Q. Two beams of light having intensities $I\ $and $4I\ $interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\pi/2$ at point A and $\pi$ at point $B$.Then the difference between the resultant intensities at $A$ and $B$ is
A) $2I$
B) $4I$
C) $5I$
D) $7I$
Explanation
(b) Here, $I_{1} = I,\ I_{2} = 4I,\ \theta_{1} = \frac{\pi}{2},\ \theta_{2} = \pi$
$I_{A} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\theta_{1}$
$\ \ \ \ \ = I + 4I + 2\sqrt{1 \times 4I}\cos{\pi/2 =}5I$
$I_{B} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\theta_{2}$
$\ \ \ \ \ = \ I + 4I + 2\sqrt{I \times 4I}\cos\pi$
$= 5I - 4I = I$
$\therefore\ \ I_{A} - I_{B} = 5I - I = 4I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.