Question

In standard $YDSE$ ($D>>d>>\lambda$) with identical slits $S_1$ and $S_2$, light reaching at point $A$ on the screen opposite to slit $S_2$ has an intensity $I$. It was also found that when only one of the two slits $S_1$ and $S_2$ was illuminated by same light beam, the intensity at $A$ is still $I$. Now when a third slit $S_3$ of four times the slit width(of $S_1$), is made as shown $(S_1{S}_2=S_2{S}_3=d)$and all the three slits are illuminated, then the intensity of light reaching $A$ is $nI$, where $n$ is

Answer 1

$I$ is the intensity of light from source. It reaches slits $S_1$ and $S_2$ which are at a phase difference of ${120}^{\circ }$ as the intensity of light reaching $A$ is $I$.Since the width of slit $S_3$ is 4 times that of the other two slits, intensity coming out of $S_3$ is $4I$. Also, $S_3$ & $S_1$ are in phase with each other. The resultant phase diagram is

$\theta ={120}^{\circ }$
$I_{total}=I+9I+2\sqrt{9I^2}cos\theta$
$=I+9I-3I$
$I_{total}=7I$