Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{4}$ '…
Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{4}$ ' is ' $I$ '. If the maximum intensity is $\mathrm{I}_0$ then the ration $\frac{\mathrm{I}_0}{\mathrm{I}}$ is $\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}=\sin 45^{\circ}\right)$
$2: 1$
$1: 4$
$1: 2$
$4: 1$
Solution
For a path difference of $f$, the intensity at point is,
$I=I_0 \cos ^2 \frac{\phi}{2}$
where, $\mathrm{I}_0$ is the maximum intensity
Here,
$\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$
Thus, $I=I_0 \cos ^2\left(\frac{\pi}{4}\right)=\frac{I_0}{2}$