In YDSE,the intensity of the central bright f | vedclass

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(B) The intensity at any point in $YDSE$ is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $I_{max}$ is the intensity of the central bright fring

Vedclass · Accepted Answer

$6 \, mW/m^2$

Vedclass · Answer

(B) The intensity at any point in $YDSE$ is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $I_{max}$ is the intensity of the central bright fringe.Given $I_{max} = 8 \, mW/m^2$ and path difference $\Delta x = \frac{\lambda}{6}$.The phase difference $\phi$ is calculated as $\phi = \frac{2\pi}{\lambda} 	imes \Delta x = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3}$.Substituting the values into the intensity formula:$I = 8 \cos^2(\frac{\pi/3}{2}) = 8 \cos^2(\frac{\pi}{6})$.Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we have $\cos^2(\frac{\pi}{6}) = \frac{3}{4}$.Therefore,$I = 8 	imes \frac{3}{4} = 6 \, mW/m^2$.

In $YDSE$,the intensity of the central bright fringe is $8 \, mW/m^2$. What will be the intensity at a path difference of $\frac{\lambda}{6}$?

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