In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is 7 λ 4 . The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is:

Question

In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is 7 λ 4 . The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is:, 1 2, 3 4, 1 3, 1 4

MARKS App · Accepted Answer

Given the path difference is ∆ x = 7 λ 4 Hence, the phase difference can be found as follows: ϕ = 2 π λ ∆ x = 2 π λ × 7 λ 4 = 7 π 2 The formula to calculate the intensity is given by I = I max cos 2 ϕ 2 . . . 1 From equation (1), it follows that I I max = cos 2 ϕ 2 = cos 2 7 π 2 × 2 = cos 2 7 π 4 = cos 2 2 π - π 4 = cos 2 π 4 = 1 2

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