Application of Newton's Ring

APPLICATION OF NEWTON'S RINGS

• Determination of wavelength of the light wave

Let us set the Newton's ring experiment in such a condition that concentric circular rings are appeared in the field of view of the travelling microscope i.e. Circular bright and dark fringes are seen with the center dark. In this case, the radius of the \(n^{th}\) dark ring with the help of travelling microscope is given by the relation

r = \(\sqrt{n\lambda R}\)

So, diameter of the \(n^{th}\) ring is; \(D_n\) = 2 \(r_n\)

$$or,\;\;\; ;D_n=2\sqrt{n\lambda R}$$

$$so,\;\;D^2_n=4n\lambda R\dotsm(1)$$

Now, with the help of travelling microscope, diameter of the (\(n + m)^{th}\) dark ring are determined. Let, it be \(D_{n+m}\). Which are given by

Then, \(D^2_{n+m} = 4(n+m)\lambda R\dotsm(2)\)

Subtracting (1) from (2), we get

$$D^2_{n+m}-D^2_n=4(n+m)\lambda R-4n\lambda R$$

$$=4n\lambda R+4m\lambda R-4n\lambda R$$

$$4m\lambda R$$

$$or,\;\;\;\lambda=\frac{D^2_{n+m}-D^2_n}{4m\lambda R}\dotsm(3)$$

The equation (3) is used to determine the wavelength of light in lab.

Suppose the diameter of the \(5^{th}\) ring and the \(15^{th}\) ring is determined. Then, m = 15 – 5 = 10

$$\therefore\;\;\lambda=\frac{D^2_{n+m}-D^2_n}{4m\lambda R}$$

The radius of the curvature of the lower surface of the ring is determined with the help of a spherometer but more accurately by Boy's method. Hence, the wavelength of a given monochromatic source of light can be determined.

• Determination of refractive index of thin transparent medium

In this case also, let us set the Newton's ring experiment in such a condition that concentric circular rings are appeared in the field of view of the travelling microscope i.e. Circular bright and dark fringes are seen with the center dark. In this case, the radius of the \(n^{th}\) dark ring with the help of travelling microscope is given by the relation

r = \(\sqrt{n\lambda R}\)

So, diameter of the \(n^{th}\) ring is; \(D_n\) = 2 \(r_n\)

$$or,\;\;\; ;D_n=2\sqrt{n\lambda R}$$

$$so,\;\;D^2_n=4n\lambda R\dotsm(1)$$

Now, with the help of travelling microscope, diameter of the (\(n + m)^{th}\) dark ring are determined. Let, it be \(D_{n+m}\). Which are given by

Then, \(D^2_{n+m} = 4(n+m)\lambda R\dotsm(2)\)

Subtracting (1) from (2), we get

$$D^2_{n+m}-D^2_n=4(n+m)\lambda R-4n\lambda R$$

$$=4n\lambda R+4m\lambda R-4n\lambda R$$

$$4m\lambda R$$

$$or,\;\;\;\lambda=\frac{D^2_{n+m}-D^2_n}{4m\lambda R}\dotsm(3)$$

Figure;

The experiment is performed when there is an air film between the plano convex lens and the optically plane glass plate. Now, the plano convex lens and glass plate are kept in the beaker without disturbing the arrangements as shown in the figure above. Then, the liquid (transparent medium) whose refractive index is to be determined is poured into the beaker. In this case, air film enclosed by the plano convex lens and glass plate is replaced by given liquid.

Then, set the arrangement in such a way that the fringes are observed in the field of view. Here, radius of the \(n^{th}\) dark ring is given by

\(r'_n=\sqrt{\frac{n\lambda R}{\mu}}\) [ where, \(r'_n\) is the radius of Newton's ring after introducing new medium and cos r = 1]

Now, the condition of dark fringes is given by

$$2 \mu t cos r+\frac{\lambda}{2}=(2n+1)\frac{\lambda}{2}$$

$$or,\;\;\;2\mu \biggl(\frac{(r'_n)^2}{2R}\biggr)= n \lambda$$

Since, \(r'_n = \frac{D'_n}{2}\)

Then, \(\frac{\biggl(D'_n\biggr)^2}{4} = \frac{n\lambda R}{\mu}\)

$$or,\;\;\biggl(D'_n\biggr)^2=\frac{4n\lambda R}{\mu}\dotsm(4)$$

And, \(\biggl(D'_{n+m}\biggr)^2 = \frac{4(n+m)\lambda R}{\mu}\dotsm(5)\)

Now, subtracting equation (4) from (5), we get

$$\biggl(D'_{n+m}\biggr)^2-\biggl(D'_n\biggr)^2=\frac{4(n+m)\lambda R}{\mu}-\frac{4n\lambda R}{\mu}$$

$$=\frac{4m\lambda R}{\mu}$$

$$or,\;\;\mu=\frac{4m\lambda R}{\biggl(D'_{n+m}\biggr)^2-\biggl(D'_n\biggr)^2}$$

\(\therefore\) from equation (3), \( \mu=\frac{D^2_{n+m}-D^2_n}{\biggl(D'_{n+m}\biggr)^2-\biggl(D'_n\biggr)^2}\dotsm(6)\)

Equation (6) is the required expression to determine the refractive index of thin transparent film by Newton's ring method.

GRAPHICAL METHOD

The diameters of the dark rings are determined for various orders, varying from the \(n^{th}\) ring to the \((n+m)^{th}\) ring, first with air as the medium and then with the liquid. A graph is plotted between \(D^2_{n+m}\) along the y-axis and m along the x-axis, where m = 0, 1, 2, 3, …,etc. The ratio of the slopes of the two lines (air and liquid ), gives the refractive index of the liquid

GRAPHICAL FIGURE: Determination of refractive index by Graphical method; From P.B. Adhikari page-77

NEWTON'S RING WITH LIGHT

With monochromatic light, Newton's rings are alternatively dark and bright. The diameter of the ring \(D_n = 2\sqrt{\frac{n\lambda R}{\mu}}\) depends upon the wavelength of light used and is directly proportional to the square root of the wavelength. If, therefore, the light is not monochromatic but consist of two or more wavelength. As a result, if white light is used, only a few rings will be observed and these will be highly coloured except the central spot, which even in this case will be dark. Only the first few rings are visible and after that due to overlapping of the rings of different colours, the ring can not be viewed.