Question

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 $\frac{\pi }{2}$ at point $A$ and $2\pi$ at point $B$. Then the difference between the resultant intensities at $B$ and $A$ will be

• A. $4I$
• B. $3I$
• C. $I$
• D. $5I$

Answer 1

The correct option is A $4I$

Resultant amplitude of interference
$A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$
Intensity $I\propto A^2$
$\therefore$ resultant intensity is given by
$I_f=I_1+I_2+2\sqrt{I_1{I}_2\cos \varphi }$
where $\varphi$ is the phase difference between the two waves at the point.
At point A,
$I_A=I+4I+2\sqrt{I\times 4I\times \cos \frac{\pi }{2}}$
$=5I$
At point B,
$I_A=I+4I+2\sqrt{I\times 4I\times \cos 2\pi }$
$=9I$

$\therefore I_B-I_A=4I$