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(B) The intensity at any point in the interference pattern is given by $I = I_0 + I_0 + 2I_0 \cos \phi = 2I_0(1 + \cos \phi)$,where $I_0$ is the inten

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$(2n + 1)\frac{\lambda}{4}$

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(B) The intensity at any point in the interference pattern is given by $I = I_0 + I_0 + 2I_0 \cos \phi = 2I_0(1 + \cos \phi)$,where $I_0$ is the intensity of each individual slit.Peak intensity $I_{max}$ occurs when $\cos \phi = 1$,so $I_{max} = 4I_0$.We are looking for the point where the intensity is half the peak intensity: $I = \frac{I_{max}}{2} = 2I_0$.Equating the two expressions: $2I_0(1 + \cos \phi) = 2I_0 \Rightarrow 1 + \cos \phi = 1 \Rightarrow \cos \phi = 0$.This implies the phase difference $\phi = (2n + 1)\frac{\pi}{2}$.Since the phase difference $\phi = \frac{2\pi}{\lambda} \Delta x$,we have $\frac{2\pi}{\lambda} \Delta x = (2n + 1)\frac{\pi}{2}$.Solving for path difference $\Delta x$: $\Delta x = (2n + 1)\frac{\lambda}{4}$.

In the Young's double slit experiment using a monochromatic light of wavelength $\lambda$,the path difference (in terms of an integer $n$) corresponding to any point having half the peak intensity is:

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