What happens to the fringe width in Young's double-slit experiment if it is performed in glycerine instead of air?

In Young's double-slit experiment,the separation between the two slits is $3 \, mm$ and they are illuminated by light of wavelength $480 \, nm$. The screen is at a distance of $2 \, m$ from the plane of the slits. Find the separation between the $8^{th}$ bright fringe and the $3^{rd}$ dark fringe with respect to the central bright fringe.

The path difference between two identical light waves at a point $Q$ on the screen is $3 \mu m$. If the wavelength of the waves is $5000 \text{ Å}$,then at point $Q$ there is:

For sustained interference fringes in a double-slit experiment,the essential condition$(s)$ is/are:
$1$. Sources must be coherent.
$2$. The intensities of the two sources must be equal.
Here,the correct option$(s)$ is/are:

In Young's double slit experiment,the two slits act as coherent sources of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same setup,the two slits are sources of equal amplitude $A$ and wavelength $\lambda$ but are incoherent. The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is:

In Young's double slit experiment,the wavelength of red light is $7800\,\mathring{A}$ and that of blue light is $5200\,\mathring{A}$. The value of $n$ for which the $n^{th}$ bright band due to red light coincides with the $(n + 1)^{th}$ bright band due to blue light is: