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

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 \(\sin \theta \approx \tan \theta \approx \frac{x}{D}=\frac{\beta}{3 D}=\frac{\lambda D}{d \times 3 D}=\frac{\lambda}{3 d}\)
\(\begin{aligned} & \therefore \Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} d \sin \theta=\frac{2 \pi}{\lambda} \times \frac{d \times \lambda}{3 d}=\frac{2 \pi}{3}=\Delta \phi \\ & I=4 I_0 \cos ^2\left(\frac{\Delta \phi}{2}\right)=4 I_0 \cos ^2\left(\frac{2 \pi}{3} \times \frac{1}{2}\right) \\ & =4 I_0 \cos ^2\left(\frac{\pi}{3}\right)=4 \times I_0 \times \frac{1}{4}\end{aligned}\)
\(\therefore I=I_0=2 \times 10^{-2} \mathrm{~W} \mathrm{~m}^{-2}\)