$A$ Young's double-slit experimental setup is immersed in water of refractive index $1.33$. It has a slit separation of $1 \ mm$ and the distance between the slits and the screen is $1.33 \ m$. If the wavelength of incident light on the slits is $6300 \ \mathring{A}$,then the fringe width on the screen is:

In a Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves:

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?

The distance of the $5^{\text{th}}$ dark fringe from the center is $4 \, mm$. If $D = 2 \, m$ and $\lambda = 600 \, nm$,then the distance between the slits is (in $mm$):

Two slits separated by a distance of $1 \,mm$ are illuminated with light of wavelength $6.5 \times 10^{-7} \,m$. The interference fringes are observed on a screen placed at $1 \,m$ from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (in $\,mm$)

For Young's double slit experiment,two statements are given below :
Statement $I:$ If the screen is moved away from the plane of the slits,the angular separation of the fringes remains constant.
Statement $II:$ If the monochromatic source is replaced by another monochromatic source of larger wavelength,the angular separation of the fringes decreases.
In the light of the above statements,choose the correct answer from the options given below :