1. Intensity in Young's double slit experiment

If two coherent waves of the same amplitude interfere with phase difference $\phi$, then the resultant intensity is

$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos \phi$

Since the two waves have the same amplitude, their individual intensities are equal:

$I_1 = I_2 = I_0$

So,

$I = 2I_0(1+\cos\phi)$

Using the identity

$1+\cos\phi = 2\cos^2\left(\frac{\phi}{2}\right),$

we can also write

$I = 4I_0\cos^2\left(\frac{\phi}{2}\right)$

2. Intensity at point $P$

At point $P$, phase difference is

$\phi_P = \frac{\pi}{3}$

Hence,

$I_P = 2I_0\left(1+\cos\frac{\pi}{3}\right)$

Since

$\cos\frac{\pi}{3} = \frac{1}{2},$

we get

$I_P = 2I_0\left(1+\frac{1}{2}\right) = 2I_0\cdot \frac{3}{2} = 3I_0$

3. Intensity at point $Q$

At point $Q$, phase difference is

$\phi_Q = \frac{\pi}{2}$

Hence,

$I_Q = 2I_0\left(1+\cos\frac{\pi}{2}\right)$

Since

$\cos\frac{\pi}{2} = 0,$

we get

$I_Q = 2I_0(1+0) = 2I_0$

4. Required ratio

Therefore,

$I_P : I_Q = 3I_0 : 2I_0 = 3:2$

5. Option check

• A: $2:3$ ❌
• B: $1:3$ ❌
• C: $3:1$ ❌
• D: $3:2$ ✅

So the correct option is D.