Resolving power of microscopes

Resolving power of a microscope: Definition

If two points of an object are so close that their diffraction discs overlap each other, we cannot see those points separately. The resolving power of a microscope tells us how far apart points can be seen separately. Resolving power of a microscope is a function of refractive index.

Resolving range 

The resolution range of an optical instrument is equal to the minimum angular distance between two point objects at which their images can be seen separately by the optical instrument, where λ is the wavelength of the light used, and d is the diameter of the aperture of the objective lens.

Resolving power 

The resolving power of an optical instrument is the minimum distance between two objects at which the optical instrument can form images of both objects separately. It is represented by D, and its unit is a metre or centimetre.

Resolution microscope definition

Celestial objects are often seen through binoculars. These bodies can be millions of miles away from each other, but the direction of the light coming from them can be almost the same. Consider two object, S₁ and S₂, which is being tried to be seen through a microscope. The first images of these two are being formed at the focus plane of the objective. These images are in the form of a diffraction disc. The direction of light coming from S₁ and the direction of light coming from S₂ makes an angle d with each other. There is an angular separation of dθ between these stars to the observer. If the centres of their diffraction discs are at a distance x from each other, then from the figure,

The radius of this diffraction disc,

r = 1.22λf/a

where λ is the wavelength of light, and a is the diameter of the objective. These discs may look different, if x > r, ie

Fdθ > 1.22λf/a

or dθ > 1.22λ/a

This minimum value of the angular gap is called the resolution limit or resolution of the microscope, and its inverse is called the resolving power. Therefore,

resolving power = a/1.22λ

The discriminative power of a microscope depends on the diameter of the objective. The larger the diameter, the greater the resolving power. Also, due to the larger diameter, the objective can capture more light, and the image becomes brighter.

The resolving power is inversely proportional to the wavelength, i.e. using light of a shorter wavelength will yield more resolving power.

Resolving Power of Microscopes

Microscopes are used to see nearby objects. The resolving power of a microscope is also determined by its resolving range (inversely proportional).

The resolution limit of a microscope is the shortest distance between two nearby objects when the images formed by the microscope are properly differentiated.

The smaller this distance, the higher the resolving power of the microscope.

Formation of an image of two nearby objects, P and Q, by microscope

In the figure, two adjacent objects, P and Q, are placed in front of the objective AB of the microscope, whose images p’ and q’ are formed by the objective.

This image is the maximum obtained as a result of the circular aperture Fresnel diffraction.

Let θ be the angle subtended by objects p and q at the objective of the microscope.

According to Rayleigh’s criterion of the marginal resolution, the minimum point of the image P’ should be at Q’, and the minimum of the point of the image Q’ should be at P’.  

In other words, if the angular semi-breadth of each major maxim is 𝛟 = θ. 

If the principal maxima of object p are p’, 

Then, PA + AP’ = PB + BP’ …… eq. 1

Similarly, if the principal maximum of object q is q’ 

Then, QA + AQ’ = QB + BQ’……… eq. 2

Now, for the first minima of the image P’ to be at the point Q’, it is necessary that the path difference between the light waves arriving from A and B at the first minimum Q’ in the object P is equal to λ so that, 

PA + AQ ‘ =  PB + BQ ‘ – λ….…eq 3

From eq2 and eq3,

(QA – PA ) = ( QB – PB ) + λ….…. eq4

Now, if ∠APB = 2α, at object P by the objective of a microscope, then the interior angle at object Q will also be about 2α because both the objects P and Q are very close. If the shortest distance between objects P and Q is Xmin, they are said to be properly differentiated.

From diagram, 

QA – PA ≈ PB – QB ……… eq5

     Xmin ≈ Sinα

Putting the value in equation 4,

XminSinα = -XminSinα + λ

2XminSinα =  λ

Xmin = λ/ 2Sinα …….eq6

In the calculation of the expression, it is assumed that the aperture is rectangular.

Since the aperture is circular, so on applying the correction for the circular aperture,

Xmin = 1.22λ/2Sinα 

         = 0.61λ /Sinα  …….eq7

If the space of refractive index H is filled in place of air between the objects and the microscope, the effective wavelength of the incident light will be λ/H, and the resolution range of the microscope Xmin = 0.61 λ/2HSinα.

In this expression, 2HSinα is the numerical aperture D of the microscope.

The resolving power of the microscope is Xmin = 1.22λ/ numerical aperture.

The resolving power depends on the aperture of the objective and the wavelength of light.

Conclusion

The microscope is a very powerful tool for viewing smaller objects. The resolving power of a microscope tells us how far apart points can be seen separately. Resolving power of a microscope is a function of refractive index. The resolution limit of a microscope is the shortest distance between two nearby objects when the images formed by the microscope are properly differentiated. The resolving power of the microscope is Xmin = 1.22λ/ numerical aperture.

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