In Young's double slit experiment, the angular width of a fringe is found to be $0.2^\circ$ on a screen placed $1 \text{ m}$ away. The wavelength of light used is $600 \text{ nm}$. If the entire apparatus is immersed in water of refractive index $4/3$, the angular width of the fringe will be: (in $^\circ$)

In a double-slit experiment performed in air, the angular width of a fringe is found to be $0.15^{\circ}$ on a screen placed $80 \ cm$ away. The wavelength of light used is $490 \ nm$. What is the angular width of the fringe if the entire apparatus is immersed in a medium of refractive index $\frac{5}{3}$ (in $^{\circ}$)?

In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

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:

What happens to the fringe width in Young's double-slit experiment when it is performed in water instead of air?

In Young's double slit experiment,the slits are $0.5\, mm$ apart and interference pattern is observed on a screen placed at a distance of $1.0\, m$ from the plane containing the slits. If the wavelength of the incident light is $6000\ \mathring A$,then the separation between the third bright fringe and the central maxima is......$mm$.