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

$A$ light of wavelength $500 \, nm$ is incident on a Young's double-slit. The distance between the slits and the screen is $D = 1.8 \, m$ and the distance between the slits is $d = 0.4 \, mm$. If the screen moves with a speed of $4 \, m/s$,with what speed will the first maxima move? (in $mm/s$)

The figure shows a schematic diagram of Young's Double Slit Experiment. Identify the correct statement$(s)$ if the source slit $S$ is moved closer to the slits $S_1$ and $S_2$,i.e.,the distance $\ell$ decreases.

Two slits are made $1\, mm$ apart and the screen is placed $1\, m$ away. The fringe separation when monochromatic light of wavelength $500\, nm$ is used is

In Young's double slit interference experiment,the slit separation is made $3$ fold. The fringe width becomes

In a double slit experiment,the distance between slits is increased $10$ times whereas their distance from the screen is halved,then what is the fringe width?