Using the geometry of the double slit experiment, derive the expression for fringe width of interference bands.

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

[Diagram: two narrow slits S₁ and S₂ separated by d, illuminated by monochromatic light; screen at distance D from the slits; O on the screen equidistant from S₁ and S₂ (central bright fringe); point P at distance x from O; S₁P and S₂P drawn; perpendiculars S₁M and S₂N dropped on the screen.]

Let S₁ and S₂ be two coherent sources separated by distance \(d\), and the screen be at distance \(D\) from them (\(D \gg d\)). Consider a point P on the screen at a distance \(x\) from the central point O.

From the figure, with S₁M and S₂N perpendicular to the screen:

\((S_2P)^2 = D^2 + \left(x + \dfrac{d}{2}\right)^2\), \((S_1P)^2 = D^2 + \left(x - \dfrac{d}{2}\right)^2\)

\((S_2P)^2 - (S_1P)^2 = 2xd\)

\((S_2P - S_1P)(S_2P + S_1P) = 2xd\). Since \(D \gg x, d\), \(S_2P + S_1P \approx 2D\).

Path difference: \(\Delta = S_2P - S_1P = \dfrac{xd}{D}\).

Bright fringes: \(\Delta = n\lambda\), so \(x_n = \dfrac{n\lambda D}{d}\), \(n = 0, 1, 2, \ldots\)

Dark fringes: \(\Delta = (2m - 1)\dfrac{\lambda}{2}\), so \(x_m = (2m-1)\dfrac{\lambda D}{2d}\), \(m = 1, 2, 3, \ldots\)

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

\(W = x_{n+1} - x_n = \dfrac{(n+1)\lambda D}{d} - \dfrac{n\lambda D}{d} = \dfrac{\lambda D}{d}\).

Similarly for dark fringes, \(x_{m+1} - x_m = \dfrac{\lambda D}{d}\). Thus all bright and dark fringes are equally spaced, and \(W \propto \lambda\), \(W \propto D\), \(W \propto 1/d\).