Question

In Young's double slit experiment using monochromatic light of wavelength '$\lambda$', the maximum intensity of light at a point on the screen is $K$ units. The intensity of light at a point where the path difference is $\frac{\lambda}{3}$ is

• A. $\frac{K}{4}$
• B. $\frac{K}{2}$
• C. $K$
• D. $\frac{3K}{4}$

Hint:

Memorize the direct shortcut ratios for symmetric wave interference: a path difference of $\frac{\lambda}{2}$ gives zero intensity ($0$), a path difference of $\frac{\lambda}{3}$ gives quarter intensity ($\frac{K}{4}$), and a path difference of $\frac{\lambda}{4}$ gives half intensity ($\frac{K}{2}$). This bypasses the trigonometric steps entirely!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a Young's Double Slit Experiment (YDSE) configuration where the maximum peak intensity on the screen is $K$. We need to compute the resultant light intensity at a specific spatial location where the relative path difference between the interfering wavefronts is exactly $\Delta x = \frac{\lambda}{3}$.

Step 2: Key Formula or Approach:
1. Relate the path difference $\Delta x$ to the corresponding phase difference $\phi$: $$\phi = \frac{2\pi}{\lambda} \cdot \Delta x$$ 2. Use the standard formula for the resultant intensity of two coherent waves with equal initial intensity $I_0$: $$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right) = K \cos^2\left(\frac{\phi}{2}\right)$$ where $K = 4I_0$ represents the maximum possible constructive interference intensity.

Step 3: Detailed Explanation:
First, calculate the phase difference $\phi$ corresponding to the given path difference $\Delta x = \frac{\lambda}{3}$: $$\phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3} \text{ radians} \quad (120^\circ)$$ Now, substitute this angle into the intensity modulation formula: $$I = K \cos^2\left(\frac{\frac{2\pi}{3}}{2}\right) = K \cos^2\left(\frac{\pi}{3}\right)$$ We know that $\cos\left(\frac{\pi}{3}\right) = \cos(60^\circ) = \frac{1}{2}$. Evaluating the squared term gives: $$I = K \left(\frac{1}{2}\right)^2 = \frac{K}{4}$$

Step 4: Final Answer:
The intensity of light at that point is $\frac{K}{4}$, which corresponds to option (A).