Question

In a Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light In a Young’s double-slit experiment, using mono chromatic light of wave length λ, the intensity of light at a point on the screen is I₀, where the path difference between the interfering waves is λ. The path difference between the interfering waves at a point where the intensity is \(\frac{I_o}{2}\) , will be:at a point on the screen is I0​, where the path difference between the interfering waves is λ. The path difference at a point where the intensity is 2I0​​ will be:

• A. \(\frac{\lambda}{2}\)
• B. \(\frac{\lambda}{4}\)
• C. λ
• D. 2λ

Answer 1

The Correct Option is B

Given:

• Intensity at path difference Δx = λ is I₀
• Find path difference when intensity is I₀/2

Key relationships:

I = 4I₁cos²(φ/2)

where φ = phase difference = (2π/λ)Δx

At Δx = λ:

φ = (2π/λ)λ = 2π

I₀ = 4I₁cos²(π) = 4I₁ (since cos(π) = -1)

When I = I₀/2:

I₀/2 = 4I₁cos²(φ/2)

2I₁ = 4I₁cos²(φ/2)

cos²(φ/2) = 1/2 ⇒ cos(φ/2) = ±1/√2

φ/2 = π/4, 3π/4,... ⇒ φ = π/2, 3π/2,...

Using φ = (2π/λ)Δx:

For φ = π/2: Δx = (λ/2π)(π/2) = λ/4

For φ = 3π/2: Δx = (λ/2π)(3π/2) = 3λ/4

The smallest positive path difference where intensity is I₀/2 is λ/4.

Correct option: (1) λ/4