Question

A double slit experiment is immersed in water of refractive index 1.33. The slit separation is 1 mm and the distance between slit and screen is 1.33 m. The slits are illuminated by a light of wavelength \(6300 \, \text{Å}\). The fringe width is

• A. \(4.9 \times 10^{-4} \, \text{m}\)
• B. \(6.3 \times 10^{-4} \, \text{m}\)
• C. \(8.6 \times 10^{-4} \, \text{m}\)
• D. \(5.8 \times 10^{-4} \, \text{m}\)

Hint:

When the entire setup is immersed in a medium, the wavelength decreases by factor μ, but the geometry (d, D) remains the same. The fringe width β ∝ λ, so β becomes β₀/μ. Here β₀ = λ₀ D / d = 6.3e-7 * 1.33 / 1e-3 = 8.379e-4 m, divided by 1.33 gives 6.3e-4 m.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Young’s double slit experiment is immersed in water (refractive index μ = 1.33). Given: slit separation d = 1 mm = 10⁻³ m, screen distance D = 1.33 m, wavelength in air λ₀ = 6300 Å = 6300 × 10⁻¹⁰ m = 6.3 × 10⁻⁷ m. Need fringe width β in water.

Step 2: Key Formula or Approach:
In a medium of refractive index μ, the wavelength becomes λ = λ₀/μ. Fringe width β = λ D / d = (λ₀ D) / (μ d).

Step 3: Detailed Explanation:
\[ β = \frac{6.3 \times 10^{-7} \times 1.33}{1.33 \times 10^{-3}} = \frac{6.3 \times 10^{-7}}{10^{-3}} = 6.3 \times 10^{-4} \, \text{m}. \] (Note: The factor 1.33 cancels because D = 1.33 m exactly.)

Step 4: Final Answer:
Option (B) \(6.3 \times 10^{-4} \, \text{m}\).