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In Young’s double slit experiment the intensity of light at a point on the screen where the path difference is $λ$ is K. The intensity of light at a point where the path difference is $\frac{λ}{3}$ [ $λ$ is the wavelength of light used] is

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K/4

K/3

K/2

answer is A.

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Detailed Solution

$When path difference = λ, then$

$Phase difference Δ ϕ = \frac{2 π}{λ} \times λ = 2 π$

$Intensity I = 4 I_{0} \cos^{2} (\frac{Δ ϕ}{2})$

$K = 4 I_{0} \cos^{2} (\frac{2 π}{2}) = 4 I_{0}$

$When path difference = \frac{λ}{3}, then$

$Phase difference Δ ϕ = \frac{2 π}{λ} \times \frac{λ}{3} = \frac{2 π}{3}$

$Intensity I = 4 I_{0} \cos^{2} (\frac{Δ ϕ}{2})$

$I' = 4 I_{0} \cos^{2} (\frac{π}{3}) = I_{0} = \frac{K}{4}$

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