Question

In the arrangement shown in figure, slits $S_1$ and $S_4$ are having a variable separation Z. Point O on the screen is at the common perpendicular bisector of $S_1{S}_2$ and $S_3{S}_4$.
When $Z=\frac{\lambda D}{2d^{\prime }}$ the intensity measured at O is $I_0$. The intensity at O when $Z=\frac{2\lambda D}{d}$ is

• A. $I_0$
• B. $2I_0$
• C. $3I_0$
• D. $4I_0$

Answer 1

The correct option is B $2I_0$


Intensity at $I_{S3}=I^{\prime }+I^{\prime }+2\sqrt{I^{\prime }}\sqrt{I^{\prime }}cos\theta$ ...(1)
here,$\theta =$phase difference=$k\times \Delta x$
where$k=\frac{2\pi }{\lambda },\Delta x=\frac{\lambda D}{4d}$

then,
$\theta =\frac{2\pi }{\lambda }\times \frac{d}{D}\times \frac{\lambda D}{4d}\theta =\frac{\pi }{2}$
so, $I_{S3}=I^{\prime }+I^{\prime }+2\sqrt{I^{\prime }}\sqrt{I^{\prime }}cos\frac{\pi }{2}=2I^{\prime }$

in the given confriguration $S_3=S_4$, because of coherent sources.
so, $I_{S4}=2I^{\prime }$

resultant intensity from $S_3$ and $S_4$
$I_0=I_{S3}+I_{S4}+2\sqrt{I_{S3}}\sqrt{I_{S3}}=2I^{\prime }+2I^{\prime }+2\sqrt{2I^{\prime }}\sqrt{2I^{\prime }}=8I^{\prime }.....\left(2\right)$

Case-2:
$Z_{initial}=\frac{\lambda d}{2D}$
$Z_{final}=\frac{2\lambda d}{D}$
so,by the factor $4$ the intensity changes.

hence, new phase difference,$\theta =4\times \frac{\pi }{2}=2\pi$
Intensity from source $I_{S3}=I^{\prime }+I^{\prime }+2\sqrt{I^{\prime }}\sqrt{I^{\prime }}cos2\pi I_{S3}=4I^{\prime }$
so, as $I_{S4}=4I^{\prime }$
Noe the resultant intensity:
$I_0=I_{S3}+I_{S4}+2\sqrt{I_{S3}}\sqrt{I_{S3}}=4I^{\prime }+4I^{\prime }+2\sqrt{4I^{\prime }}\sqrt{4I^{\prime }}=16I^{\prime }....\left(3\right)$

On comparing eq. (2) and eq.(3),
$I_{initial}=8I^{\prime }=I_0$
$I_{initial}=16I^{\prime }=I$
so, $I=2I_0$