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 $\pi/2$ 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. $7I$

Answer 1

Answer: (B)

Here, $I_{1} = I, I_{2}=4I, \phi_{1} = \frac{\pi}{2}, \phi_{2} = \pi $
$ I_{A} = I_{1}+I_{2}+2\sqrt{I_{1}I_{2}} cos\phi_{1} $
$= I + 4I +2\sqrt{I\times4I} cos \frac{\pi}{2} = 5I $
$ I_{B} = I_{1}+I_{2} +2\sqrt{I_{1}I_{2}}cos \phi_{2} $
$ = I +4I +2\sqrt{I \times4I} cos \pi $
$= 5I -4I = I $
$\therefore I_{A}-I_{B} = 5I - I = 4I $