In Young's double slit experiment,if the widths of the slits are in the ratio $4 : 9$,the ratio of the intensity at maxima to the intensity at minima will be

In the figure, Young's double-slit experiment is shown. $Q$ is the position of the first bright fringe on the right side of $O$. $P$ is the $11^{th}$ fringe on the other side, as measured from $Q$. If the wavelength of the light used is $6000 \times 10^{-10} \text{ m}$, then $S_1B$ will be equal to:

In a double slit experiment,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. The fringe width:

In the Young's double slit experiment with sodium light,the slits are $0.589 \ m$ apart. The angular separation of the third maximum from the central maximum will be (given $\lambda = 589 \ nm$):

In $YDSE$,the $Y$-coordinates of the central maxima and the $10^{th}$ maxima are $2 \, cm$ and $5 \, cm$ respectively. When the $YDSE$ apparatus is immersed in a liquid of refractive index $1.5$,what will be the corresponding $Y$-coordinates?

In a Young's double-slit experiment,the light beam consists of two wavelengths $6500 \, \mathring A$ and $5200 \, \mathring A$. The distance between the slits is $2 \, mm$ and the distance between the plane of the slits and the screen is $120 \, cm$. Find the distance of the third bright fringe from the central maximum for the wavelength $6500 \, \mathring A$ in $mm$.