Question 7 of 8intermediate🔧 ApplyNumerical3 marks

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

Correct Answer

K/4 units

Exercise: EXERCISES | Q: 10.5 | (Page Page 19)
For More Understanding

Explanation

This question requires understanding the relationship between path difference, phase difference, and resultant intensity in interference patterns. The context confirms that constructive interference occurs when path difference equals nλ, giving maximum intensity 4I₀. Using the phase-intensity relationship, we can find intensity at any path difference.

Solution Steps

  1. Step 1: At path difference λ, we have constructive interference. From the context, when path difference = nλ, resultant intensity = 4I₀. Given intensity = K, therefore K = 4I₀, which gives I₀ = K/4.

  2. Step 2: Path difference λ/3 corresponds to phase difference φ = (2π/λ) × (λ/3) = 2π/3.

  3. Step 3: The resultant intensity in interference is I = 4I₀cos²(φ/2). Substituting φ = 2π/3: I = 4I₀cos²(π/3) = 4I₀ × (1/2)² = 4I₀ × 1/4 = I₀.

  4. Step 4: Since I₀ = K/4, the intensity at path difference λ/3 is K/4 units.

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