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(B) The intensity at any point on the screen is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_0$ is the intensity of each individual wave and $\phi$ is

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$K/2$

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(B) The intensity at any point on the screen is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference.The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = (2\pi / \lambda) 	imes \Delta x$.Case $1$: When path difference $\Delta x = \lambda$,the phase difference is $\phi = (2\pi / \lambda) 	imes \lambda = 2\pi$.Substituting this into the intensity formula: $I = 4I_0 \cos^2(2\pi / 2) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$. Given this intensity is $K$,we have $K = 4I_0$.Case $2$: When path difference $\Delta x = \lambda / 4$,the phase difference is $\phi = (2\pi / \lambda) 	imes (\lambda / 4) = \pi / 2$.Substituting this into the intensity formula: $I' = 4I_0 \cos^2((\pi / 2) / 2) = 4I_0 \cos^2(\pi / 4) = 4I_0 (1 / \sqrt{2})^2 = 4I_0 (1 / 2) = 2I_0$.Since $K = 4I_0$,then $2I_0 = K / 2$. Therefore,the intensity is $K / 2$.

In the Young's double slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\lambda / 4$ will be:

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