In a Young's double-slit experiment,the fring | vedclass

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(A) The intensity at any point in a Young's double-slit experiment is given by $I = I_{\max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase differe

Vedclass · Accepted Answer

$\frac{\lambda D}{4d}$

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(A) The intensity at any point in a Young's double-slit experiment 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 maximum intensity,we have $\frac{I_{\max}}{2} = I_{\max} \cos^2(\frac{\phi}{2})$.This simplifies to $\cos^2(\frac{\phi}{2}) = \frac{1}{2}$,which means $\cos(\frac{\phi}{2}) = \frac{1}{\sqrt{2}}$.Therefore,$\frac{\phi}{2} = \frac{\pi}{4}$,which gives the phase difference $\phi = \frac{\pi}{2}$.The path difference $\Delta x$ is related to the phase difference by $\Delta x = \frac{\lambda}{2\pi} \phi$.Substituting $\phi = \frac{\pi}{2}$,we get $\Delta x = \frac{\lambda}{2\pi} 	imes \frac{\pi}{2} = \frac{\lambda}{4}$.The distance $x$ from the central point is given by $x = \frac{\Delta x \cdot D}{d}$.Substituting the value of $\Delta x$,we get $x = \frac{\lambda D}{4d}$.

In a Young's double-slit experiment,the fringe pattern is observed on a screen placed at a distance $D$. The slits are separated by $d$ and are illuminated by light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to half the maximum intensity.

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