Using analytical method, obtain an expression for the fringe width of two interfering waves.

Answer: Fringe width \(X = \dfrac{\lambda D}{d}\)

Let S₁ and S₂ be two coherent monochromatic sources of wavelength \(\lambda\), separated by distance \(d\), and let the screen be at distance \(D\) from them (\(D \gg d\)). O is the point on the screen on the perpendicular bisector of S₁S₂. Consider a point P on the screen at distance \(y\) from O.

The feet of the perpendiculars drawn from S₁ and S₂ to the screen are at distances \(y - d/2\) and \(y + d/2\) from P. Hence

Since \(D \gg y, d\), \(S_2P + S_1P \approx 2D\). Hence the path difference is

Bright fringes: \(\Delta = n\lambda\) \((n = 0, 1, 2, \dots)\) \(\Rightarrow y_n = \dfrac{n\lambda D}{d}\).

Dark fringes: \(\Delta = (2n - 1)\dfrac{\lambda}{2}\) \(\Rightarrow y_n' = \dfrac{(2n-1)\lambda D}{2d}\).

Fringe width is the distance between two consecutive bright (or dark) fringes:

The same result is obtained from consecutive dark fringes. So bright and dark fringes are equally spaced, and \(X \propto \lambda\), \(X \propto D\) and \(X \propto 1/d\).