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

• A. 7
• B. 5
• C. 6
• D. 4

Answer 1

The correct option is A 7


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