In a Young's double-slit experiment,a glass slab of thickness $1.2 \, \mu m$ and refractive index $1.5$ is placed in front of one slit. Another slab of thickness $t$ and refractive index $2.5$ is placed in front of the other slit. If the position of the central fringe remains unchanged,then the thickness $t$ is equal to $...... \, \mu m$.

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,the wavelength $\lambda$ of the light source is $400 \ nm$,the slit separation $d$ is $20 \ \mu m$,and the individual slit width $a$ is $4 \ \mu m$. Considering the interference of light from the two slits and the diffraction of light through each slit,find the number of bright interference fringes that lie within the central peak of the diffraction envelope.

In the figure shown,if a parallel beam of white light is incident on the plane of the slits,then the distance of the white spot on the screen from $O$ is [Assume $d << D, \lambda << d$].

In the setup shown in the figure,the two slits,$S_1$ and $S_2$,are not equidistant from the slit $S$. The central fringe at $O$ is then

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 . . . . . . .