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Newton's Rings-Refractive index of liquid
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## Aim:

1. To revise the concept of interference of light waves in general and thin-film interference in particular.
2. To set up and observe Newton’s rings.
3. Find the refractive index of given liquid.

## Thin film interference:

A film is said to be thin when its thickness is about the order of one wavelength of visible light which is taken to be 550 nm.  When light is incident on such a film, a small portion gets reflected from the upper surface and a major portion is transmitted into the film. Again a small part of the transmitted component is reflected back into the film by the lower surface and the rest of it emerges out of the film. These reflected beams reunite to produce interference.  Also the transmitted beams too interfere.  This type of interference that takes place in thin films is called interference by division of amplitude.

Θ1    angle of incidence at medium 1 to medium 2 boundary.
Θ2    angle of refraction at medium 1 to medium 2 boundary.
Θ3    angle of refraction at medium 2 to medium 3 boundary.
r12    reflected light from medium 1 to medium 2 boundary.
r23    reflected light from medium 2 to medium 1 boundary.
r21    reflected light from medium 2 to medium 3 boundary.
t21    transmitted light from medium 2 to medium 1 boundary.
t23    transmitted light from medium 2 to medium 3 boundary.
d       thickness of the film.

In the above figure the rays r12 and t21 interfere and results in a constructive or destructive interference depending on their path differences, given as,

constructive interference

destructive interference

Where,     refractive index of the  medium 2 and     the order of interference.
The transmitted light from t23 can also interfere and result in constructive or destructive interference.

## To find the refractive index of liquid:

The experiment is performed when there is an air film between the plano-convex lens and the optically plane glass plate.The diametre of the mth and the (m+p)th dark rings are determined with the help of a travelling microscope.

For air

As shown in figure arrange the lens with glass plate. Pour one or two drops of liquid whose refractive index is to be determined without disturbing the arrangement. Now the air film between the lens and glass plate is replaced by the liquid. The diameters of m+pth and mth rings are determined.

For liquids,

,for dark rings

For normal incidence cosr =1,so

But

Rearranging the above equation ,we get

$D_{_{m}}^{^{2}}=\frac{4m\lambda R}{\mu }$

We have $D_{m+p}^{2}-D_{m}^{2}=4p\lambda R$

for liquids,$D_{m+p}^{'2}-D_{m}^{'2}=\frac{4p\lambda R}{\mu }$

From these two euations the refractive index of the given liquids is given by

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