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In a Young's double slit experiment, the maximum intensity of light on the screen is I₀ . Match the path differences given in List-I with their corresponding intensities given in List-II. List-I (Path Difference) List-II (Intensity) (A) 6 (I) 0 (B) 4 (II) I₀ 4 (C) 3 (III) I₀ 2 (D) 2 (IV) 3I₀ 4 Choose the correct answer from the options given below:

Options

  1. A(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  2. B(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  3. C(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  4. D(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Correct answer

D. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Step-by-step solution

Phase difference is related to path difference x by = 2 x . The intensity I at a point on the screen is given by I = I₀ ^2 ( 2 ) . For (A) x = 6 , = 3 . I = I₀ ^2 ( 6 ) = I₀ ( 3 2 )^2 = 3I₀ 4 . This matches (IV). For (B) x = 4 , = 2 . I = I₀ ^2 ( 4 ) = I₀ ( 1 2 )^2 = I₀ 2 . This matches (III). For (C) x = 3 , = 2 3 . I = I₀ ^2 ( 3 ) = I₀ ( 1 2 )^2 = I₀ 4 . This matches (II). For (D) x = 2 , = . I = I₀ ^2 ( 2 ) = 0 . This matches (I). Therefore, the correct matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I). Answer:

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