Path difference = nλ (constructive), (n+½)λ (destructive). Examples.

Superposition of Waves — Constructive and Destructive Interference

Question

Two coherent sources of light are placed such that the path difference at a point P is 2.5λ. Determine whether the interference at P is constructive or destructive. Also, if the path difference is changed to 3λ, what type of interference occurs?

Solution — Step by Step

Constructive interference happens when the two waves arrive in phase — crest meets crest, trough meets trough. This requires the path difference to be a whole number multiple of wavelength.

\Delta x = n\lambda, \quad n = 0, 1, 2, 3, \ldots

Destructive interference happens when the waves arrive completely out of phase — crest meets trough, and they cancel. This needs a half-integer multiple of wavelength.

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

We write $2.5\lambda = \frac{5\lambda}{2} = \left(2 + \frac{1}{2}\right)\lambda$ .

This fits the form $(n + \frac{1}{2})\lambda$ with $n = 2$ , so interference at P is destructive. The resultant amplitude is zero (assuming equal amplitudes).

Here $\Delta x = 3\lambda = n\lambda$ with $n = 3$ .

This is exactly a whole-number multiple, so the interference is constructive. The two waves reinforce each other fully.

Path difference $= 2.5\lambda$ → Destructive interference (dark fringe in YDSE)
Path difference $= 3\lambda$ → Constructive interference (bright fringe)

Why This Works

Think about what “path difference” physically means. One wave travels a longer distance to reach point P. Every time that extra distance equals one full wavelength, the lagging wave has completed one full extra cycle — so it’s back in step with the leading wave. They add up constructively.

When the extra distance is exactly half a wavelength, the lagging wave is at its crest exactly when the leading wave is at its trough. They cancel perfectly. This is destructive interference.

This is the entire physical basis of Young’s Double Slit Experiment (YDSE). The bright fringes sit where $\Delta x = n\lambda$ , and the dark fringes sit where $\Delta x = (n+\frac{1}{2})\lambda$ . The fringe pattern is just a map of these path differences across the screen.

Alternative Method — Using Phase Difference

Path difference and phase difference are connected by:

\delta = \frac{2\pi}{\lambda} \cdot \Delta x

For $\Delta x = 2.5\lambda$ :

\delta = \frac{2\pi}{\lambda} \times 2.5\lambda = 5\pi

Since $5\pi$ is an odd multiple of $\pi$ , the waves are in anti-phase → destructive.

For $\Delta x = 3\lambda$ :

\delta = \frac{2\pi}{\lambda} \times 3\lambda = 6\pi

$6\pi$ is an even multiple of $\pi$ (equivalently, a multiple of $2\pi$ ) → waves are in phase → constructive.

Quick check: if $\Delta x / \lambda$ is a whole number → constructive. If it’s a half-integer (like 0.5, 1.5, 2.5) → destructive. You can do this in two seconds without writing the full condition.

Common Mistake

Many students write the destructive condition as $\Delta x = n\lambda/2$ and conclude that $\Delta x = \lambda/2, \lambda, 3\lambda/2, 2\lambda \ldots$ are all destructive. This is wrong. $\Delta x = \lambda$ and $2\lambda$ are constructive — they also satisfy $n\lambda/2$ with even $n$ . The correct form is $(2n+1)\lambda/2$ , which picks out only the odd multiples: $\lambda/2, 3\lambda/2, 5\lambda/2, \ldots$ This exact confusion appeared in JEE Main 2023 where one option was deliberately set up to trap students using the incorrect $n\lambda/2$ form.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Phase Difference

Common Mistake

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