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(B) The resultant intensity $I_R$ in an interference pattern is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase diffe

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$K/2$

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(B) The resultant intensity $I_R$ in an interference pattern is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference.Since it is a double-slit experiment,let $I_1 = I_2 = I_0$. Then $I_R = 2I_0 + 2I_0 \cos \phi = 4I_0 \cos^2(\phi/2)$.For a path difference $\Delta x = \lambda$,the phase difference $\phi = (2\pi / \lambda) 	imes \Delta x = (2\pi / \lambda) 	imes \lambda = 2\pi$.The intensity $K = 4I_0 \cos^2(2\pi / 2) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$.Now,for a path difference $\Delta x' = \lambda / 4$,the phase difference $\phi' = (2\pi / \lambda) 	imes (\lambda / 4) = \pi / 2$.The new intensity $I' = 4I_0 \cos^2(\phi' / 2) = 4I_0 \cos^2(\pi / 4) = 4I_0 (1 / \sqrt{2})^2 = 4I_0 (1 / 2) = 2I_0$.Since $K = 4I_0$,we have $2I_0 = K / 2$.Therefore,the intensity is $K / 2$.

In a Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $K$. What will be the intensity at a point where the path difference is $\lambda / 4$?

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