In two different Young's double-slit experiments,the fringe width is the same when the ratio of wavelengths is $1:2$. If the ratio of the distance between the slits in the two cases is $2:1$,then the ratio of the distance between the slits and the screen in the two experiments is:

In Young's double slit experiment,the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to

The graph shows the variation of fringe width $(X)$ versus the distance of the screen from the plane of the slits $(D)$ in Young's double-slit experiment (keeping other parameters constant,where $d$ is the distance between the slits). The wavelength of light used can be calculated as:

In Young's double slit experiment,$62$ fringes are seen in the visible region for sodium light of wavelength $5893 \ \mathring{A}$. If violet light of wavelength $4358 \ \mathring{A}$ is used in place of sodium light,then the number of fringes seen will be:

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 slits in a Young's double-slit experiment have equal widths, and the source is placed symmetrically relative to the slits. The intensity at the central fringe is $I_0$. If one of the slits is closed, the intensity at this point will be