Newtons Ring Experiment 18-Amazing Viva Questions

Last updated on Wednesday, May 22nd, 2024

Newton’s Ring Experiment Viva an Introduction

Newton’s ring experiment viva include many concepts of interference, you can say it includes all the basic fundamental in this experiment.

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Objective

Determination of the wavelength of a monochromatic source by using Newton’s Ring experiment

The key idea behind Newtons ring experiment is the thin film formation between a plane-convex lens and a glass plate. Due to this thin film of air a path difference occurs in the waves which reflect from the lower surface of the lens and the top surface of the glass plate. As a result of it, they superimpose and develop the interference pattern. The intensity depends on the nature of the pattern means it is constructive or destructive.

Apparatus

• Monochromatic light source; sodium lamp
• Plano-convex lens
• Glass plates of uniform thickness
• Convex lens
• And obviously a traveling microscope with clean scale and eye piece

Theory of Newton’s Rings

The Newton’s riings are formed as a result of interference between reflected light waves that comes from the upper and lower surface of thin air film. This thin air film formed between the plano convex lens and the flat glass plate.

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To explain it in more precise way, see below the figure 2., its 2nd picture. In this picture an incident light ray coming from the monochromatic source strikes at two points one is upper surface and the second one at lower surface of the thin film.

NOTE: To understand this concept you have to recall the topic of Interference by reflected and transmitted light rays from a thin transparent film.

Can you tell me about the path difference between these two rays, how much it would be? See, carefully and remember this point, where this light ray strike we are considering only that thickness of the thin film. So now implement the concept of path difference between these two reflected rays from the thin film. You can see black and red light reflected rays.

You know when light reflect from the denser medium, we observe the λ/2 additional change  in to the optical path of the reflected light ray compare to the second one. In the simple language when light reflect from the denser medium there occurs a phase change of π.

Here, we will use the path difference between these two reflected light waves using the wedge shape thin film where effective path difference is

Δ = 2 μ t cos ( r + θ) – λ/2             ….    (1)

Here μ is the refractive index of the thin film medium (now medium is air).

t is the thickness of the thin film at the point where incident light strikes and get reflects.

r is the angle of refraction in the thin film.

θ is the angle of wedge and λ operating wavelength that is from sodium lamp.

In the figure 2, picture (1) you can see that light are falling perpendicularly on the thin film, so how much you expect refraction angle will be? Right zero, because angle of incidence is zero to the normal at point O.

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At any Point P the Path Difference

Also, we are considering of large focal length plano convex lense so it will develop a very small wedge angle. so you can see the path difference factor

cos ( r + θ) = cos (0+θ) = cos (0) = 1 because theta is very small.Therefore the effective path difference will be

Δ= 2 μ t – λ/2                      ——(2)

Okay, where thickness is zero? At the point of contact of planoconvex lense and the plate that is the point P.

So net path difference at the point P will be;

Δ= λ/2          ———– (3)

You know here path difference is equal to the λ/2  which satisfy the condition of minimum intensity. This is the reason you see central spot black in the Newton’s ring experiment.

The condition of maximum intensity  

Δ= 2 μ t – λ/2   = 2m. λ/2   here m= 0,1,2….

or   2 μ t = 2m.λ/2 + λ/2 = (2m + 1) λ/2    ;here m= 0,1,2….

or   2 μ t = (2n – 1) λ/2   ;here n= 1,2,3….     ……… (4)

The condition of minimum Intensity

for dark fringes

or  2 μ t – λ/2  = (2n – 1) λ/2   ;here n= 1,2,3….

or   2 μ t = n λ  ……… (5)

From the above two equations (4) and (5), there is no change in the thickness.

How to use this concept?

See in the Newton’s rings experiment, this thickness would be constant if we measure the distance of this point from the center and draw a circle of this radius. So on this locus, light rays are reflecting. And you are  getting interference pattern.

Now the question is does for each variation in the thickness you will get only one fringe or fringes respectively?

Yes! number of fringes with respect to the thickness. This is the reason that you observe the circular ring for a constant thickness. If thickness decreses or increases by some point you will get fringes accordingly. I am not sure at this point you are clear or not. But I hope that you have observed the point of view (reflections from each point of that circle).

Diameters of Dark and Bright Newtons Ringes

To derive this result we will use second figure and the third part of it. Where a geometrical picture can be seen. In this picture you can see a glass plate E, a planoconvex lense L on that. This you can see in the form of a circle but don’t imagine it vertically right now, this is flat on the glass plate as you put in the experiment.

Let R is the radius of the curved surface, (now see it like a spherical ball) of the lens and t is its thickness of the thin film at any point P.

Now using the property of the circle,

we can write

AB x AP = OA x AF                  ———–(6)

As you can see that r is radius of the circle (Newton’s circular ring).

So AB = AP =r

While OA is equivalent to the thickness “t”.

Now you have to focus on the blue circle that is sphere you can imagine from which we cut the plano convex lens, right. So its Radius we have assumed R so diameter will be 2R. The distance AF will be (2R-t).

So, now putting the values, in circle equation, we get

r²= t (2R-t) = 2R t -t²

As you already know that radius is much larger than the thickness so  t² will be much smaller as compared to the 2Rt. So we will ignore this term and as a result you can see, this will be

r²= t (2R-t) = 2R t

or t = r²/2R                 ———–(7)

Substituting this value in the maxima and minima conditions for the Newton’s Rings;

For maxima or bright fringes

Put the value of t here,

2 μ t = (2n – 1) λ/2   ;here n= 1,2,3…

=> 2 μ. r²/2R = (2n – 1) λ/2

r² = (2n – 1) λ R /2μ

This denotes the radius of the n^th bright fringe, so can be written as

If suppose D_n is the diameter, then we will solve the above equation for diameter,

for the (n+m) ^th fringe the diameter

Similarly we can derive for the dark fringes.

For minima or dark fringes

or   2 μ t = n λ

2 μ. r²/2R = n λ

or we can write

for the n^th rings

For the nth fringe diameter

Measurement of Wavelength of Sodium Light By Newton’s Rings

From the formula;

The diameter of the nth dark ring is;

and the diameter for the (n+q)th dark fringe is;

Subtracting the first equation from the second one, we get

Next method is

If we plot a graph between and n, it will be a straight line and its slope value put in the below equation, by measuring the radius of curvature of the planoconvex lens.

Newtons Ring Experiment Viva

Here some questions are asked to make experiments working more clear in terms of physical concepts. If you find difficult to give the answer to any question you can ask through the comment box of this post.

1Q. What is the superposition principle?

In simpler terms, it means that when two or more waves are present in a system, the resultant wave at any given point is the sum of the individual waves at that point.

2Q. What do you mean by an interference of two or more than two waves?

Interference is a phenomenon that occurs when two or more waves meet or interact with each other. When waves interact, their amplitudes can either add together constructively or cancel each other out destructively, depending on their relative phase and amplitude.

Constructive interference occurs when the peaks and troughs of two or more waves are in phase, resulting in a larger amplitude at the point of interaction. This results in an increase in the intensity of the resulting wave.

Destructive interference, on the other hand, occurs when the peaks of one wave align with the troughs of another wave, resulting in the cancellation of the waves’ amplitudes at the point of interaction. This leads to a decrease in the intensity of the resulting wave.

3Q. What is the relation between path difference and phase difference?

Path difference refers to the difference in distance traveled by two waves from their respective sources to a given point. In other words, it is the difference in the length of the path that the waves have taken to reach the point of observation. Path difference is usually expressed in terms of the wavelength of the wave.

Phase difference, refers to the difference in the phase of two waves at a given point in space and time. Phase is a measure of the position of a wave in its cycle. If the phase difference between two waves is zero, it means that they are in phase, and their peaks and troughs line up, leading to constructive interference. If the phase difference is 180 degrees, they are out of phase, leading to destructive interference.

The relationship between path difference and phase difference depends on the wavelength of the wave. For example, if the path difference between two waves is equal to one wavelength, the waves will be in phase, and constructive interference will occur. If the path difference is equal to half a wavelength, the waves will be out of phase, and destructive interference will occur.

4Q. When two waves interfere, the intensity distributes uniformly. How do you see the output of intensity distribution as a result of interference?

When two waves interfere, the resulting intensity distribution can be calculated by adding the intensities of the individual waves at each point in space. The resulting intensity distribution can be visualized using an interference pattern, which is a pattern of bright and dark fringes that form as a result of constructive and destructive interference.

The specific interference pattern that is formed depends on the phase relationship between the two waves, the amplitude of the waves, and the distance between the sources of the waves.

In the case of interference between two coherent sources of light, such as a double-slit experiment, the resulting interference pattern appears as a series of bright and dark fringes. The bright fringes correspond to regions of constructive interference, where the waves are in phase and add together to produce a larger amplitude. The dark fringes correspond to regions of destructive interference, where the waves are out of phase and cancel each other out.

The spacing of the fringes depends on the wavelength of the light and the distance between the sources, with smaller distances and shorter wavelengths resulting in narrower fringe spacing.

The uniform distribution of intensity between the fringes is due to the interference between waves from nearby sources, which produces a continuous variation in the phase relationship between the waves across the entire region of observation. This leads to a gradual variation in the intensity of the light, resulting in a smooth and uniform background between the fringes.

5Q. Why do you use the plane-convex lens?

In the Newton’s rings experiment, a convex lens is placed in contact with a flat glass plate, creating a thin film of air between the lens and the plate. When light is incident on the lens, some of it is reflected off the surface of the air film and interferes with the light that is transmitted through the film. The resulting interference pattern can be observed as a series of concentric rings, known as Newton’s rings.

A plane-convex lens is commonly used in the Newton’s rings experiment because it produces a nearly uniform curvature of the air film, which leads to a more uniform interference pattern. This is because the curvature of the air film is determined by the curvature of the lens, and a plane-convex lens has a constant curvature across its surface.

6Q. What is the approximate focal length of the plano-convex lens?

The approximate focal length of a plano-convex lens can be calculated using the lens maker’s formula:

1/f = (n – 1) * [(1/r₁) – (1/r₂)]

where:
f = focal length of the lens
n = refractive index of the lens material
r₁ = radius of curvature of the curved surface
r₂ = radius of curvature of the flat surface

For a plano-convex lens, the radius of curvature of the flat surface is considered to be infinite, so the formula simplifies to:

1/f = (n – 1) * (1/r₁)

Assuming a typical refractive index of 1.5 for the lens material, and using a radius of curvature of 10 cm for the curved surface (which is a common value for plano-convex lenses), the focal length would be:

1/f = (1.5 – 1) * (1/0.1 m) = 5 m^-1

f = 0.2 m = 20 cm

Therefore, the approximate focal length of a plano-convex lens with a curved surface radius of 10 cm and a refractive index of 1.5 would be approximately 20 cm. However, it is important to note that the exact value of the focal length can vary based on factors such as the exact curvature of the lens and the refractive index of the material.

7Q. How do you obtain thin film and what is the role of it?

In the case of the Newton’s rings experiment, a thin film is created by placing a flat glass plate in contact with a convex lens. The surface of the lens and the glass plate form a thin film of air between them, which acts as a partially reflecting mirror. When monochromatic light is incident on the lens, it is partially reflected from the surface of the air film and partially transmitted through the film. The reflected light interferes with the transmitted light, creating an interference pattern of concentric rings known as Newton’s rings.

The role of the thin film in the Newton’s rings experiment is to act as a partially reflecting mirror and create the interference pattern. The thickness of the air film determines the path difference between the reflected and transmitted light waves, which in turn determines the phase difference and the resulting interference pattern.

8Q. Can a thin film be of the air?

Yes

9Q. Is it true that only through the monochromatic light you can observe Newton’s ring?

Yes, it is true that Newton’s rings are observed only when monochromatic light is used. This is because Newton’s rings are formed due to the interference between light waves of the same wavelength, or color. When white light is used, it contains a range of wavelengths or colors, which interfere with each other in a complex manner and do not produce a clear and visible interference pattern of rings.

10Q. Alright, the thin film which developed here is of uniform thickness or it gradually decreases towards the center, where the plane-convex lens (its convex surface is placed) and glass plate contacts. Why is the central spot dark in Newton’s ring?

In the Newton’s ring experiment, the central spot appears dark because at this point, the path difference between the reflected and transmitted rays is zero. The reflected and transmitted rays interfere with each other, and at the center, they meet in phase with each other, resulting in complete destructive interference. This means that the amplitude of the resulting wave at the center is zero, and hence the spot appears dark.

11Q. Why do you put a thin glass plate at 45 degrees to the light source?

It reflects light at 90 degree and on the other side light ray transmitted straight

12Q. The circular ring formed of various diameters in this experiment. This is the result of light reflection from upper and lower surfaces or is a result of the refraction?

The circular ring pattern observed in the Newton’s rings experiment is primarily the result of interference between the light waves reflected from the upper and lower surfaces of the air film formed between the lens and the glass plate. When a monochromatic light source is used, the reflected light waves interfere constructively or destructively depending on the thickness of the air film at a particular point, leading to the formation of bright and dark circular rings.

13Q. Here you observe dark and bright circular rings through the eyepiece, the reason behind it is a monochromatic source or interference conditions?

Both

14Q. How you can determine the wavelength of a monochromatic source by using this experiment in which you just observe the diameter and number of the rings? Is there any formula for wavelength determination?

Yes, it is possible to determine the wavelength of a monochromatic light source using the Newton’s Rings experiment, by measuring the diameter of the rings and the order of the ring. The formula used for this purpose is:

λ = (r_m^2 – r_n^2)/(2nR)

where λ is the wavelength of the light, r_m and r_n are the radii of the mth and nth order rings, respectively, n is the order of the ring, and R is the radius of curvature of the plano-convex lens.

—–See above derivation———-

This denotes the radius of the n^th bright fringe, so can be written as

If suppose D_n is the diameter, then we will solve the above equation for diameter,

for the (n+m) ^th fringe the diameter

15Q. What do you understand from the least count of any measuring device?

The least count of a measuring device is the smallest measurement that can be made using that device. It is also known as the resolution of the device. For example, if a ruler has a least count of 1 mm, it means that the smallest distance that can be measured using that ruler is 1 mm.

16Q. Can you determine the radius of curvature of a spherical surface from this experiment? Is there any formula for this purpose?

Yes, it is possible to determine the radius of curvature of a spherical surface using Newton’s rings experiment. The formula used for this purpose is:

R = (r_m^2 – r_n^2)/(2λn)

where R is the radius of curvature of the spherical surface, r_m and r_n are the radii of the mth and nth order rings, respectively, λ is the wavelength of the light source, and n is the order of the ring.

To use this formula, we need to measure the radii of at least two different order rings (such as m and n) and calculate the difference in their squares. We also need to know the wavelength of the light source. Then, by substituting these values into the formula, we can calculate the radius of curvature of the spherical surface.

———See above question also—-

If suppose D_n is the diameter, then we will solve the above equation for diameter,

17Q. What type of relationship do you see between the diameter and number of rings? Are they directly proportional or inversely proportional?

In the Newton’s rings experiment, there is an inverse relationship between the diameter of the rings and their order number. That is, as the order number of the ring increases, the diameter of the ring decreases.

18Q. Where do you can use this experiment? What are the possible applications?

The Newton’s rings experiment has several practical applications in science and engineering, including:

Testing the quality of optical components: The experiment can be used to test the quality of optical components such as lenses, mirrors, and prisms. By analyzing the interference pattern formed by the rings, one can determine the surface flatness, curvature, and other optical properties of the component.

Measuring the wavelength of light: The experiment can be used to measure the wavelength of monochromatic light sources. By using the formula for the radius of the nth order ring and measuring the diameter of the rings, one can calculate the wavelength of the light.

Studying thin film interference: The experiment provides a simple way to observe and study thin film interference, which is important in many fields of science and engineering, such as surface science, nanotechnology, and material science.

Calibration of micrometers: The experiment can be used to calibrate the least count of micrometers or other precision measuring devices. By measuring the diameter of the rings with the micrometer, one can determine its least count and accuracy.

Teaching tool: The experiment is often used as a teaching tool in optics and physics courses to illustrate concepts such as interference, diffraction, and thin film interference.

Brainstorming newtons ring experiment viva

1. Can you please tell me that incident light is falling normal to the plane but fringes are forming in circular form, why?
2. Can we use LASER as a monochromatic source in this experiment?
3. If your answer is yes, then fringe will be circular or straight line or there will be only a spot?
4. This question may be asked in this way, why we have used extended source in this experiment?
5. What do you mean by equispaced fringes in the newton’s rings experiment setup?
6. Can we use diffracted LASER beams instead of putting a glass plate at 45 degrees in the Newton’s ring physics experiment ?
7. Dark and bright circular fringes formed in the wedge shape thin film, do you think it will be formed if we use uniform thickness thin film?
8. Can you see the fringes obviously circular from the transmitted light rays in the same experiment?
9. Which parameters we can determine using this experiment?
10. Instead of Sodium lamp can we use normal white light bulb (this is also extended source), if your answer is yes then how fringes pattern will appear to you?
11. Can we determine the refractive index of the water by using this experiment?
12. How Sodium lamp works?

I am sure these questions will help you (Newtons ring experiment viva) to understand better it. If you have any question you may please ask through email or contact us page.

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