JEE Main 2023 — Wave Optics Question with Solution

In a Young's double slit experiment, the intensity at a point on the screen is given by the formula :

$$I=4I_0{\cos }^2\left(\frac{\pi d\sin \theta }{\lambda }\right)$$

where $I_0$ is the intensity at the center of the pattern, $d$ is the distance between the slits, $\theta$ is the angle between the line joining the point to the center of the pattern and the line passing through the center of the pattern and the slits, and $\lambda$ is the wavelength of the light used.

The phase difference between the waves from the two slits at a point on the screen is given by :

$$\Delta \varphi =\frac{2\pi d\sin \theta }{\lambda }$$

For a phase difference of $\pi /3$ between the waves from the two slits at point P, we have :

$$\frac{\Delta \varphi }{\pi }=\frac{2d\sin \theta }{\lambda }=\frac{1}{3}$$

For a phase difference of $\pi /2$ between the waves from the two slits at point Q, we have :

$$\frac{\Delta \varphi }{\pi }=\frac{2d\sin \theta }{\lambda }=\frac{1}{2}$$

Solving for $\sin \theta$ in both cases, we get:

$$\sin {\theta }_P=\frac{\lambda }{6d},\sin {\theta }_Q=\frac{\lambda }{4d}$$

Substituting these values in the formula for intensity, we get :

$$\frac{I_P}{I_Q}=\frac{4{\cos }^2\left(\frac{\pi }{6}\right)}{4{\cos }^2\left(\frac{\pi }{4}\right)}=\frac{3}{2}$$

Therefore, the ratio of intensities at points P and Q is 3 : 2