Question

Define resolving power of microscope. Write its formula. How will it vary, when
(a) Wavelength of incident light is decreased?
(b) Aperture of the objective lens is decreased?

Answer 1

Hint: In order to see objects that are too small for the naked eye, microscopes are used. For magnifying the object, we use circular openings like lenses. Whenever we use a circular opening, like our eye's lens or iris. A circular pattern of diffraction formed. Two such diffraction patterns, which are very close to each other, must be distinguished. A microscopes resolving power gives us that limit.

Complete solution:
The microscope's resolving power is given by the inverse of the distance between two lines or points, which can be clearly resolved using the microscope.

Let's figure out why we need that amount. When we see an object through a circular opening, as shown in the following figure, it creates a diffraction pattern. The concentric rings are called the rings of Airy.

We need to ensure that these rings do not overlap with each other when two objects are very close to each other.

Creating a blurry picture of the two objects. The Rayleigh criterion gave rise to this. In order to find the resolving power of microscopes, we can use this criterion. It is given by the inverse of the distance that can be resolved between two objects. We can write the Abbe's criterion.

Resolving power $=\dfrac{1}{\Delta d}=\dfrac{2 n \sin \theta}{\lambda}$
Where, $\dfrac{1}{\Delta d}$ is the resolving power of the microscope
$n$ is the refractive index separating the object and aperture.

The wavelength of the light used for observation.

The term $n \sin \theta$ is also called Numerical Aperture (N.A.) of the microscope. Hence, we can write,
$=\dfrac{1}{\Delta d}=\dfrac{2 \mathrm{~N} \cdot \mathrm{A}}{\lambda}$

(a) From equation (1) we can say that the resolving power of the microscope will increase if the wavelength of incident light is decreased.

(b) From equation (1) we can again say that the resolving power of the microscope will decrease if we decrease the aperture of the objective lens.

Note:
Rayleigh's criterion is given by, $\theta=1.22 \dfrac{\lambda}{d}$ Where,
$\theta$ is the half angle subtended by the first minimum at the source
$\lambda$ is the wavelength of light that is being used for measurement
$d$ is the distance between the two objects
This is an important criterion for utilising telescopes or microscopes to observe two fully resolved objects.