Q.In Young's double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is $K$ units. What is the intensity of light at a point where path difference is $\lambda/3$?

Concept: Double Slit Interference — intensity depends on the phase difference between the two waves.

Reasoning:

1. At path difference $\lambda$, phase difference $\delta = 2\pi$. The resultant intensity is maximum: $I_{\text{max}} = K$.

2. For two equal slits, intensity at any point is $I = I_0 + I_0 + 2I_0\cos\delta = 4I_0\cos^2(\delta/2)$. Since $I_{\text{max}} = 4I_0 = K$, we have $I_0 = K/4$. …

The core idea in Young’s double-slit experiment is that two coherent sources produce an interference pattern whose intensity at any point is determined by the phase difference between the two waves arriving there. The phase difference $\phi$ is directly proportional to the path difference $\Delta x$:

When two waves of equal amplitude $A$ (and hence equal individual intensity $I_0$) superpose, the resultant intensity is given by:

This is the fundamental formula — it tells you that intensity varies smoothly from maximum ($4I_0$) when $\phi = 0, 2\pi, 4\pi, \dots$ to zero when $\phi = \pi, 3\pi, \dots$.

Now let’s apply this to the given data.

1. Find $I_0$ in terms of $K$ At a point where path difference is $\lambda$, the phase difference is:

Then:

This intensity is given as $K$ units. So:

2. Find intensity for path difference $\lambda/3$ Phase difference:

Then: …

Method: Phasor Addition (or Wave Superposition) Method

This is the cleanest approach for intensity problems in Young's double-slit interference.

Step 1: Recall the intensity formula

For two coherent sources, the resultant intensity $I$ at a point with phase difference $\phi$ is:

Using the identity $1 + \cos\phi = 2\cos^2(\phi/2)$:

where $I_0$ is the intensity from each slit alone.

Step 2: Relate path difference to phase difference

Step 3: Apply for path difference $= \lambda$ (given $I = K$)

Path difference $= \lambda \implies \phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$

Given $I = K$, we get:

Step 4: Apply for path difference $= \lambda/3$ …

Common Mistakes in Double Slit Interference (Path Difference → Intensity)

This is a classic exam question from wave optics. Let's break down the common mistakes and how to avoid them.

✗ Mistake 1: Confusing Path Difference with Phase Difference

What students do wrong:

They directly plug path difference ($\Delta x$) into the intensity formula without converting to phase difference ($\Delta \phi$).

Correct approach:

The intensity at any point is given by:

where $\Delta \phi = \frac{2\pi}{\lambda} \cdot \Delta x$

How to avoid:

Always convert path difference to phase difference first using:

✗ Mistake 2: Forgetting that $I_0$ is the Maximum Intensity

What students do wrong:

They treat $K$ as $I_0$ (the intensity from one slit alone) instead of the intensity at a specific path difference.

Correct approach:

When $\Delta x = \lambda$, we have:

So:

Thus, $K = I_0$ (the maximum intensity).

How to avoid:

Read carefully: "intensity at path difference $\lambda$ is $K$ units" — this is the maximum intensity, not the intensity from one slit.

✗ Mistake 3: Using Wrong Formula for Intensity

What students do wrong:

Using $I = I_0 \cos(\Delta \phi)$ or $I = I_0 \sin^2(\Delta \phi/2)$.

Correct formula:

For two coherent sources of equal amplitude:

where $I_0$ is the maximum intensity (at $\Delta \phi = 0, 2\pi, 4\pi, \dots$).

How to avoid:

Memorise the cos² form — it comes from $I \propto (A_1 + A_2)^2$ with equal amplitudes.

✗ Mistake 4: Arithmetic Error in Final Step

What students do wrong:

After finding $\Delta \phi = \frac{2\pi}{3}$ for $\Delta x = \lambda/3$, they incorrectly compute: …