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 :

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

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

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

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

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

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

$$\sin\theta_P = \frac{\lambda}{6d},\quad \sin\theta_Q = \frac{\lambda}{4d}$$

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