The ratio of intensities at two points $\mathrm{P}$ and $\mathrm{Q}$ on the screen in a Young's double slit experiment where phase difference between two waves of same amplitude are $\pi / 3$ and $\pi / 2$, respectively are

<p>In a Young&#39;s double slit experiment, the intensity at a point on the screen is given by the formula :</p> <p>$I = 4I_0\cos^2\left(\frac{\pi d\sin\theta}{\lambda}\right)$</p> <p>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.</p> <p>The phase difference between the waves from the two slits at a point on the screen is given by :</p> <p>$\Delta \phi = \frac{2\pi d\sin\theta}{\lambda}$</p> <p>For a phase difference of $\pi/3$ between the waves from the two slits at point P, we have :</p> <p>$\frac{\Delta \phi}{\pi} = \frac{2d\sin\theta}{\lambda} = \frac{1}{3}$</p> <p>For a phase difference of $\pi/2$ between the waves from the two slits at point Q, we have :</p> <p>$\frac{\Delta \phi}{\pi} = \frac{2d\sin\theta}{\lambda} = \frac{1}{2}$</p> <p>Solving for $\sin\theta$ in both cases, we get:</p> <p>$\sin\theta_P = \frac{\lambda}{6d},\quad \sin\theta_Q = \frac{\lambda}{4d}$</p> <p>Substituting these values in the formula for intensity, we get :</p> <p>$$\frac{I_P}{I_Q} = \frac{4\cos^2\left(\frac{\pi}{6}\right)}{4\cos^2\left(\frac{\pi}{4}\right)} = \frac{3}{2}$$</p> <p>Therefore, the ratio of intensities at points P and Q is 3 : 2</p>