Discuss Young’s double-slit experiment and show that the width of bright and dark fringes are equal.

Young's double-slit experiment:

Let us consider S to be the monochromatic source of light. A and B are two slits that act as the coherent sources, d be the distance between them. Let a screen is placed at a distance D from the coherent sources, C be the central point on the screen, which is equidistance from A and B. The path difference of the wave reaching at C from A and B is zero.

Hence, a bright fringe of maximum intensity is observed. from central bright fringes, alternative dark and bright fringes of equal width are formed on either side of it as shown in the figure. To find the nth bright or dark fringe, let us consider a point P which lies at a distance x from C.

Here, \(PR=\left(x-\frac d2\right)\)

and, \(PQ=\left(x+\frac d2\right)\)

Now, \(AP^2=\left(x-\frac d2\right)^2+D^2\)

And, \(BP^2=\left(x+\frac d2\right)^2+D^2\)

So, \(BP^2-AP^2=\left(x+\frac d2\right)^2-\left(x+\frac d2\right)^2\)

∴ \(BP-AP=\frac{2xd}{BP+AP}\)

If the point P lies very near to C, then

BP ≅ AP ≅ D

∴ \(BP-AP=\frac{2xd}{2D}\)

i.e., path difference = \(\frac{dx}D\)

Since, phase difference = \(\frac{2\mathrm\pi}\lambda\) x (Path difference)

= \(\frac{2\mathrm\pi}\lambda\cdot\left(\frac{xd}D\right)\) ------ (i)

The point P may be bright or dark fringes depending upon the path difference (BP - AP).

Case (I): Bright fringes (or Constructive interference)

When the path difference between the two paths is an integral multiple of wavelength (λ), the bright fringes are obtained. 

i.e., Path difference = nλ = 1, 2, 3, ...

Or, \(\frac{\displaystyle xd}{\displaystyle D}=n\lambda\)

∴ \(x=\frac{n\lambda D}d\)

When n = 0, it gives the zero-order which indicates central bright fringe, and n = 1, 2, 3, 4, ... gives the 1^st, 2^nd, 3^rd, ... order bright fringes, and so on

Thus, for, n = 1, \(x_1=\frac{\lambda D}d\)

for, n =2, \(x_2=\frac{2\lambda D}d\)

for, n = 3, \(x_3=\frac{3\lambda D}d\)

And, for, n = n, \(x_n=\frac{n\lambda D}d\)

The distance between two consecutive bright fringes called fringe width (β), is given by

\(\beta=x_2-x_1=\frac{\lambda D}d\)

Case (II): Dark fringes (or Destructive interference)

When the path difference between two paths is half of the odd integral wavelength, then the dark fringes are obtained.

i.e., path difference = \(\left(\frac{2n+1}2\right)\lambda\), n = 0, 1, 2, 3, ...

Or, \(\frac{xd}D=\left(\frac{2n+1}2\right)\lambda\)

∴ \(x=\left(\frac{2n+1}2\right)\frac{\lambda D}d\)

This equation gives the distance of the dark fringes form the point C

for, n = 0, \(x_0=\frac{\lambda D}{2d}\)

for, n = 1, \(x_1=\frac{3\lambda D}{2d}\)

for, n = 2, (x_2=\frac{5\lambda D}{2d}\)

for, n = n, \(x_n=\left(\frac{2n+1}2\right)\frac{\lambda D}d\)

The distance between two consecutive dark fringes called fringe width (β), is given by

\(\beta=x_2-x_1=\left(\frac{2n+1}2\right)\frac{\lambda D}d\)

Thus, both dark and bright fringes are of equal width

This shows that the fringe width β is:

1. Directly proportional to the wavelength of the light used.
   i.e., β ∝ λ
   
   
2. Directly proportional to the distance between the screen from the coherent sources.
   i.e.,  β ∝ D
   
   
3. Inversely proportional to the distance between the slits.
   i.e, \(\beta\propto\frac1d\)
   
   
4. Independent of the order of the fringes. So, all the interference fringes are of equal width.
   The intensity distribution in Young's double-slit experiment is given as