Question

The objective lens of a compound microscope produces magnification of $10$ . In order to get an overall magnification of $100$ when image is formed at $25\,cm$ from the eye, the focal length of the eye lens should be :(in cm)
A) $4$
B) $10$
C) $\dfrac{{25}}{9}$
D) $9$

Answer 1

To solve this question, you need to realize that the net magnification of a compound microscope is nothing but the product of magnification due to objective lens and magnification due to eye lens. Thus, you can write:
$M = {M_o} \times {M_e}$

Complete step by step answer:
As described in the hint section of the solution to this question, we need to use the fact and formula:
$M = {M_o} \times {M_e}$
We can see that the value of magnification of the compound microscope should be $100$
The value of magnification due to objective lens is $10$
What’s not given is the value of magnification due to the eye lens. So, we need to find it.
Luckily, it is a very generalized result that the magnification due to eye lens in a compound microscope can be given as:
${M_e} = 1 + \dfrac{D}{{{f_e}}}$
Where, $D$ is the distance of nearest vision, which is numerically given to be $25\,cm$ in the question.
${f_e}$ is the focal length of the eye lens, which is not given in the question and we have to find its value.
Now, we can confidently substitute this value in the formula
$M = {M_o} \times {M_e}$
Substituting ${M_e} = 1 + \dfrac{D}{{{f_e}}}$ , we get:
$M = {M_o}\left( {1 + \dfrac{D}{{{f_e}}}} \right)$
After transposing, we get the formula for ${f_e}$ as:
${f_e} = \dfrac{{{M_o}D}}{{M - {M_o}}}$
Substituting in the values, we get:
${f_e} = \dfrac{{10 \times 25}}{{100 - 10}} \\\ {f_e} = \dfrac{{25}}{9}\,cm \\\$
Hence, we can see that the option (C) is correct as it matches the value that we found out upon solving the question.

Note: Many students add the magnifications of the respective lenses instead of multiplying them to find the net magnification of the lens and thus, reach at a wrong answer. Another mistake is that they do not use the correct formula for the magnification due to the eye lens as there are many confusing similar formulae for different microscopes and telescopes present in the same topic.