Conditions for constructive and destructive interference — derive

Question

Derive the conditions for constructive and destructive interference of light waves. Express conditions in terms of path difference and phase difference.

Solution — Step by Step

Consider two coherent sources $S_1$ and $S_2$ emitting light waves of the same frequency $f$ , same wavelength $\lambda$ , and constant phase difference (this is the requirement for coherence).

Let a point $P$ be at distances $r_1$ from $S_1$ and $r_2$ from $S_2$ .

The waves arriving at $P$ from both sources are:

y_1 = A\sin(\omega t) \quad \text{(from } S_1\text{)}

y_2 = A\sin(\omega t + \phi) \quad \text{(from } S_2\text{, with phase difference } \phi\text{)}

The phase difference $\phi$ arises because of the difference in path length: $\Delta r = |r_2 - r_1|$ .

One complete wavelength ( $\lambda$ ) corresponds to a phase of $2\pi$ radians.

Therefore, a path difference of $\Delta r$ corresponds to a phase difference:

\phi = \frac{2\pi}{\lambda} \Delta r

where $\Delta r = r_2 - r_1$ is the path difference.

Constructive interference occurs when the two waves arrive in phase — crest meets crest, trough meets trough. The resultant amplitude is maximum ( $2A$ ), and intensity is maximum ( $4I_0$ where $I_0$ is the intensity of each wave).

For waves to be in phase, the phase difference must be a multiple of $2\pi$ :

\phi = 0, 2\pi, 4\pi, \ldots = 2n\pi \quad (n = 0, 1, 2, \ldots)

Substituting $\phi = \frac{2\pi}{\lambda}\Delta r$ :

\frac{2\pi}{\lambda}\Delta r = 2n\pi

\boxed{\Delta r = n\lambda \quad (n = 0, 1, 2, \ldots)}

Condition for constructive interference: Path difference must be an integer multiple of the wavelength.

Destructive interference occurs when the two waves arrive exactly out of phase — crest meets trough. The resultant amplitude is minimum (0 for equal-amplitude waves), and intensity is minimum (0).

For waves to be exactly out of phase, the phase difference must be an odd multiple of $\pi$ :

\phi = \pi, 3\pi, 5\pi, \ldots = (2n-1)\pi \quad (n = 1, 2, 3, \ldots)

Or equivalently: $\phi = (2n+1)\pi$ for $n = 0, 1, 2, \ldots$

Substituting:

\frac{2\pi}{\lambda}\Delta r = (2n-1)\pi

\boxed{\Delta r = \frac{(2n-1)\lambda}{2} = \left(n - \frac{1}{2}\right)\lambda \quad (n = 1, 2, 3, \ldots)}

Or: $\Delta r = \lambda/2, 3\lambda/2, 5\lambda/2, \ldots$

Condition for destructive interference: Path difference must be an odd multiple of $\lambda/2$ .

Condition	Phase difference ( $\phi$ )	Path difference ( $\Delta r$ )	Effect
Constructive	$0, 2\pi, 4\pi, \ldots$	$0, \lambda, 2\lambda, \ldots$	Maximum intensity
Destructive	$\pi, 3\pi, 5\pi, \ldots$	$\lambda/2, 3\lambda/2, \ldots$	Zero intensity (for equal amplitudes)

Why This Works

The mathematics of interference follows directly from the principle of superposition — when two waves overlap, the resultant displacement is the algebraic sum of both displacements at each point.

When two waves of the same amplitude are in phase (path difference = $n\lambda$ ), they add: $A + A = 2A$ . Intensity $\propto A^2$ , so $I_{max} = 4A^2 = 4I_0$ .

When they are out of phase by $\pi$ (path difference = $(n-1/2)\lambda$ ), they cancel: $A + (-A) = 0$ . $I_{min} = 0$ .

Energy is conserved — the energy “missing” from dark fringes appears in bright fringes. Interference redistributes energy; it doesn’t create or destroy it.

Alternative Method

Using the resultant amplitude formula directly: for two waves with phase difference $\phi$ :

I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi

For $I_1 = I_2 = I_0$ :

I = 4I_0\cos^2(\phi/2)

Maximum: $\cos^2(\phi/2) = 1 \Rightarrow \phi/2 = 0, \pi, 2\pi, \ldots \Rightarrow \phi = 0, 2\pi, \ldots$ (constructive)
Minimum: $\cos^2(\phi/2) = 0 \Rightarrow \phi/2 = \pi/2, 3\pi/2, \ldots \Rightarrow \phi = \pi, 3\pi, \ldots$ (destructive)

Same conditions as derived above.

For JEE Main, interference questions appear in almost every shift (Wave Optics chapter). The key formula to memorize: $I = 4I_0\cos^2(\delta/2)$ where $\delta$ is the phase difference. Constructive when $\delta = 2n\pi$ ( $n$ = integer), destructive when $\delta = (2n+1)\pi$ . Path difference conditions come directly from $\phi = 2\pi\Delta r/\lambda$ . Also memorize fringe width: $\beta = \lambda D/d$ for Young’s Double Slit Experiment (YDSE).

Common Mistake

Students often mix up the conditions: they write “constructive when $\Delta r = (2n+1)\lambda/2$ ” (which is actually destructive). The correct rule: constructive = integer multiples of $\lambda$ (complete waves fit in the path difference); destructive = half-integer multiples of $\lambda$ (the waves are shifted by half a wavelength = out of phase). A memory device: Constructive = Complete wavelengths (both “C” words); Destructive = $\lambda/2$ (half wavelength) (D and Half → Destructive brings Half the wave).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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