Derive expression for fringe width in Young's double slit experiment

Question

Derive the expression for fringe width in Young’s double slit experiment. State the conditions for constructive and destructive interference.

Solution — Step by Step

Two narrow slits $S_1$ and $S_2$ are separated by distance $d$ , at distance $D$ from a screen. The slits are illuminated by coherent light of wavelength $\lambda$ .

Let point P on the screen be at height $y$ above the central axis. The distances from $S_1$ and $S_2$ to point P are $r_1$ and $r_2$ respectively.

We need to find the path difference $\Delta = r_2 - r_1$ in terms of $y$ , $d$ , and $D$ .

Using the geometry (with the approximation $d \ll D$ , so the slits are effectively on the same vertical line):

r_1^2 = D^2 + \left(y - \frac{d}{2}\right)^2

r_2^2 = D^2 + \left(y + \frac{d}{2}\right)^2

r_2^2 - r_1^2 = (r_2 - r_1)(r_2 + r_1) = \left(y + \frac{d}{2}\right)^2 - \left(y - \frac{d}{2}\right)^2 = 2yd

Since $r_1 \approx r_2 \approx D$ (for small angles):

(r_2 - r_1)(2D) \approx 2yd

\Delta = r_2 - r_1 = \frac{yd}{D}

Constructive interference (bright fringe): When the path difference is an integer multiple of wavelength (waves arrive in phase):

\Delta = n\lambda \quad n = 0, \pm 1, \pm 2, \ldots

\frac{y_n d}{D} = n\lambda \Rightarrow y_n = \frac{n\lambda D}{d}

Destructive interference (dark fringe): When path difference is a half-integer multiple of wavelength (waves arrive exactly out of phase):

\Delta = \left(n + \frac{1}{2}\right)\lambda \quad n = 0, \pm 1, \pm 2, \ldots

y_n = \frac{(2n+1)\lambda D}{2d}

Fringe width ( $\beta$ ) is the distance between any two consecutive bright fringes (or dark fringes).

Position of $n$ th bright fringe: $y_n = \frac{n\lambda D}{d}$

Position of $(n+1)$ th bright fringe: $y_{n+1} = \frac{(n+1)\lambda D}{d}$

\beta = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}

\boxed{\beta = \frac{\lambda D}{d}}

The fringe width is constant (independent of $n$ ) — fringes are equally spaced.

Why This Works

The key physics: light from two coherent sources creates an interference pattern because the phase difference between the two waves at any point depends on the path difference. Where the path difference is a whole number of wavelengths, the waves reinforce (bright fringe). Where it’s a half-integer number of wavelengths, they cancel (dark fringe).

The formula $\beta = \lambda D/d$ tells us:

Larger $\lambda$ → larger fringe width (red light gives broader fringes than blue)
Larger $D$ (further screen) → wider fringes (makes sense geometrically — the angle between fringes is fixed, so further away gives bigger spacing)
Larger $d$ (wider slit separation) → narrower fringes (more compressed pattern)

This is one of the most experimentally elegant results in optics — a simple geometry argument linking macroscopic spacing to light’s wavelength.

Alternative Method

Using angular approach: the condition for bright fringe is $d\sin\theta = n\lambda$ . For small angles, $\sin\theta \approx \tan\theta = y/D$ . So $y = n\lambda D/d$ . Fringe width = distance between $n=0$ and $n=1$ : $\beta = \lambda D/d$ . Same result, cleaner for visualising the geometry.

Common Mistake

A very common mistake is using the wrong formula for path difference. Some students write $\Delta = d\cos\theta$ instead of $\Delta = d\sin\theta$ . The correct formula: the path difference between the two rays is $d\sin\theta$ (where $\theta$ is the angle from the central axis), which equals $yd/D$ for small angles. The cosine appears in a different context (phase difference in a different geometry).

JEE frequently asks: “What happens to fringe width if the setup is immersed in water?” In water, the wavelength changes to $\lambda' = \lambda/n$ (where $n$ = refractive index of water ≈ 1.33). Since $\beta = \lambda D/d$ , fringe width decreases by a factor of $n$ . Fringe width in water = $\beta/1.33$ ≈ 0.75 times the air value.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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