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(A) The path difference is given by $\Delta x = \frac{7 \lambda}{4}$.The phase difference $\phi$ is related to the path difference by the formula $\ph

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(A) The path difference is given by $\Delta x = \frac{7 \lambda}{4}$.The phase difference $\phi$ is related to the path difference by the formula $\phi = \frac{2 \pi}{\lambda} \Delta x$.Substituting the value of $\Delta x$: $\phi = \frac{2 \pi}{\lambda} 	imes \frac{7 \lambda}{4} = \frac{7 \pi}{2}$.The intensity $I$ at any point is given by $I = I_{\max} \cos^2\left(\frac{\phi}{2}ight)$.Therefore,the ratio $\frac{I}{I_{\max}} = \cos^2\left(\frac{7 \pi}{2 	imes 2}ight) = \cos^2\left(\frac{7 \pi}{4}ight)$.Using the property $\cos(2 \pi - 	heta) = \cos(	heta)$,we get $\cos^2\left(2 \pi - \frac{\pi}{4}ight) = \cos^2\left(\frac{\pi}{4}ight)$.Since $\cos\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$,we have $\left(\frac{1}{\sqrt{2}}ight)^2 = \frac{1}{2}$.

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