Question

Two beams of light having intensity $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi }{2}$ at a point $A$ and $2\pi$ at point $B$. The difference between the resultant intensities at $B$ and $A$ will be,

• A. $2I$
• B. $3I$
• C. $4I$
• D. $5I$

Answer 1

The correct option is C $4I$

Given:
$I_1=I$
$I_2=4I$
Phase difference at $A$, ${\varphi }_A=\frac{\pi }{2}$
Phase difference at $B$, ${\varphi }_B=2\pi$

The resultant intensity due to interference is given by,
$I_r=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

Now, at point $A$,
$I_A=I+4I+2\sqrt{I\times 4I}\cos \frac{\pi }{2}$
$\therefore I_A=5I$

At point $B$,
$I_B=I+4I+2\sqrt{I\times 4I}\cos 2\pi$
$\therefore I_B=9I$

So, $I_B-I_A=9I-5I=4I$

Hence, $\left(C\right)$ is the correct answer.