Question

The total magnification produced by a compound microscope is $20$. The magnification produced by the eyepiece is $5$. The microscope is focussed on a certain object. The distance between the objective and eye-piece is observed to be $14 cm$. If the least distance of distinct vision is $20 cm$, calculate the focal length of the objective and the eye-piece.

Answer 1

The focal length of the eyepiece is obtained by dividing least distance of distinct vision and magnification produced by the eyepiece. The focal length of objective is obtained by the formula of length of microscope i.e., microscope i.e., the sum of focal length of objective, image distance from the objective and focal length of the eyepiece.

Complete answer:
In a compound microscope, two converging lens are arranged coaxially. The one which is facing the object is called an objective and the one close to the eye is called eyepiece. The objective has a smaller focal length than eyepiece. The separation between the objective and the eyepiece is called the length of the microscope (L) i.e., $L = 14cm$ .

Let the overall magnification be M and it is given by $M = \left( {\dfrac{{{v_o}}}{{{f_o}}}} \right)\left( {\dfrac{D}{{{f_e}}}} \right)$ where ${v_o}$ is the image distance from the objective, ${f_o}$ is the focal length of the objective, D is the least distance and ${f_e}$ is the focal length of eyepiece. The magnification due to eyepiece is $5$ i.e., $\dfrac{D}{{{f_e}}} = 5cm$ . We have assumed that object is kept very closer to focus i.e., ${u_e} = {f_e}$
$\dfrac{{20}}{{{f_e}}} = 5$
$\Rightarrow{f_e} = 4cm$ .
And $M = \left( {\dfrac{{{v_o}}}{{{f_o}}}} \right)\left( {\dfrac{D}{{{f_e}}}} \right)$
$\Rightarrow \left[ {{u_o} = {f_o}} \right]$
$\Rightarrow 20 = \left( {\dfrac{{{v_o}}}{{{f_o}}}} \right) \times 5$
$\Rightarrow \dfrac{{{v_o}}}{{{f_o}}} = 4cm$
$\Rightarrow {v_o} = 4{f_e}$

Length of the compound microscope $\left( L \right) = {f_o} + {v_o} + {f_e}$
$14 = {f_o} + {v_o} + {f_e}$
$\Rightarrow 14 = {f_o} + 4{f_o} + {f_e}$
$\Rightarrow 14 = 5{f_o} + 4$
$\Rightarrow 14 - 4 = 5{f_o}$
$\Rightarrow {f_o} = \dfrac{{10}}{5}$
$\therefore{f_o} = 2cm$

The focal length of the eyepiece is 4 cm and focal length of objective is 2 cm.

Note: The formula of magnifying power of eyepiece has distance of the object from the eyepiece in its denominator but the object is placed very close to a microscope in this case. Thus, the focal length of the eyepiece can be written in place of object distance from the eyepiece. So, kindly don’t get confused why focal length of eyepiece is written in place of object distance from the eyepiece.