In Young's double-slit experiment,frac{d}{D} | vedclass

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(A) The intensity at any point $P$ is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_0$ is the intensity of each individual source.Given that the intensi

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(A) The intensity at any point $P$ is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_0$ is the intensity of each individual source.Given that the intensity at $P$ is $I_0$,we have $I_0 = 4I_0 \cos^2(\phi/2)$.This simplifies to $\cos^2(\phi/2) = 1/4$,so $\cos(\phi/2) = 1/2$.Therefore,$\phi/2 = \pi/3$,which gives the phase difference $\phi = 2\pi/3$.The path difference $\Delta$ is related to the phase difference by $\Delta = \frac{\lambda}{2\pi} \phi$.Substituting $\phi = 2\pi/3$,we get $\Delta = \frac{\lambda}{2\pi} 	imes \frac{2\pi}{3} = \frac{\lambda}{3}$.Also,the path difference is $\Delta = \frac{xd}{D}$.Equating the two,$\frac{xd}{D} = \frac{\lambda}{3}$.Given $\frac{d}{D} = 10^{-4}$ and $\lambda = 6000 	imes 10^{-10} \, m$,we have $x 	imes 10^{-4} = \frac{6000 	imes 10^{-10}}{3}$.$x = 2000 	imes 10^{-6} \, m = 2 	imes 10^{-3} \, m = 2 \, mm$.

In Young's double-slit experiment,$\frac{d}{D} = 10^{-4}$. The intensity at point $P$ on the screen is equal to the intensity of one of the sources. If the wavelength of the light used is $\lambda = 6000 \, \mathring{A}$,what is the distance of point $P$ from the central bright fringe in $mm$?

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