Question

Describe the bi prism experiment to find the wavelength of the monochromatic light. Draw the necessary ray diagram. The width of plane incident wave front is found to be doubled on refraction in denser medium.

Answer 1

Bi prism experiment is designed by Fresnel for illustrating interference property of light waves. In this experiment, two coherent sources of light are passed through a combined system of two prisms, i.e. bi prism, to obtain interference pattern. The fringe width of the interference patterns is given by the formula:-
$\beta =\dfrac{\lambda D}{d}$

Complete step by step solution:
Bi prism experiment is almost similar to young’s double slit experiment, with a slight difference which is, we are using a bi prism to create two coherent sources of monochromatic light from a single source. Light rays from these two virtual sources undergo interference and an interference pattern of bright and dark fringes can be obtained on a screen placed at specific distance.
We must know that a bi prism is made by combining two thin identical prisms base to base. The angles of prisms will be very small ranging as the acute angle of the prism is about $0.5{}^\circ$ and the obtuse angle is about $179{}^\circ$.

Now, if we place a monochromatic light source in front of the bi prism, two virtual coherent sources are formed. The light rays from these two virtual sources are assumed to be interfering at points P and Q as shown in the figure.
So, from YDSE experiment, we know that the fringe width is given by the expression,
$\beta =\dfrac{\lambda D}{d}$
Here,
$\lambda$ is the wavelength of the light wave.
$D$ is the distance between the light source and screen
And d is the distance between two virtual sources.
Now, we can see from the figure that, the distance between the light source and the screen is equal to D = (a+b).
Let us assume the angle of prism as $\alpha$ and the refractive index of prism as $\mu$ .
Then,
$d=2a(\mu -1)\alpha$
So, to find the fringe width, we know it is given as,
$\begin{aligned} & \beta =\dfrac{\lambda D}{d} \\\ & \beta =\dfrac{\lambda (a+b)}{2a(\mu -1)\alpha } \\\ & \Rightarrow \lambda =\dfrac{2a(\mu -1)\alpha \beta }{(a+b)} \\\ \end{aligned}$
If the fringe width of an interference pattern due to a biprism is given, we can find the wavelength of the light.
Also, we can see that $\lambda \propto \mu$,
That means if the refractive index is larger, wavelength will also be larger. That is why the width of plane incident wave front is found to be doubled on refraction in a denser medium.

Note:
The purpose of Fresnel’s biprism experiment and young’s double slit experiment is to obtain an interference pattern on the screen but the methods involved are different. The difference is in the production of coherent sources. In YDSE, two sources are produced by passing light through two narrow slits whereas in biprism, it is two virtual sources formed by refraction.