In the Young's double slit experiment, the intensity at a point on the screen, where the path difference is…
In the Young's double slit experiment, the intensity at a point on the screen, where the path difference is $\lambda(\lambda=$ wavelength $)$ is $\beta$. The intensity at a point where the path difference is $\lambda / 3$, will be $\left.\cos \frac{\pi}{3}=1 / 2\right]$
$\beta$
$\beta / 2$
$\frac{\beta}{4}$
$\beta / 8$
Solution
The intensity is given by
$\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}$ Maximum intensity $\beta=4 \mathrm{I}_0$ when $\phi=0$ when path difference is $\frac{\lambda}{3}, \phi=\frac{2 \pi}{3}$
$\therefore \quad I=\beta \cos ^2 \frac{2 \pi}{3}=\beta\left(-\frac{1}{2}\right)^2=\frac{\beta}{4}$