In young’s double slits experiment a parallel monochromatic light beam of wavelength 500 nm falls on the slits at angle ${\theta }_0$  with the normal to the plane of the slits. A transparent sheet of refractive index 1.5 and thickness 19.75  $\mu m$  is placed in front of the slit  $S_2$ .The intensity on the axis of the screen O is I and maximum fringe intensity on the screen is  $I_0$. The value of $\frac{I}{I_0}=X$  and the minimum distance from the origin O where the intensity is maximum is Ymm. ( Given  $\sin {\theta }_0=\frac{1}{50}$, the distance between the slits $S_1 and S_2 = 0.5mm,$ the distance between the slits and the screen =1m)