Young's double slit experiment is performed with monochromatic light. $A$ thin film is introduced in front of one of the slits.

Interference fringes were produced in Young's double slit experiment using light of wavelength $5000 \, Å$. When a film of material $2.5 \times 10^{-3} \, cm$ thick was placed over one of the slits, the fringe pattern shifted by a distance equal to $20$ fringe widths. The refractive index of the material of the film is

In a Young's double-slit experiment,let $A$ and $B$ be the two slits. $A$ thin film of thickness $t$ and refractive index $\mu$ is placed in front of $A$. Let $\beta =$ fringe width. The central maximum will shift:

In $YDSE$,the source $S$ placed symmetrically with respect to the slits $S_1$ and $S_2$ is now moved parallel to the plane of the slits so that it is closer to the upper slit $S_1$,as shown. Then,

In a Young's double-slit experiment,light of wavelength $4800 \, \mathring A$ is used. If both slits are covered by transparent plates of the same thickness $t$ having refractive indices $\mu_1 = 1.5$ and $\mu_2 = 1.8$,and the central bright fringe shifts to the position of the fifth bright fringe,then the thickness of the plates is ....... $\mu m$.

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 $D$.