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

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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 $I = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.Given that the intensity is half of the peak intensity,$I = \frac{I_{max}}{2}$.Substituting this into the equation: $\frac{I_{max}}{2} = I_{max} \cos^2(\frac{\phi}{2})$.This simplifies to $\cos^2(\frac{\phi}{2}) = \frac{1}{2}$,which means $\cos(\frac{\phi}{2}) = \pm \frac{1}{\sqrt{2}}$.Thus,$\frac{\phi}{2} = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \dots$,which implies $\phi = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$.In general,$\phi = (2n+1) \frac{\pi}{2}$.Since the path difference $\Delta x = \frac{\lambda}{2\pi} \phi$,we substitute $\phi$:$\Delta x = \frac{\lambda}{2\pi} 	imes (2n+1) \frac{\pi}{2} = (2n+1) \frac{\lambda}{4}$.

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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