Q.Derive an expression for bandwidth of interference fringes in Young's double slit experiment.

Computing the path difference at a general point on the screen in Young's double slit setup and applying the conditions for bright and dark fringes gives the positions of successive fringes, whose spacing (the bandwidth) works out to $\beta=\lambda D/d$.

Setup

Two narrow, coherent slits $S_1$ and $S_2$, separated by a small distance $d$, are illuminated by monochromatic light of wavelength $\lambda$ from a single source, so that $S_1$ and $S_2$ act as coherent sources. A screen is placed parallel to the plane of the slits at a distance $D$ from them, with $D \gg d$. Let $O$ be the point on the screen on the perpendicular bisector of $S_1S_2$, and let $P$ be a point on the screen at distance $x$ from $O$.

Path difference at P

Using the right-angled triangles formed by $S_1$, $S_2$, and $P$ (with $S_1S_2=d$ and the perpendicular screen distance $D$):

$S_2P^2 - S_1P^2 = \left[D^2+\left(x+\frac{d}{2}\right)^2\right] - \left[D^2+\left(x-\frac{d}{2}\right)^2\right] = 2xd$

so $(S_2P-S_1P)(S_2P+S_1P) = 2xd$. Since $D\gg d$ and $D\gg x$, we may approximate $S_2P+S_1P \approx 2D$, giving the path difference:

$\delta = S_2P - S_1P = \frac{2xd}{2D} = \frac{xd}{D}$

Condition for bright and dark fringes

Bright fringe (constructive interference): occurs when the path difference is an integral multiple of the wavelength:

$\delta = n\lambda \quad (n=0,\pm1,\pm2,\ldots) \Rightarrow \frac{x_n d}{D} = n\lambda \Rightarrow x_n = \frac{n\lambda D}{d}$

Dark fringe (destructive interference): occurs when the path difference is an odd multiple of $\lambda/2$:

$\delta = \left(n+\frac{1}{2}\right)\lambda \Rightarrow x_n' = \left(n+\frac{1}{2}\right)\frac{\lambda D}{d}$

Fringe width (bandwidth)

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