Object distance for the objective lens is denoted by  $u_{1}$

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$

$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$

$\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$

$u_{1}=-2.5cm$

Now for the Magnifying power –

$m=\frac{v_{0}}{|u_{0}|}(1+\frac{d}{f_{e}}) $

$=\frac{10}{2.5}(1+\frac{25}{6.25}) $

$=20$

Hence, the magnifying power of the microscope is 20.

(b) The final Image formed is at infinity.

Therefore image distance for the eyepiece is

 $v_{2}=\infty$

Object distance for the eyepiece is denoted by $ u_{2}$

With respect to the lens formula, we have the relation:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$

$\frac{1}{\infty}-\frac{1}{u_{2}}=\frac{1}{6.25}$

$u_{2}=-6.25cm$

Now, image distance for the objective lens becomes –

 $v_{1} = d + u_{2}=15 -6.25 = 8.75 cm$

Object distance for the objective lens=u₁

From the lens formula, we have –

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$

$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$

$\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$

$u_{1}=-\frac{17.5}{6.75}=-2.59cm$

Magnitude of the object distance, $|u_{1}|$=2.59 cm

The magnifying power of a compound microscope is given by the relation –

m=$\frac{v_{1}}{|u_{1}|}(\frac{d’}{|u_{2}|})$

=$\frac{8.75}{2.59}\times\frac{25}{6.25}$

=13.51

Therefore, 13.51 is the magnifying power of the microscope.