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MHT CET20244 May 2024Evening ShiftPhysicsWave OpticsActual

A parallel beam of light of intensity I₀ is incident on a glass plate, 25 % of light is reflected by upper surface and 50 % of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is

Options

  1. A[ 1 2 + 3 8 1 2 - 3 8 ]^2
  2. B[ 1 4 + 3 8 1 2 - 3 8 ]^2
  3. C5 8
  4. D8 5

Correct answer

A. [ 1 2 + 3 8 1 2 - 3 8 ]^2

Step-by-step solution

Given that, 25 % of total intensity of incident light is reflected from upper surface. This implies, if intensity of incident light is I ₀ , the intensity of light reaching the lower surface of plate will be 3 4 I ₀ . As 50 % of this intensity is reflected, the final intensity of light emerging from glass plate will be 3 8 I ₀ . aligned I₁ & = I₀ 4 & I₂= 3 8 I₀ aligned Now, I_ I_ = ( I₁ + I₂ )^2 ( I₁ - I₂ )^2 = ( 1 2 + 3 8 1 2 - 3 8 )^2

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