The two light beams having intensities $$I$$ and $$9I$$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $$\frac{\pi}{2}$$ at point $$P$$ and $$\pi$$ at point $$Q$$. Then the difference between the resultant intensities at $$P$$ and $$Q$$ will be :

We are given two beams with intensities $$I$$ and $$9I$$, and we need to find the difference in their resultant intensities at points $$P$$ and $$Q$$, where the phase difference is $$\frac{\pi}{2}$$ and $$\pi$$ respectively.

Recall that when two waves of intensities $$I_1$$ and $$I_2$$ interfere with a phase difference $$\delta$$, the resultant intensity is given by:
$$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta$$

At point $$P$$, since $$\delta = \frac{\pi}{2}$$, one finds
$$I_P = I + 9I + 2\sqrt{I \times 9I}\cos\frac{\pi}{2}$$
Because $$\cos\frac{\pi}{2}=0$$, it follows that
$$I_P = 10I + 0 = 10I$$

Similarly, at point $$Q$$ where $$\delta = \pi$$, we have
$$I_Q = I + 9I + 2\sqrt{I \times 9I}\cos\pi$$
Using $$\cos\pi = -1$$ and $$\sqrt{9I^2} = 3I$$ yields
$$I_Q = 10I + 2(3I)(-1) = 10I - 6I = 4I$$

Therefore, the desired difference is
$$I_P - I_Q = 10I - 4I = 6I$$

The correct answer is Option B: $$6I$$.