In a young’s double slit experiment set up source S of wavelength $\lambda =500nm$ illuminates two symmetrically located slits $S1$  and  $S1$ The source S oscillates about its shown position parallel to the screen according to the  equation  $y=\left(0.5mm\right)\sin \left(\pi t\right).$ Where  t is time in second. Distances are as marked in the figure.  least value of time $\left(t\right)$ at which the intensity becomes maximum at a point on the screen that is exactly in front of slit $S1$ is equal to $\left(\frac{1}{P}\right)s$ Find P 

Displacement of the source at time  t is $Y=\left(0.5mn\right)\sin \left(\pi t\right)$
Path difference of the two waves reaching at P is  $\Delta x=\frac{yd}{D}+\frac{y\text{'}d}{D\text{'}}$
 For central maximum   $\Delta x=0$
 $\Rightarrow y\text{'}=-\frac{D\text{'}}{D}y=-\left(\frac{2}{1}\right)0.5\sin \pi t$

$\therefore y\text{'}=-\sin \left(\pi t\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}in\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}mm}$
 

 (b) $y\text{'}\frac{d}{2}$  for a point exactly in front of  $S_1$
 $\therefore \Delta x=\frac{yd}{D}+\frac{\frac{d}{2}d}{D\text{'}}$
 For maximum intensity  $\Delta x=n\lambda$ 
$\frac{yd}{D}+\frac{\frac{d^2}{2}}{D\text{'}}=n\lambda$ 
   $\Rightarrow 0.5\sin \pi t+\frac{{\left(1\right)}^2}{4}=n\times 500\times {10}^{-6}\times {10}^3$
$\Rightarrow 0.5\sin \pi t+0.25=0.5n$   
   $\Rightarrow \sin \left(\pi t\right)=\frac{0.5n-0.25}{0.5}$
 For minimum ‘t’ we have $n=1$
$$\begin{gathered} \therefore \sin \left(\pi t\right)=\frac{0.5-0.25}{0.5}=0.5 \\ \Rightarrow \pi t=\frac{\pi }{6}\Rightarrow t=\frac{1}{6}s \end{gathered}$$