Question

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 $15cm$. 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 $\left(25cm\right)$, and
(b) at infinity? What is the magnifying power of the microscope in each case?

Answer 1

Focal length of the objective lens $f_1=2cm$

Focal length of the eyepiece,$f_2=6.25cm$

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

Least distance of distinct vision,$d^{\prime }=25cm$

Image distance for the eyepiece,$v_2=-25cm$

bject distance for the eyepiece = $u_2$

According to the lens formula

$\frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2}$

$\frac{1}{6.25}=\frac{1}{-25}-\frac{1}{u_2}$

$u_2=-5cm$

Image distance for the objective lens $v_1=d+u_2=15-5=10cm$

According to the lens formula

$\frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u_1}$

$\frac{1}{2}=\frac{1}{10}-\frac{1}{u_1}$

$u_2=-2.5cm$

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

$m=\frac{v_1}{\left|u_1\right|}(1+\frac{d^{\prime }}{f_2})$

$m=\frac{10}{2.5}(1+\frac{25}{6.25})=20$

b)The final image is formed at infinity

Image distance for the eyepiece $v_2=\infty$

According to the lens formula

$\frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2}$

According to the lens formula

$\frac{1}{6.25}=\frac{1}{\infty }-\frac{1}{u_2}$

$u_2=-6.25cm$

Image distance for the objective lens $v_1=d+u_2=10-6.25=8.75$

Object distance for the objective len =$u_2$

According to the lens formula

$\frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u_1}$

$\frac{1}{2}=\frac{1}{8.75}-\frac{1}{u_1}$

$u_2=-2.59cm$

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

$\frac{v_1}{\left|u_1\right|}\left(\frac{d^{\prime }}{\left|u_2\right|}\right)=13.51$