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

In Young's experiment,the ratio of maximum to minimum intensities of the fringe system is $4:1$. The amplitudes of the coherent sources are in the ratio: (in $:1$)

The width of one of the two slits in a Young's double slit experiment is $4$ times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: (in $: 1$)

In a Young's double slit experiment, the slits are $2 \,mm$ apart and are illuminated with a mixture of two wavelengths ${\lambda _1} = 750 \,nm$ and ${\lambda _2} = 900 \,nm$. The minimum distance from the common central bright fringe on a screen $2 \,m$ from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is.....$mm$

The fringe width in an interference pattern is $X$. The distance between the sixth dark fringe from one side of the central bright band to the fourth bright fringe on the other side is: (in $X$)

In an interference pattern,the $(n + 4)^{th}$ order blue bright fringe and the $n^{th}$ order red bright fringe coincide at a point. If the wavelengths of red and blue light are $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$ respectively,then the value of $n$ is . . . . . .