Question

An angular magnification (magnifying power) of $30{\text{X}}$ is desired using an objective of focal length $1.25\,{\text{cm}}$ and an eyepiece of focal length $5\,{\text{cm}}$ . How will you set up the compound microscope?

Answer 1

First of all, we will find out the angular magnification of the eyepiece using a formula. After that we will find angular magnification for the objective lens. Then we will use lens formula for both the eyepiece and the objective lens. We will manipulate accordingly and find the result.

Complete step by step solution:
In the given question, we are supplied the following data:
The desired angular magnification that is what we call magnifying power is of $30{\text{X}}$. The focal length of the objective is given as $1.25\,{\text{cm}}$ .The focal length of the eyepiece is given as $5\,{\text{cm}}$ .We are asked to find out how we will set up the compound microscope.

To begin with, we can write the total magnification of the given compound microscope is $30$ .We know that the least distance to get a distinct vision is $25\,{\text{cm}}$ .Let us proceed to solve the problem as given below: We know the relation which relates angular magnification of the eyepiece with its focal length is given below:
${m_{\text{e}}} = 1 + \dfrac{d}{{{f_{\text{e}}}}}$ …… (1)
Where,
${m_{\text{e}}}$ indicates the magnification of the eyepiece.
$d$ indicates the least distance of distinct vision.
${f_{\text{e}}}$ indicates the focal length of the eyepiece.

Now, we substitute the required values in the equation (1) and we get:
${m_{\text{e}}} = 1 + \dfrac{d}{{{f_{\text{e}}}}} \\\ \Rightarrow {m_{\text{e}}} = 1 + \dfrac{{25}}{5} \\\ \Rightarrow {m_{\text{e}}} = 1 + 5 \\\ \Rightarrow {m_{\text{e}}} = 6$
Therefore, the angular magnification of the eyepiece is found to be $6$ .

As we know, there is a formula which gives the magnifying power of the compound microscope, which is given below:
$m = {m_{\text{o}}} \times {m_{\text{e}}}$ …… (2)
Where,
$m$ indicates the magnifying power of the microscope.
${m_{\text{o}}}$ indicates the angular magnification of the objective lens.
${m_{\text{e}}}$ indicates the angular magnification of the eyepiece.

Now, we substitute the required values in the equation (2) and we get:
$m = {m_{\text{o}}} \times {m_{\text{e}}} \\\ \Rightarrow 30 = {m_{\text{o}}} \times 6 \\\ \Rightarrow {m_{\text{o}}} = 5$
Therefore, the angular magnification of the objective lens is found to be $5$ .

Again, we know:
${m_{\text{o}}} = \dfrac{{ - {v_{\text{o}}}}}{{{u_{\text{o}}}}} \\\ \Rightarrow 5 = \dfrac{{ - {v_{\text{o}}}}}{{{u_{\text{o}}}}} \\\ \Rightarrow {v_{\text{o}}} = - 5{u_{\text{o}}}$

We will the lens formula now, to determine ${v_{\text{o}}}$ and ${u_{\text{o}}}$ .
$\dfrac{1}{{{v_{\text{o}}}}} - \dfrac{1}{{{u_{\text{o}}}}} = \dfrac{1}{{{f_{\text{o}}}}} \\\ \Rightarrow \dfrac{1}{{ - 5{u_{\text{o}}}}} - \dfrac{1}{{{u_{\text{o}}}}} = \dfrac{1}{{1.25}} \\\ \Rightarrow - \left( {\dfrac{{1 + 5}}{{5{u_{\text{o}}}}}} \right) = \dfrac{1}{{1.25}} \\\ \Rightarrow {u_{\text{o}}} = - 1.5\,{\text{cm}}$
Since, we have,
${v_{\text{o}}} = - 5{u_{\text{o}}} \\\ \Rightarrow {v_{\text{o}}} = - 5 \times 1.25 \\\ \Rightarrow {v_{\text{o}}} = - 7.5\,{\text{cm}}$
From the above calculation, we can say that the object should be placed at a distance of $1.5\,{\text{cm}}$ from the objective lens to get the desired magnification.

Now, we have the distance of the image for the eyepiece is $- 25\,{\text{cm}}$ .
Again, by using the lens formula, we get:
$\dfrac{1}{{{v_{\text{e}}}}} - \dfrac{1}{{{u_{\text{e}}}}} = \dfrac{1}{{{f_{\text{e}}}}} \\\ \Rightarrow \dfrac{1}{{ - 25}} - \dfrac{1}{{{u_{\text{e}}}}} = \dfrac{1}{5} \\\ \Rightarrow \dfrac{1}{{{u_{\text{e}}}}} = \dfrac{1}{{ - 25}} - \dfrac{1}{5} \\\ \Rightarrow {u_{\text{e}}} = - 4.17\,{\text{cm}}$

Now, we can get the separation between the objective lens and the eyepiece:
$\left| {{u_{\text{e}}}} \right| + \left| {{v_{\text{o}}}} \right| \\\ \Rightarrow \left| { - 4.17} \right| + \left| { - 7.5} \right| \\\ \Rightarrow 4.17 + 7.5 \\\ \therefore 11.67{\text{cm}}$

Hence, the distance of separation between the objective lens and the eyepiece should be $11.67{\text{cm}}$.

Note: In order to solve this problem, one must have a strong knowledge microscope and telescope operation. Most of the students seem to have a confusion regarding the angular magnification of the objective lens and the eyepiece, whose magnitudes are not given. We can simply get those values by using the formula. After we get those values, then only we can proceed. It is important to remember that the focal length of a convex lens is always positive. Taking a negative value can ruin the whole solution.