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

We need to find the ratio of intensities at points P and Q on the screen in Young’s double slit experiment, where the phase differences are $$\frac{\pi}{3}$$ and $$\frac{\pi}{2}$$ respectively.

In Young’s double slit experiment, when two coherent waves of equal amplitude $$a$$ interfere with a phase difference $$\phi$$, the resultant intensity at that point is given by $$I = 4I_0\cos^2\left(\frac{\phi}{2}\right)$$, where $$I_0$$ is the intensity due to a single slit (i.e., $$I_0 \propto a^2$$). This formula follows from the superposition principle: if the two waves are $$y_1 = a\sin(\omega t)$$ and $$y_2 = a\sin(\omega t + \phi)$$, then the resultant amplitude is $$A = 2a\cos(\phi/2)$$, and since intensity is proportional to the square of amplitude, we obtain $$I = 4I_0\cos^2(\phi/2)$$.

We begin by calculating the intensity at point P where the phase difference is $$\phi_P = \frac{\pi}{3}$$. Substituting into the formula gives $$I_P = 4I_0\cos^2\left(\frac{\pi/3}{2}\right) = 4I_0\cos^2\left(\frac{\pi}{6}\right)$$. We know that $$\cos\left(\frac{\pi}{6}\right) = \cos 30° = \frac{\sqrt{3}}{2}$$, and therefore $$I_P = 4I_0 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 4I_0 \times \frac{3}{4} = 3I_0$$.

Next, we calculate the intensity at point Q where the phase difference is $$\phi_Q = \frac{\pi}{2}$$. Substituting into the same formula yields $$I_Q = 4I_0\cos^2\left(\frac{\pi/2}{2}\right) = 4I_0\cos^2\left(\frac{\pi}{4}\right)$$. Since $$\cos\left(\frac{\pi}{4}\right) = \cos 45° = \frac{1}{\sqrt{2}}$$, it follows that $$I_Q = 4I_0 \times \left(\frac{1}{\sqrt{2}}\right)^2 = 4I_0 \times \frac{1}{2} = 2I_0$$.

Finally, the ratio of the intensities is $$\frac{I_P}{I_Q} = \frac{3I_0}{2I_0} = \frac{3}{2}$$, which shows that $$I_P : I_Q = 3 : 2$$. The correct answer is Option 4: 3 : 2.