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 light emitted from the source has $\lambda = 6.5 \times 10^{-7} \, m$ and the distance between the two slits is $1 \, mm$. The distance between the screen and the slits is $1 \, m$. The distance between the third dark fringe and the fifth bright fringe will be ......... $mm$.

In Young's double slit experiment,the $y$-coordinates of the central maxima and the $10^{th}$ maxima are $2 \, cm$ and $5 \, cm$ respectively. When the $YDSE$ apparatus is immersed in a liquid of refractive index $\mu = 1.5$,what will be the corresponding $y$-coordinates?

In Young's double-slit experiment,if the amplitudes of the interfering waves are not equal,then . . . . .

In Young's double slit experiment,when the wavelength used is $6000 \ \mathring{A}$ and the screen is $40 \ cm$ from the slits,the fringes are $0.012 \ cm$ wide. What is the distance between the slits in $cm$?

$A$ student is asked to measure the wavelength of monochromatic light. He sets up the apparatus as shown in the figure. $S_1, S_2, S_3$ are narrow parallel slits,$L$ is a sodium lamp,and $M$ is a microscope eyepiece. The student fails to observe interference fringes. Your first advice to him will be: