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Resolving power of microscopes of telescopes or any optical instrument is its ability to resolve two nearly placed objects as separate objects. There is a theoretical upper limit to the resolving power of any optical instrument due to overlapping diffraction patterns created by two nearly placed objects. The resolving power of any instrument depends upon various factors. We can do modifications to have a better resolution or higher resolving power. In this article, we will learn about the resolving power of microscopes and telescopes, the limit of resolution, and Rayleigh’s criteria.
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Key takeaways: Resolving power, diffraction limit, limit of resolution, resolving power of microscope, resolving power of telescope, Airy discs
What is Resolving Power?
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Resolving power refers to the ability of an optical instrument to form separate distinguishable images of two objects placed at a small angular distance between them. If two objects are placed too close to each other beyond the limit of resolving power of the optical instrument, the image of the two objects will appear as one single object. Resolving power has no unit, it is a dimensionless quantity.

Resolving Power
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Limit of Resolution
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The limit of resolution is the angular or linear distance between two objects that are just resolved. You can think of it as the same as resolving power. In fact, the reciprocal of the limit of resolution gives us the resolving power. So, a lower value of the limit of resolution would mean a higher resolving power of the objective lens.
What is Diffraction Limit?
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A point object is when imaged through an aperture (a circular opening), a diffraction pattern is obtained rather than a point image. When the object size is comparable to the wavelength of the light, the diffraction pattern is more prominent. The diffraction pattern appears as concentric rings fainting out as we go further from the center. These are called Airy discs or Airy patterns.
As the concentric rings fade out as we go further, there is a point where two Airy discs or patterns are no more distinguishable from each other. This is called the diffraction limit.
Rayleigh’s Criterion
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Rayleigh’s criterion gives us the distance between two objects just resolved. It states that two images are said to be just resolved when the diffraction pattern of one image is over the first minimum diffraction of the second image.

Rayleigh Criterion
The first minimum diffraction pattern is observed at θ = 1.22λ/D
Rayleigh’s criterion gives us the formula for minimum angular distance between two objects just resolvable,
θ = 1.22λ/D
Where D is the circular aperture
And, λ is the wavelength.
Resolving Power of Telescope
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A telescope is an instrument used to observe distant objects. It usually consists of lenses, curved mirrors, or a combination of them to help us observe distant objects as if we are looking at them from a closer distance. It is usually used to observe different planetary objects.
When two distant stars are too close to each other, they impart a very low angular separation. The resolving power of the telescope determines if the stars will appear as separate stars to us.

Resolving Power of Telescope
The resolving power of a telescope is given by,
Resolving power = 1/\(\Delta\)θ = d/1.22 λ
Where,
\(\Delta\)θ = Angular separation between the two objects just resolved
d = diameter of the lens
And, λ = wavelength of light
Thus, telescopes with a larger d or larger diameter of the lens have a better resolving power. It also depends on the wavelength of light.
Example: Calculate the resolving power of a telescope whose aperture is 150cm for the wavelength of light 5.7 x 10-7 m.
Solution: Resolving power = D/1.22λ
D = 1.5m
λ = 5.7 x 10-7 m
Therefore, RP = 1.5/5.7 x 10-7 m = 0.263 x 107
Resolving Power of Microscope
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A microscope is an instrument used to observe microscopic specimens that are too small for the naked eye. It comprises of two lenses; objective lens and eyepiece or ocular lens. The resolving power of a microscope depends on the angle subtended by the diameter of the objective lens at the focus of the microscope, and the refractive index of the medium between the lens and microscopic specimen.
As we already discussed above, the resolving power is the reciprocal of the distance between two objects that are just resolved.

Microscope Resolving Power
So, the resolving power of a microscope is given by,
Resolving power = 1/\(\Delta\)d = 2a/λ
Where,
a = numerical aperture
λ = wavelength
Numerical aperture is often given on the lens body, it can also be calculated using the formula
a = n sinβ
Where,
n = refractive index of the medium,
and 2β = angle subtended by the diameter of the objective lens at the focus.
From the equation above, we can deduce that resolving power is higher when the numerical aperture of the lens is higher. So, by replacing air with a medium of higher refractive index (n) we can improve the resolving power of the microscope.
This is exactly what is used in the ‘oil immersion objective’. Here, a drop of suitable oil is placed on the glass slide containing the specimen to replace air with the oil that has a higher refractive index than air.
Example: You have a microscope whose numerical aperture is 1.3. Find out what should be the minimum distance between two microscopic specimens to be resolvable. (Wavelength = 580 nm)
Solution: Numerical aperture, a = 1.3 (given)
We know, d = λ/2a
=> d = 580/1.3 = 446nm
So, two microscopic specimens must be at least 446 nm apart to be resolvable by the microscope.
Resolving Power of Human Eye
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Like any man-made optical instrument, our eyes have a resolving power too. You can easily understand this by performing a simple experiment. Print an image with vertical lines each separated by 1mm. Look at it from a distance, say 5 meters, you will no longer be able to distinguish one line from another, it will appear gray.
The resolving power of the eyes, just like any other optical instrument, depends on the diameters of the lens, pupil in this case, and the wavelength which is equal to the wavelength of visible light.

Resolving Power of human eye
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Things to Remember
- Resolving power refers to the ability of an optical instrument to create separate images for two objects placed close to each other.
- The lower the distance between the two objects just resolved by the instrument, the higher is the resolving power.
- The limit of resolution is the distance between two objects just resolved. Resolving power is given by the reciprocal of the limit of resolution.
- The diffraction limit is the point where two Airy discs or patterns are no more distinguishable from each other.
- Rayleigh’s criterion states that two objects are said to be just resolved when the diffraction pattern of one object overlaps the first minimum diffraction pattern of the second object.
It is given by, θ=1.22λ/D
- The resolving power of a telescope determines its ability to distinguish between twin stars or two closely spaced stars at a distance.
The resolving power of a telescope is given by 1/\(\Delta\)θ = d/1.22 λ
- The resolving power of a microscope determines how well it is capable of distinguishing two microscopic specimens placed close to each other.
The resolving power of a microscope is given by 1/\(\Delta\)d = 2a/λ
- The resolving power of the human eye is determined by pupil size and the wavelength of light.
Sample Questions
Ques. Two convex lenses of the same focal length but of apertures A1 and A2 (A1 < A2), are used as the objective lenses in two astronomical telescopes having identical eyepieces. What is the ratio of their resolving power? Which telescope will you prefer and why? Give reason. (2011, 2 marks)
Ans. Resolving power of telescope = A/1.22λ
Therefore, R is directly proportional to A.
So, the ratio of resolving powers of two telescopes, R1/R2 = A1/A2
As given in the question, A1 < A2,
Therefore, R1 < R2
The larger the aperture, the higher is the resolving power. So, the telescope with aperture A2 will be preferred over the telescope with aperture A1.
Ques. Define the resolving power of a telescope. Write any two advantages of a reflecting telescope over a refracting telescope. (2010, 3 marks)
Ans. The resolving power of a telescope is the ability of the telescope to distinguish between two distant objects or light sources with very low angular separation.
Two advantages of reflecting telescope over refracting telescope:
(i) Reflecting telescopes use mirrors as objective instead of a lens, as a result, there is no chromatic aberration.
(ii) Refracting telescopes often suffer from spherical aberrations, whereas, reflecting telescopes are free from any spherical aberration.
Ques. An astronomical telescope has an objective lens of diameter 150 mm and a focal length of 4.00 m. The eyepiece has a focal length of 25 mm. Calculate the magnifying and resolving power of the telescope. (wavelength = 6000 A for yellow Color). (2010, 4 marks)
Ans. the diameter of the objective lens = 150 x 10-3m
f0 = 4m, fe = 25 x 10-3m, and D = 0.25 m
Therefore, magnifying power = - f0/fe [1 + (D/fe)]
= - 4/0.025 (1+0.25/0.025)
= - 4000/25 (1+ 10)
= -160 X 11
= 1760
And, resolving power = d/1.22λ
= 0.25/ 1.22(6x10-7)
= 0.34 x 106
Ques. Describe two ways by which the resolving power of a compound microscope can be increased. (3 marks)
Ans. Resolving power of a microscope is given by 2n sinβ/λ
Where, n = refractive index,
λ = wavelength of light
From the equation above, we can see that resolving power is directly proportional to the refractive index and indirectly proportional to the wavelength of light used.
So we can modify these two factors to increase the resolving power of the microscope.
(i) The resolving power of the microscope can be increased by increasing the refractive index. We can do this by using oil of a higher refractive index in place of air.
(ii) the resolving power can also be increased by using UV rays, or gamma ray or X-ray because the wavelength of UV rays, gamma, and an x-ray is smaller than that of visible light. But usually, biological specimens are susceptible to these types of rays, so one can subject the specimen to UV rays for a short span.
Ques. What is the resolving power of a 30 cm diameter telescope at a wavelength of 600 nm? (5 marks)
Ans. Resolving power of telescope = d/1.22λ
d = diameter of the objective lens of the telescope
And, λ = wavelength
d = 0.15m
λ = 600 x 10-9 m
Therefore, R = 0.15 / 600x10-9
= 15 x 10-7 /600
= 2.5x 10-9 rad
Ques. How can you increase the resolving power of a telescope? (3 marks)
Ans. The resolving power of a telescope is given by d/1.22 λ
Now, wavelength in the case of a telescope is fixed, this cannot be changed. So, one can increase the resolving power of a telescope by increasing the diameter of the objective. The diameter of the objective is directly proportional to the resolving power of the telescope. So, if you can increase the diameter by 2 times, the resolving power will be twice.
Ques. Two nearby celestial objects are separated by angular separation of 6 x 10-7. You are provided with a telescope of aperture 100cm. Would you be able to observe these two celestial objects? If not then what changes should you make to be able to observe them.(wavelength = 540nm) (5 marks)
Ans. We know that for a telescope to resolve two objects they must be separated by an angular distance, θ = 1.22 λ/D
λ = 540nm = 5.4 x 10-7 m
D = 100cm = 1m
So, θ = 1.22 x 5.4 x 10-7/ 1 = 6.588 x 10-7
Therefore, the telescope can just resolve two objects when the angular distance between them is at least 6.588 x 10-7. But as given in the question, the celestial bodies are only 6 x 10-7 apart. So the telescope provided is not enough to observe them.
To observe the two celestial objects, one must use a telescope of a larger aperture than 1m.
θ = 1.22 λ/D
=> D = 1.22 λ/θ
=> D = 1.22 x 5.4 x 10-7/ 6 x 10-7
=> D = 1.098 m = 109.8 cm
One would need a telescope of at least 110cm (approx.) to be able to observe the celestial objects in question.
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