A compound microscope consists of an objective lens of focal length 2.0 cm

Solution:

Focal length of the objective lens, f1 = 2.0 cm

Focal length of the eyepiece, f2 = 6.25 cm

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

(a) Least distance of distinct vision, $d^{\prime}=25 \mathrm{~cm}$

∴ Image distance for the eyepiece, v2 = −25 cm

Object distance for the eyepiece = u2

 

According to the lens formula, we have the relation:

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

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

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

$\therefore u_{2}=-5 \mathrm{~cm}$

Image distance for the objective lens, $v_{1}=d+u_{2}=15-5=10 \mathrm{~cm}$

Object distance for the objective lens = u1

According to the lens formula, we have the relation:

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

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

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

$\therefore u_{1}=-2.5 \mathrm{~cm}$

Magnitude of the object distance, $\left|u_{1}\right|=2.5 \mathrm{~cm}$

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

$m=\frac{v_{1}}{\left|u_{1}\right|}\left(1+\frac{d^{\prime}}{f_{2}}\right)$

$=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=4(1+4)=20$

Hence, the magnifying power of the microscope is 20.

(b) The final image is formed at infinity.

$\therefore$ Image distance for the eyepiece, $v_{2}=\infty$

Object distance for the eyepiece = u2

 

According to the lens formula, we have the relation:

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

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

$\therefore u_{2}=-6.25 \mathrm{~cm}$

Image distance for the objective lens, $v_{1}=d+u_{2}=15-6.25=8.75 \mathrm{~cm}$

Object distance for the objective lens = u1

According to the lens formula, we have the relation:

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

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

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

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

Magnitude of the object distance, $\left|u_{1}\right|=2.59 \mathrm{~cm}$

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

$m=\frac{v_{1}}{\left|u_{1}\right|}\left(\frac{d^{\prime}}{\left|u_{2}\right|}\right)$

$=\frac{8.75}{2.59} \times \frac{25}{6.25}=13.51$

Hence, the magnifying power of the microscope is 13.51.