Young's double-slit experiment is performed in air and then performed in water. How does the fringe width change?

$A$ Young's double slit interference arrangement with slits $S_1$ and $S_2$ is immersed in water (refractive index $\mu_w = 4/3$) as shown in the figure. The positions of maxima on the surface of water are given by $x^2 = p^2 m^2 \lambda^2 - d^2$,where $\lambda$ is the wavelength of light in air (refractive index $\mu_a = 1$),$2d$ is the separation between the slits,and $m$ is an integer. The value of $p$ is

In Young's double slit experiment,white light is used. The separation between the slits is $b$. The screen is at a distance $d$ $(d >> b)$ from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are

The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

In Young's double-slit experiment,the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

Young's double-slit experiment is performed with a wavelength of $550 \,nm$. If the distance between the two slits is $1.10 \,mm$ and the distance between the screen and the slits is $1 \,m$,then the distance between two consecutive bright or dark fringes is .....$mm$.