In a Young's double slit experiment,the source illuminating the slits is changed from blue to violet. The width of the fringes

$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 \,mm$, and the distance between the slit and the screen is $1.33 \,m$. The slits are illuminated by light of wavelength $6300 \,Å$. The fringe width is:

In a Young's double slit experiment,$12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600 \ nm$ is used. If the wavelength of light is changed to $400 \ nm$,the number of fringes observed in the same segment of the screen is:

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 the Young's double slit experiment,the distance between the slits varies in time as $d(t) = d_{0} + a_{0} \sin \omega t$; where $d_{0}$,$\omega$,and $a_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,$I/I_0$ is equal to