$A$ plate of refractive index $1.6$ is introduced in the path of light from one of the slits in Young's double slit experiment,then:

Monochromatic light of wavelength $500 \, nm$ is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index $= 1.5$), the central maximum is shifted to a position previously occupied by the $4^{th}$ bright fringe. The thickness of the glass plate is ..................... $\mu m$.

In Young's double slit arrangement,water is filled in the space between the screen and the slits. Then:

When a thin plastic film of refractive index $\mu = 1.45$ is placed in the path of one of the interfering waves,the central fringe shifts by a distance equal to $5$ fringes. If the wavelength of light used is $5890 \, \mathring{A}$,find the thickness of the film.

$A$ double slit experiment is performed with light of wavelength $500\, nm$. $A$ thin film of thickness $2\, \mu m$ and refractive index $1.5$ is introduced in the path of the upper beam. The location of the central maximum will

In a standard $YDSE$ setup,a small transparent slab of thickness $t$ and refractive index $\mu = 1.5$ is placed along the path $AS_2$ (as shown in the figure). Given that the slab thickness $t = d/4$,where $d$ is the slit separation,and the distance from the source $A$ to the slits is not explicitly needed for the shift calculation,find the position of the central maxima on the screen relative to $O$. Assume the distance between the slits and the screen is $D$.