Path Difference for Half Peak Intensity in Youngs Double Slit Experiment

In Young's double-slit experiment, the resultant intensity $I$ at a point on the screen where the phase difference between the two interfering waves is $\phi$ is given by: $I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right)$

where $I_{\text{max}}$ is the peak intensity.

According to the given condition, the intensity at the point drops to half of the peak intensity: $I = \frac{I_{\text{max}}}{2}$

Substituting this into the intensity formula: $\frac{I_{\text{max}}}{2} = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right)$

$\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{2}$

$\cos\left(\frac{\phi}{2}\right) = \pm\frac{1}{\sqrt{2}}$

The general solution for the phase difference $\phi$ in terms of an integer $n$ is: $\frac{\phi}{2} = (2n + 1)\frac{\pi}{4}$

$\phi = (2n + 1)\frac{\pi}{2}$

The relationship between phase difference $\phi$ and path difference $\Delta x$ for light of wavelength $\lambda$ is: $\phi = \frac{2\pi}{\lambda} \Delta x$

Equating the two expressions for $\phi$: $\frac{2\pi}{\lambda} \Delta x = (2n + 1)\frac{\pi}{2}$

Solving for the path difference $\Delta x$: $\Delta x = (2n + 1)\frac{\lambda}{4}$

Thus, the correct option is (B).