In Young's double-slit experiment,the fringe width is $12 \ mm$. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$,then the fringe width becomes (in $mm$):

In a Young's double slit experiment,${I_o}$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ distant $x$ from the centre will be

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is (in $ \text{ mm}$)

The correct curve between fringe width $\beta$ and the distance between the slits $(d)$ is:

Young's double slit experiment is first performed in air and then in a medium other than air. It is found that the $8^{th}$ bright fringe in the medium lies where the $5^{th}$ dark fringe lies in air. The refractive index of the medium is nearly:

In a double-slit experiment, the angular width of a fringe for sodium light $(\lambda = 5890 \, \mathring{A})$ is $0.20^o$. If the fringe width increases by $10\%$, what is the change in the wavelength?