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In the Young's double slit experiment, the intensity of light passing through each of
the two double slits is $2 \times 10^{-2} \mathrm{Wm}^{-2}$. The screen-slit distance is very large in comparison with slit-slit distance. The fringe width is $\beta$. The distance between the central maximum and a point $P$ on the screen is $x=\frac{\beta}{3}$. Then, the total light intensity at the point is
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the two double slits is $2 \times 10^{-2} \mathrm{Wm}^{-2}$. The screen-slit distance is very large in comparison with slit-slit distance. The fringe width is $\beta$. The distance between the central maximum and a point $P$ on the screen is $x=\frac{\beta}{3}$. Then, the total light intensity at the point is
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Verified Answer
The correct answer is:
$2 \times 10^{-2} \mathrm{Wm}^{-2}$
Given, $x=\frac{\beta}{3}$
$I_0=2 \times 10^{-2} \mathrm{Wm}^{-2}$
for YDSE where $D \gg d, \Delta x=d \sin \theta$
Also $\sin \theta \approx \tan \theta \approx \frac{x}{D}=\frac{\beta}{3 D}=\frac{\lambda D}{d \times 3 D}=\frac{\lambda}{3 d}$
$\begin{aligned} & \therefore \Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} d \sin \theta=\frac{2 \pi}{\lambda} \times \frac{d \times \lambda}{3 d}=\frac{2 \pi}{3}=\Delta \phi \\ & I=4 I_0 \cos ^2\left(\frac{\Delta \phi}{2}\right)=4 I_0 \cos ^2\left(\frac{2 \pi}{3} \times \frac{1}{2}\right) \\ & =4 I_0 \cos ^2\left(\frac{\pi}{3}\right)=4 \times I_0 \times \frac{1}{4}\end{aligned}$
$\therefore I=I_0=2 \times 10^{-2} \mathrm{~W} \mathrm{~m}^{-2}$
$I_0=2 \times 10^{-2} \mathrm{Wm}^{-2}$
for YDSE where $D \gg d, \Delta x=d \sin \theta$
Also $\sin \theta \approx \tan \theta \approx \frac{x}{D}=\frac{\beta}{3 D}=\frac{\lambda D}{d \times 3 D}=\frac{\lambda}{3 d}$
$\begin{aligned} & \therefore \Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} d \sin \theta=\frac{2 \pi}{\lambda} \times \frac{d \times \lambda}{3 d}=\frac{2 \pi}{3}=\Delta \phi \\ & I=4 I_0 \cos ^2\left(\frac{\Delta \phi}{2}\right)=4 I_0 \cos ^2\left(\frac{2 \pi}{3} \times \frac{1}{2}\right) \\ & =4 I_0 \cos ^2\left(\frac{\pi}{3}\right)=4 \times I_0 \times \frac{1}{4}\end{aligned}$
$\therefore I=I_0=2 \times 10^{-2} \mathrm{~W} \mathrm{~m}^{-2}$
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