Interference fringes are produced on the screen by using two light sources of intensities I and 9I. The phase difference between the beams is $\pi / 2$ at point $P$ and $\pi$ at point $Q$ on the screen. The difference between the resultant intensities at points $P$ and $Q$ is $\left(\cos 90^{\circ}=0, \cos \pi=-1\right)$

The interference of two light sources with intensities $I_1$ and $I_2$ and phase difference $\phi$ produces a resultant intensity given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi$.

With $I_1 = I$ and $I_2 = 9I$, the resultant at point P where $\phi_P = \frac{\pi}{2}$ is $I_P = I + 9I + 2\sqrt{I \cdot 9I} \cos\frac{\pi}{2} = 10I$ since $\cos\frac{\pi}{2} = 0$.

At point Q where $\phi_Q = \pi$, $\cos\pi = -1$, so $I_Q = I + 9I + 2\sqrt{I \cdot 9I}(-1) = 10I - 6I = 4I$.

The difference in intensities is $I_P - I_Q = 10I - 4I = \boxed{6I}$.