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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Light Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Light Waves

Two transparent slabs of equal thickness but distinct refractive...

Two transparent slabs of equal thickness but distinct refractive indices μ1\mu_1 and μ2\mu_2 are joined side by side to create a composite slab. This slab is positioned immediately after the double slit in a Young's experiment, ensuring that light from one slit passes through one material and light from the other slit passes through the other material. What must be the minimum thickness of the slab to produce a minimum at the point P0P_0, which is equidistant from the slits?

Options

  1. Aλ2∣μ1−μ2∣\dfrac{\lambda}{2|\mu_1 - \mu_2|}
  2. Bλ4∣μ1−μ2∣\dfrac{\lambda}{4|\mu_1 - \mu_2|}
  3. C2λ∣μ1−μ2∣\dfrac{2\lambda}{|\mu_1 - \mu_2|}
  4. Dλ∣μ1−μ2∣\dfrac{\lambda}{|\mu_1 - \mu_2|}

Correct answer

A. λ2∣μ1−μ2∣\dfrac{\lambda}{2|\mu_1 - \mu_2|}

Step-by-step solution

Assume the thickness of the slabs is tt.

The optical path length for light from the first slit to P0P_0 is S1P0+(μ1−1)tS_1P_0 + (\mu_1 - 1)t.

The optical path length for light from the second slit to P0P_0 is S2P0+(μ2−1)tS_2P_0 + (\mu_2 - 1)t.

Since P0P_0 is equidistant from the slits, S1P0=S2P0S_1P_0 = S_2P_0.

The path difference at P0P_0 is:

Δx=∣(S1P0+(μ1−1)t)−(S2P0+(μ2−1)t)∣=∣μ1−μ2∣t\Delta x = |(S_1P_0 + (\mu_1 - 1)t) - (S_2P_0 + (\mu_2 - 1)t)| = |\mu_1 - \mu_2|t

For a minimum intensity at P0P_0, the path difference must be an odd multiple of λ2\dfrac{\lambda}{2}:

Δx=(2n−1)λ2\Delta x = (2n - 1)\dfrac{\lambda}{2}

For the minimum thickness, taking n=1n = 1 gives Δx=λ2\Delta x = \dfrac{\lambda}{2}.

Equating the path differences:

∣μ1−μ2∣t=λ2|\mu_1 - \mu_2|t = \dfrac{\lambda}{2}

t=λ2∣μ1−μ2∣t = \dfrac{\lambda}{2|\mu_1 - \mu_2|}
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