Newtons Ring Experiment Viva Concepts

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Table of Contents

Newtons Ring Experiment Viva an Introduction

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

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.

READ ALSO: Michelson Interferometer Experiment Viva for LASER wavelength

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.

At any Point P the Path Differnce

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 t²will 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

$\large r_{n}^{2}= \frac{(2n-1)\lambda R}{2 \mu}$

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

$\large D_{n}^{2}=\frac{2(2n-1) \lambda R}{\mu}$

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

$\large D_{n+m}^{2}=\frac{2(2n+2m-1) \lambda R}{\mu}$

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

$\large r^{2}=\frac{n \lambda R}{\mu}$

for the n^th rings

$\large r_{n}^{2}=\frac{n \lambda R}{\mu}$

For the nth fringe diameter

$\large D_{n}^{2}=\frac{4 n \lambda R}{\mu}$

Measurement of Wavelength of Sodium Light By Newton’s Rings

From the formula;

The diameter of the nth dark ring is;

$\large D_{n}^{2}=4 n \lambda R$

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

$\large D_{n+q}^{2}=4 (n+q) \lambda R$

Subtracting the first equation from the second one, we get

$\large D_{n+q}^{2}-D_{n}^{2}=4p \lambda R\\ \\ \lambda= \frac{D_{n+q}^{2}-D_{n}^{2}}{4pR}$

Next method is

If we plot a graph between $\large D_{n}^{2}$ 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.

$\large \frac {D_{n}^{2}}{n}=4 \lambda R$

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?

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

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

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

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

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

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

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

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

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?

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

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?

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

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?

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

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

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

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

Brainstorming newtons ring experiment viva

Can you please tell me that incident light is falling normal to the plane but fringes are forming in circular form, why?
Can we use LASER as a monochromatic source in this experiment?
If your answer is yes, then fringe will be circular or straight line or there will be only a spot?
This question may be asked in this way, why we have used extended source in this experiment?
What do you mean by equispaced fringes in the newton’s rings experiment setup?
Can we use diffracted LASER beams instead of putting a glass plate at 45 degrees in the Newton’s ring physics experiment ?
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?
Can you see the fringes obviously circular from the transmitted light rays in the same experiment?
Which parameters we can determine using this experiment?
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?
Can we determine the refractive index of the water by using this experiment?
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.

List of all Other Experiments

Hall Effect experiment	Magnetic Susceptibility of the FeCl3
Michelson Morley experiment	Charge to Mass Ratio by Thomson
Stewart and Gee’s	Melde Experiment
Attenuation losses	Semiconductor Diode
Planck’s Constant	Magnetic Susceptibility by Quincke Method
Hall Effect Experiment	Magnetic Susceptibility
Michelson Interferometer	Newtons Ring Experiment
viva questions on newton rings	newton’s ring viva questions
viva questions for newton’s rings experiment

Newtons Ring Experiment 18-Amazing Viva Questions

Newtons Ring Experiment Viva an Introduction

Objective

Apparatus

Theory of Newton’s Rings