Name: Brewster’s Law
Uploaded: 2019-07-19
Duration: 6 min 57 s
Description: Light is a transverse electromagnetic wave with electric field vectors vibrating in infinite directions perpendicular to the direction of propagation. Let's...

Question

Describe Young's double slit experiment with necessary theory and obtain expression for fringe width.

Accepted Answer

Young's double slit experiment gives the practical reference of interference of light that is redistribution of light energy due to superposition of light waves from two coherent sources producing constructive (when the light waves from the two coherent sources are in phase they will add up and produce bright fringe) and destructive (when the light waves from the two coherent sources are out of phase they will cancel out each other and produce dark fringe) interference pattern on the screen is referred to be interference of light.

Young's double slit experiment:

Here, $S$ is monochromatic source of light is producing spherical wavefront in all direction. As the pass through slit $S_{1}$ and $S_{2}$ they produce secondary wavefront. Since the wavefront are produces primarily by source $S$ and they have same wavelength and hey are in same phase. Hence, $S_{1}$ and $S_{2}$ are referred to be coherent source of light wave.

By the principle of superposition the secondary wavefronts from $S_{1}$ and $S_{2}$ will produce interference pattern on screen $G G'$ .

Assuming an arbitrary point on the screen $G G'$ , that is $P$ to be the point of maximum intensity or maxima.

Then, $S_{2} P - S_{1} P = n λ$ ........................... $(1)$

Here, $n$ is a natural number that is $n = 1, 2, 3, . . .$

Now, ${(S_{2} P)}^{2} - {(S_{1} P)}^{2} = [D^{2} + {(x + \frac{d}{2})}^{2}] - [D^{2} + {(x - \frac{d}{2})}^{2}] = 2 x d$ ........................... $(2)$

Here, $S_{1} S_{2} = d$ and $O P = x$

Then, ${(S_{2} P)}^{2} - {(S_{1} P)}^{2} = 2 x d \Rightarrow S_{2} P - S_{1} P = \frac{2 x d}{S_{2} P + S_{1} P}$ ........................... $(3)$

Here, the screen is much farther from the secondary source and the light wavefront from the secondary slits are approximately plane wavefront. Then, the light rays are assumed to be parallel. So, $D > > d$ .

If $P$ is at $O$ then, $S_{1} P = S_{2} P \approx D$

Here, $d > > λ \Rightarrow \frac{d}{λ} > > 1$ .

Then, $S_{2} P - S_{1} P = \frac{x d}{D}$ ........................... $(4)$

Now, constructive interference will result in bright fringe then, $x = x_{n} = \frac{n λ D}{d}$ ........................... $(5)$

Now, destructive interference will result in dark fringe then, $x = x_{n} = (n + \frac{1}{2}) \frac{λ D}{d}$ ........................... $(6)$

Here, the distance between two consecutive bright or dark fringe is referred to be fringe width.

Here, the fringe width will be, $β = x_{n} - x_{n + 1} = \frac{λ D}{d}$ ........................... $(7)$

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