Question

A compound microscope consists of an objective of focal length $2\,{\text{cm}}$ and an eye piece of focal length $5\,{\text{cm}}$ . When an object is kept $2.4\,{\text{cm}}$ from the objective, the final image formed is virtual and $25\,{\text{cm}}$ from the eye piece. Determine the magnifying power of this compound microscope in this set i.e. normal use.

Answer 1

First we will calculate the magnification produced by the eye piece. After that we will find the image distance produced by the objective by using the lens formula, followed by magnification produced by it. We know the product of magnifying power of eyepiece and the objective give the magnification power of the compound microscope.

Complete step by step answer:
In the given question, we are supplied with the following information:
The focal length of the objective is given as $2\,{\text{cm}}$ .The focal length of the eyepiece is given as $5\,{\text{cm}}$ .The object distance is given as $2.4\,{\text{cm}}$ from the objective.The nature of the final image is virtual. We are asked to find the magnifying power of this compound microscope in this set i.e. normal use.

To begin with, first we will find the magnification of the eye piece.
Let us proceed to solve the problem.
The formula which gives the magnification for the eyepiece, is given below:
${M_{\text{e}}} = 1 + \dfrac{D}{{{f_{\text{e}}}}}$ …… (1)
Where,
${M_{\text{e}}}$ indicates the magnification for the eyepiece.
$D$ indicates the distance of the final image produced.
${f_{\text{e}}}$ indicates the focal length of the eye piece.

Substituting the required values in the equation (1) we get:
${M_{\text{e}}} = 1 + \dfrac{D}{{{f_{\text{e}}}}} \\\ \Rightarrow {M_{\text{e}}} = 1 + \dfrac{{25}}{5} \\\ \Rightarrow {M_{\text{e}}} = 6$
Therefore, the magnification produced by the eyepiece is $6$ .

Now, we apply the lens formula to find the image distance produced by the objective, which is given by:
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ …… (2)
Where,
$v$ indicates the distance of the image.
$u$ indicates the distance of the object.
$f$ indicates the focal length of the lens.

Now, we substitute the required values in the equation (2) and we get:
\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{2} + \dfrac{1}{{ - 2.4}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{2} - \dfrac{1}{{2.4}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{2} - \dfrac{{10}}{{24}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{12}} \\\ \Rightarrow v = 12\,{\text{cm}} \\\
Therefore, the image distance of the objective is found to be $12\,{\text{cm}}$ .
So, we will now the magnification produced by it:
{M_{\text{o}}} = \dfrac{v}{u} \\\ \Rightarrow {M_{\text{o}}} = \dfrac{{12}}{{2.4}} \\\ \Rightarrow {M_{\text{o}}} = 5 \\\
We know, the magnification power $M$ of compound microscope is given by:
$M = {M_{\text{o}}} \times {M_{\text{e}}} \\\ \Rightarrow M = 5 \times 6 \\\ \Rightarrow M = 30$

Hence, the magnifying power of this compound microscope in this set i.e. normal use is $30$ .

Note: While solving this problem, most of the students have confusion regarding the type of lens used in it. Here the convex lenses are used. It is important to note that the focal length of the convex lens is always positive. By not taking the appropriate sign convention, one may get irrelevant results. The focal length of the concave lens is taken to be negative.