In Young's double-slit experiment, the fringe width is $\beta$. If the entire arrangement is placed in a liquid of refractive index $n$, what does the fringe width become?

In a double slit experiment,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. The fringe width:

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\theta$ is the angle $S_1CS_2$ and $\lambda$ is the wavelength,the fringe width will be

In Young's double slit experiment,the $y$-coordinates of the central maxima and the $10^{th}$ maxima are $2 \, cm$ and $5 \, cm$ respectively. When the $YDSE$ apparatus is immersed in a liquid of refractive index $\mu = 1.5$,what will be the corresponding $y$-coordinates?

In Young's double-slit experiment,a thin mica sheet of thickness $12 \times 10^{-7} \ m$ is placed in the path of one of the interfering beams. It is observed that the central bright fringe shifts by a distance equal to the fringe width. If the wavelength of light used is $6 \times 10^{-7} \ m$,find the refractive index of the mica sheet.

If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)