A monochromatic light of `lambda=500 nm` is incident on two identical slits separated by a distance of `5xx10^(-4)m`. The interference pattern is seen on a screen placed at a distance of `1 m` from the plane of slits. A thin glass plate of thickness `1.5xx10^(-6)m` and refractive index `mu=1.5` is placed between one of the slits and the screen. Find the intensity at the centre of the screen if the intensity is `I_(0)` in the absence of the plate. Also find the lateral shift of the central maxima and number of fringes crossed through centre.

Answer

`0,1.5 mm`

Explanation

After the thin plate is placed between one of the slits and the screen, an additional path difference `[DeltaP=(mu-1)t]` is created between the rays emerging out of `S_(1) and S_(2)`. Thus
`S_(1)O-S_(2)O=(mu -1) t= 0.75xx10^(-6) m`
`((mu-1)t)/(lambda)=(3)/(2) rArr (mu-1)t=(3)/(2)lambda`
Point `O` would be an interference minimum. Thus intensity at `O` is zero.

`S_(1)P-S_(2)P=(dy_(m))/(D)`
`S_(2)P-S_(1)P=(dy_(m))/(D)-(mu-1)t`
For central maxima, the path difference between the interfering beams is zero.
`i.e. S_(2)P-S_(1)P=0`
`rArr (dy_(m))/(D)=(mu-1)t rArr y_(m)=(D)/(d)(mu-1)t=(1)/(5xx10^(-4))xx0.75xx10^(-6)=1.5 mm`
Thus, the effect of putting a thin film in front of one of the slits is to shift the central maxima towards that alit.