Solveeit Logo

Question

Question: Describe the biprism experiment to find the wavelength of the monochromatic light. Draw the necessar...

Describe the biprism experiment to find the wavelength of the monochromatic light. Draw the necessary ray diagram.

Explanation

Solution

Hint : In the biprism experiment, two coherent sources of light that are required for the interference pattern are produced with the help of two thin prisms. To find the wavelength of light we can measure the fringe width of the interference pattern that is formed from the interference of these two coherent sources.

Formula used: In this solution, we will use the following formula
Δy=λDd\Delta y = \dfrac{{\lambda D}}{d} where Δy\Delta y is the fringe width of the interference pattern, λ\lambda is the wavelength of the light, DD is the distance between the sources of light, and dd is the distance between the two coherent sources.

Complete step by step answer

When a monochromatic light source (S)(S) is placed in front of a biprism, the light will be refracted such that the biprism will interfere with itself in a manner and form an interference pattern on a screen. We can treat this phenomenon as the interference of light from two coherent sources that are placed a distance dd apart.
From the properties of interference of light waves, we know that the fringe width is
Δy=λDd\Delta y = \dfrac{{\lambda D}}{d}
As we can see in the figure, the distance between the screen and the light source is D=a+bD = a + b
If the angle of the prism is represented by α\alpha and its refractive index is μ\mu then from the angle of biprism, we can write
d=2a(μ1)αd = 2a(\mu - 1)\alpha
Substituting the value of D=a+bD = a + b and d=2a(μ1)αd = 2a(\mu - 1)\alpha in Δy=λDd\Delta y = \dfrac{{\lambda D}}{d} , we get
Δy=λ(a+b)2a(μ1)α\Delta y = \dfrac{{\lambda (a + b)}}{{2a(\mu - 1)\alpha }}
Solving for the wavelength of light, we can write
λ=2α(μ1)αΔy(a+b)\lambda = \dfrac{{2\alpha (\mu - 1)\alpha \Delta y}}{{(a + b)}}
In the above equation, we know the angle of the prism α\alpha and its refractive index μ\mu and we also know the values of aandba\,{\text{and}}\,b . The fringe width Δy\Delta y can be measured experimentally using a microscope. So, then we can determine the wavelength of light.

Note
The biprism experiment is very similar to young’s modulus where a single source of light is split using two different slips whereas, in a biprism experiment, a biprism is used to form two apparent coherent sources. The two sources of light that interfere must be coherent with each other an interference pattern will not be formed.