In Young's double-slit experiment, the fringe width is $\beta$. If the entire arrangement is placed in a liquid of refractive index $n$, what does the fringe width become?

Given below are two statements:
Statement $I$: In a Young's double slit experiment, the angular separation of fringes will increase as the screen is moved away from the plane of the slits.
Statement $II$: In a Young's double slit experiment, the angular separation of fringes will increase when a monochromatic source is replaced by another monochromatic source of higher wavelength.
In the light of the above statements, choose the correct answer from the options given below:

In a double slit experiment,at a certain point on the screen,the path difference between the two interfering waves is $\frac{1}{8}$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:

In a Young's double slit experiment,the angular width of a fringe formed on a distant screen is $1^{\circ}$. The wavelength of the light used is $6280 \; \mathring{A}$. The distance between the two coherent sources is $...........\,mm$.

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

In Young's double-slit experiment,if the separation between the slit and the screen increases,the fringe width: