Question

In Young’s double slit experiment, λ = 500nm, d = 1mm, D = 1m. Minimum distance from the central maximum for which intensity is half of the maximum intensity is:

• A. \( 2.5 \times 10^{-4}\,\text{m} \)
• B. \( 1.25 \times 10^{-4}\,\text{m} \)
• C. \( 0.625 \times 10^{-4}\,\text{m} \)
• D. 0.3125 × 10⁻4m

Hint:

Half-intensity points in YDSE occur when the phase difference is π/2.

Answer 1

The Correct Option is D

Step 1: For half maximum intensity: cos² β = (1)/(2) ⟹ β = (π)/(4)
Step 2: Path difference: β = (π d x)/(λ D)
Step 3: x = (λ D)/(4d)
Step 4: Substitute values: x = dfrac500 × 10⁻94 × 10⁻3 = 0.3125 × 10⁻4m