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(B) The intensity $I$ at any point in an interference pattern is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.Phase diff

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$k/2$

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(B) The intensity $I$ at any point in an interference pattern is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.For path difference $\Delta x = \lambda$,the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.Given intensity $I = k$ at $\Delta x = \lambda$,we have $k = I_{max} \cos^2(2\pi/2) = I_{max} \cos^2(\pi) = I_{max}(1)^2 = I_{max}$. So,$I_{max} = k$.Now,for path difference $\Delta x = \lambda/4$,the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.The intensity at this point is $I = I_{max} \cos^2(\phi/2) = k \cos^2(\pi/4) = k \cdot (1/\sqrt{2})^2 = k/2$.

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

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