Q. Two waves of intensity $I$ undergo Interference. The maximum intensity obtained is
A) $I/2$
B) $I$
C) $2I$
D) $4I$
Explanation
(d) For maximum intensity $\phi = 0{^\circ}$
$\therefore I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\phi = I + I + 2\sqrt{I\ I}\cos{0{^\circ} = 4I}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.