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(D) The intensity in a double-slit experiment is given by $I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference.Phase difference $\phi$ is re

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$K/2$

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(D) The intensity in a double-slit experiment is given by $I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference.Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = (2\pi/\lambda) \Delta x$.Case $1$: Path difference $\Delta x = \lambda$. Then $\phi = (2\pi/\lambda) 	imes \lambda = 2\pi$.Intensity $I = 4I_0 \cos^2(2\pi/2) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0 = K$.Case $2$: Path difference $\Delta x = \lambda/4$. Then $\phi = (2\pi/\lambda) 	imes (\lambda/4) = \pi/2$.Intensity $I' = 4I_0 \cos^2((\pi/2)/2) = 4I_0 \cos^2(\pi/4) = 4I_0 (1/\sqrt{2})^2 = 4I_0(1/2) = 2I_0$.Since $K = 4I_0$,then $2I_0 = K/2$.Therefore,the intensity at the point is $K/2$.

In a Young's double-slit experiment,the intensity at a point on the screen where the path difference is $\lambda$ is $K$. Find the intensity at a point where the path difference is $\lambda/4$.

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