In YDSE,a=2 , mm,D=2 , m,and lambda=500 , nm. | vedclass

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(D) The intensity at any point in $YDSE$ is given by $I = I_{0} \cos^{2} \left( \frac{\phi}{2} \right)$,where $I_{0}$ is the maximum intensity and $\p

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(D) The intensity at any point in $YDSE$ is given by $I = I_{0} \cos^{2} \left( \frac{\phi}{2} ight)$,where $I_{0}$ is the maximum intensity and $\phi$ is the phase difference.Given that the intensity $I = \frac{I_{0}}{2}$,we have:$\frac{I_{0}}{2} = I_{0} \cos^{2} \left( \frac{\phi}{2} ight)$$\cos^{2} \left( \frac{\phi}{2} ight) = \frac{1}{2}$$\cos \left( \frac{\phi}{2} ight) = \frac{1}{\sqrt{2}}$$\frac{\phi}{2} = \frac{\pi}{4} \implies \phi = \frac{\pi}{2}$.The path difference $\Delta x$ is related to the phase difference $\phi$ by $\Delta x = \frac{\lambda}{2\pi} \phi$.Substituting $\phi = \frac{\pi}{2}$,we get $\Delta x = \frac{\lambda}{2\pi} 	imes \frac{\pi}{2} = \frac{\lambda}{4}$.The distance $y$ from the central maxima is given by $y = \frac{\Delta x D}{a}$.Substituting the values $\Delta x = \frac{\lambda}{4}$,$D = 2 \, m$,$a = 2 	imes 10^{-3} \, m$,and $\lambda = 500 	imes 10^{-9} \, m$:$y = \frac{\lambda D}{4a} = \frac{500 	imes 10^{-9} 	imes 2}{4 	imes 2 	imes 10^{-3}}$$y = \frac{1000 	imes 10^{-9}}{8 	imes 10^{-3}} = 125 	imes 10^{-6} \, m = 125 \, \mu m$.

In $YDSE$,$a=2 \, mm$,$D=2 \, m$,and $\lambda=500 \, nm$. Find the distance of the point on the screen from the central maxima where the intensity becomes $50 \%$ of the central maxima (in $\mu m$).

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