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

In Young's double slit experiment,the angular width of fringes is $0.20^o$ for sodium light of wavelength $5890 \ \mathring A$. If the complete system is dipped in water,then the angular width of the fringes becomes....$^o$

In Young's double slit experiment,the distance between the sources is $1 \ mm$ and the distance between the screen and the source is $1 \ m$. If the fringe width on the screen is $0.06 \ cm$,then $\lambda = \dots \mathring{A}$.

In Young's double slit experiment,the two slits are $0.6 \; mm$ apart. The interference pattern is observed on a screen at a distance of $80 \; cm$ from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light is $\dots \; nm$.

Explain Young's double-slit experiment,its experimental arrangement,and how it produces an interference pattern.

In Young's double slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be (in $:1$)