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 with light of wavelength $\lambda ,$ the fringe pattern on the screen has a fringe width $\beta .$ When two thin transparent glass (refractive index $\mu$) plates of thickness $t_1$ and $t_2$ $(t_1 > t_2)$ are placed in the path of the two beams respectively,the fringe pattern will shift by a distance:

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index $1.56$, the central fringe shifts to the position of $7^{th}$ bright fringe, obtained with both slits uncovered. If the light source wavelength is $450 \text{ nm}$, the thickness of mica sheet is $\alpha \times 10^{-9} \text{ m}$. The value of $\alpha$ is . . . . . . .

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 ....... $\mu m$.

When one of the slits in Young's experiment is covered with a transparent sheet of thickness $3.6 \times 10^{-3} \ cm$,the central fringe shifts to a position originally occupied by the $30^{th}$ bright fringe. If $\lambda = 6000 \ \mathring{A}$,then what is the refractive index of the sheet?

Two coherent narrow slits emitting light of wavelength $\lambda$ in the same phase are placed parallel to each other at a small separation of $3 \lambda$. The light is collected on a screen $S$ which is placed at a distance $D (>> \lambda)$ from the slits. Find the smallest distance $x$ from the center $O$ such that the point $P$ is a maxima.