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(A) The intensity at any point in an interference pattern is given by $I = I_{max} \cos^2 \left( \frac{\Delta \phi}{2} \right)$,where $I_{max}$ is the

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$0.853$

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(A) The intensity at any point in an interference pattern is given by $I = I_{max} \cos^2 \left( \frac{\Delta \phi}{2} ight)$,where $I_{max}$ is the intensity at the centre of a bright fringe.The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.Given the path difference $\Delta x = \frac{\lambda}{8}$,we calculate the phase difference:$\Delta \phi = \frac{2 \pi}{\lambda} 	imes \frac{\lambda}{8} = \frac{\pi}{4}$.Now,substitute this into the intensity formula:$\frac{I}{I_{max}} = \cos^2 \left( \frac{\pi / 4}{2} ight) = \cos^2 \left( \frac{\pi}{8} ight)$.Using the value $\cos(\frac{\pi}{8}) \approx 0.9239$,we get:$\frac{I}{I_{max}} = (0.9239)^2 \approx 0.853$.

In a double slit experiment,at a certain point on the screen,the path difference between the two interfering waves is $\frac{1}{8}$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:

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