In Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{4}$…
In Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{4}$ [ $\lambda$ is wavelength of light used] is ' $I$ '. If ' $\mathrm{I}_0$ ' is the maximum intensity, then $\frac{\mathrm{I}}{\mathrm{I}_0}$ is equal to
$3: 2$
$2: 3$
$3: 4$
$1: 2$
Solution
If I' is the intensity of each wave, then resultant intensity is given by
$\mathrm{I}=4 \mathrm{I} \cos ^2 \frac{\phi}{2}$
I will have maximum value when $\cos ^2 \frac{\phi}{2}=1$
$\therefore$ maximum intensity, $\mathrm{I}_0=4 \mathrm{I}$
When path difference is $\frac{\lambda}{4}$, the phase difference, $\phi=\frac{\pi}{2}$
The resultant intensity, $I=4 I^{\prime} \cos ^2 \frac{\pi}{4}=4 I^{\prime} \times \frac{1}{2}=2 I^{\prime}$
$\therefore \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{1}{2}$