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:
In YDSE, maximum intensity is K; find resultant intensity at a point where path difference Δx = λ/3.

Step 2: Key Formula or Approach:
Phase difference φ = (2π/λ)·Δx. Intensity I = K cos²(φ/2).

Step 3: Detailed Explanation:
φ = (2π/λ)(λ/3) = 2π/3. I = K cos²(π/3) = K (1/2)² = K/4.

Step 4: Final Answer:
Intensity is K/4, matching option (A).