Question

A compound microscope consists of two convex lenses of focal length $2cm$ and $5cm$. When an object is kept at a distance of $2.1cm$ from the objective, a virtual and magnified image is formed $25cm$ from the eyepiece. Calculate the magnifying power of the microscope.

Answer 1

Hint: The magnification of a compound microscope depends on the object distance and image distance of the image formed by the objective lens, and the focal point of the eyepiece. The magnification of a compound microscope is the product of the magnifications of its objective lens and eyepiece.
Formula used:
The magnification of a microscope is given by

$M=-\dfrac{{{v}_{0}}}{{{u}_{0}}}(1+\dfrac{D}{{{f}_{e}}})$

Where ${{v}_{0}}$ and ${{u}_{0}}$ are the image distance and object distance of the image formed by the objective lens, $D$ is the least distance of distinct vision $\left( D=25cm \right)$ and ${{f}_{e}}$ is the focal length of the eyepiece.
For a lens,
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ (Lens formula)
Where $v$, $u$ and $f$ are the object distance, image distance and focal length of the lens respectively.

Complete step by step answer:
This question can be solved by finding out the image distance of the image formed by the objective from the given object distance and objective focal length using the lens formula and then plug them in the formula for the magnifying power of a compound microscope.
For a lens,

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ (Lens formula) --(1)

Where $v$, $u$ and $f$ are the object distance, image distance and focal length of the lens respectively.
The magnification of a microscope is given by
$M=-\dfrac{{{v}_{0}}}{{{u}_{0}}}(1+\dfrac{D}{{{f}_{e}}})$ --(2)

Where ${{v}_{0}}$ and ${{u}_{0}}$ are the image distance and object distance of the image formed by the objective lens, $D$ is the least distance of distinct vision $\left( D=25cm \right)$ and ${{f}_{e}}$ is the focal length of the eyepiece.
Let us analyze the problem.

For the objective,
The object distance is ${{u}_{0}}=-2.1cm$ (negative by applying proper sign conventions)
The image distance is ${{v}_{0}}$.
The focal length of the objective is ${{f}_{o}}=2cm$.
Therefore using these values in (1), we get,
$\dfrac{1}{{{v}_{0}}}-\dfrac{1}{-2.1}=\dfrac{1}{2}$
$\therefore \dfrac{1}{{{v}_{o}}}+\dfrac{1}{2.1}=\dfrac{1}{2}$
$\therefore \dfrac{1}{{{v}_{o}}}=\dfrac{1}{2}-\dfrac{1}{2.1}=0.024$
$\therefore {{v}_{o}}=\dfrac{1}{0.024}=41.67cm$ --(3)
Now, the focal length of the eyepiece is${{f}_{e}}=5cm$.
The least distance of distinct vision $D=25cm$
Hence, using these values and (3) In equation (2), we get, magnification of the compound microscope as
$M=-\dfrac{41.67}{-2.1}\left( 1+\dfrac{25}{5} \right)=-\dfrac{41.67}{-2.1}\times \left( 1+5 \right)=-\dfrac{41.67}{-2.1}\times 6\approx 119$
Therefore, the required magnification of the microscope is $119$.

Note: The magnification produced in the compound microscope is the product of the magnifications of the objective lens $\left( \dfrac{{{v}_{o}}}{{{u}_{o}}} \right)$ and the eyepiece $\left( 1+\dfrac{D}{{{f}_{e}}} \right)$ . In fact, whenever two or more lenses are combined, the total magnification is the product of the individual magnifications of the lenses.

Students often make the mistake of not applying the proper signs to the magnitudes of the variables in optics problems. Applying the proper signs according to the sign convention is absolutely essential since then, one will arrive at a completely different and wrong answer. If, after applying proper sign convention, we get a negative magnification, this implies that the final image formed is inverted. If we had not applied, we would have got a positive value of magnification which would be wrong and would not be able to tell the complete characteristics of the final image.