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In Young's double slit experiment, the ratio of intensities of light from one slit to the other is 9: 1 . If Im is the maximum intensity, what is the resultant intensity when they interfere at phase difference ?

Options

  1. AIm 9 [1+8 ^2 ( 2 ) ]
  2. BIm 4 [1+3 ^2 ( 2 ) ]
  3. CIm 4 [1+8 ^2 ( 2 ) ]
  4. DIm 2 [4+12 ^2 ( 2 ) ]

Correct answer

B. Im 4 [1+3 ^2 ( 2 ) ]

Step-by-step solution

Let the intensities of the two slits be I₁ and I₂ . Given the ratio I₁ : I₂ = 9 : 1 , we can write I₁ = 9k and I₂ = k for some constant k . The maximum intensity I_m is given by I_m = ( I₁ + I₂ )^2 = ( 9k + k )^2 = (3 k + k )^2 = (4 k )^2 = 16k . Thus, k = I_m 16 . The resultant intensity I at a phase difference is given by I = I₁ + I₂ + 2 I₁ I₂ . Substituting I₁ = 9k and I₂ = k : I = 9k + k + 2 9k^2 = 10k + 6k = 2k(5 + 3 ) . Using the identity = 2 ^2( /2) - 1 : I = 2k(5 + 3(2 ^2( /2) - 1)) = 2k(5 + 6 ^2( /2) - 3)

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