Interference fringes are produced on a screen by using two light sources of intensities ‘I’ and ‘9I’. The phase difference between the beams is \frac{\pi}{2} at point \mathrm{P} and \pi at point \mathrm{Q} on the screen. The difference between the resultant intensities at point \mathrm{P} and \mathrm{Q} is
A) 2 \mathrm{I}
B) 4 \mathrm{I}
C) 6 \mathrm{I}
D) 8 \mathrm{I}
Interference fringes are produced on a screen by using two light sources of intensities ‘I’ and ‘9I’. The phase difference between the beams is \frac{\pi}{2} at point \mathrm{P} and \pi at point \mathrm{Q} on the screen. The difference between the resultant intensities at point \mathrm{P} and \mathrm{Q} is
A) 2 \mathrm{I}
B) 4 \mathrm{I}
C) 6 \mathrm{I}
D) 8 \mathrm{I}

Answer is (C)
At point \mathrm{P}, resultant intensity of interfering wave is
\left(\mathrm{I}_{\mathrm{p}}\right)_{\mathrm{res}}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}} \cos \theta
For \theta=\frac{\pi}{2},\left(I_{\rho}\right)_{\text {res }}=I+9 I=10 \mathrm{I}
At point Q, resultant intensity of interfering wave is
\left(\mathrm{l}_{\mathrm{Q}}\right)_{\mathrm{res}}=10 \mathrm{I}-2 \cdot \sqrt{9 \mathrm{I}^{2}}=10 \mathrm{I}-6 \mathrm{I}=4 \mathrm{I} \quad for \theta=\pi
\therefore \Delta \mathrm{I}=\left(\mathrm{I}_{\mathrm{p}}\right)_{\mathrm{res}}-\left(\mathrm{I}_{\mathrm{g}}\right)_{\mathrm{res}}=10 \mathrm{I}-10 \mathrm{I}+6 \mathrm{I}=6 \mathrm{I}