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 $\lambda/3$?

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> [!TLDR]
> In Young’s double-slit interference, intensity depends on the phase difference via $I = I_0 \cos^2(\phi/2)$. For path difference $\lambda$, the phase difference is $2\pi$ and intensity is $K = 4I_0$. For path difference $\lambda/3$, phase difference is $2\pi/3$, giving intensity $I = K/4$.

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 $\Delta x$:

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

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

$$
I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)
$$

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:
   $$
   \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi
   $$
   Then:
   $$
   I = 4I_0 \cos^2\left(\frac{2\pi}{2}\right) = 4I_0 \cos^2(\pi) = 4I_0 \cdot 1 = 4I_0
   $$
   This intensity is given as $K$ units. So:
   $$
   4I_0 = K \quad \Rightarrow \quad I_0 = \frac{K}{4}
   $$

2. **Find intensity for path difference $\lambda/3$**  
   Phase difference:
   $$
   \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}
   $$
   Then: …

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 understanding — Double Slit Interference

Double Slit Interference: From Ripples to Light

The Core Idea

The Geometry

What This Tells You

Why It Matters

Why this formula?

Double Slit Interference: Why the Formula Holds

1. The Core Idea: Path Difference Creates Phase Difference

2. Finding the Path Difference

3. Condition for Constructive Interference (Bright Fringes)

4. Condition for Destructive Interference (Dark Fringes)

5. From Angle to Position on Screen

Method: Phasor Addition (or Wave Superposition) Method

Step 1: Recall the intensity formula

Step 2: Relate path difference to phase difference

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

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

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

✗ Mistake 1: Confusing Path Difference with Phase Difference

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

✗ Mistake 3: Using Wrong Formula for Intensity

✗ Mistake 4: Arithmetic Error in Final Step

Concept

Solution

Concept

Solution

Concept

Solution

Concept …

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

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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 $\lambda/3$ ?

Unlock everything free for 14 days