$A$ 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 double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ 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-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

In Young's double slit experiment,if one of the slits is closed fully,then in the interference pattern:

The ratio of intensities at two points on the screen in Young's double slit experiment when waves from the two slits have a path difference of $0$ and $\lambda/4$ ($\lambda$ is the wavelength of light used) $(\cos 0^\circ = 1, \cos \pi/2 = 0)$.

In a Young's double slit experiment,the slit separation $d$ is $0.3 \text{ mm}$ and the screen distance $D$ is $1 \text{ m}$. $A$ parallel beam of light of wavelength $600 \text{ nm}$ is incident on the slits at angle $\alpha$ as shown in the figure. On the screen,the point $O$ is equidistant from the slits and the distance $PO$ is $11.0 \text{ mm}$. Which of the following statement$(s)$ is/are correct?

The fringe width in a $YDSE$ pattern is $2.4 \times 10^{-4} \, m$ when red light of wavelength $6400 \, \mathring{A}$ is used. How much will it change if blue light of wavelength $4000 \, \mathring{A}$ is used?