Question

In a YDSE, the light of wavelength λ = 5000AA is used, which emerges in phase from two slits at distance d = 3×10⁻7m apart. A transparent sheet of thickness t = 1.5×10⁻7m and refractive index μ = 1.17 is placed over one of the slits. What is the new angular position of the central maxima of the interference pattern from the centre of the screen? Find the value of y. 

• A. \(4.9^\circ\) and \( \dfrac{D(\mu-1)t}{2d} \)
• B. \(4.9^\circ\) and \( \dfrac{D(\mu-1)t}{d} \)
• C. \(3.9^\circ\) and \( \dfrac{D(\mu+1)t}{2d} \)
• D. 2.9^∘ and (D(μ+1)t)/(d)

Hint:

In YDSE, inserting a thin sheet causes a shift of central fringe without changing fringe width.

Answer 1

The Correct Option is A

Step 1: Optical path difference introduced by the sheet: Δ = (μ-1)t
Step 2: Condition for central maxima shift: dsinθ = (μ-1)t
Step 3: Substituting values: sinθ = frac(1.17-1)×1.5×10⁻73×10⁻7 ≈ 0.085 ⟹ θ ≈ 4.9^∘
Step 4: Linear shift on screen: y = (D(μ-1)t)/(2d)