In a Young's double-slit experiment,light of wavelength $5890 \ \mathring A$ is used,and the angular fringe width on the screen is $0.20^\circ$. If the entire apparatus is immersed in water,find the new angular fringe width. (Refractive index of water $\mu = 4/3$) (in $^\circ$)

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 a double slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1\, m$ away. The wavelength of light used is $600\, nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.

In two separate set-ups of the Young's double slit experiment,fringes of equal width are observed when lights of wavelengths in the ratio $1:2$ are used. If the ratio of the slit separation in the two cases is $2:1$,the ratio of the distances between the plane of the slits and the screen in the two set-ups is

In a $YDSE$ apparatus,two identical slits are separated by $1\, mm$ and the distance between the slits and the screen is $1\, m$. The wavelength of light used is $6000\, \mathring{A}$. The minimum distance between two points on the screen having $75\%$ intensity of the maximum intensity is ......$mm$.

In a Young's double slit experiment illuminated by monochromatic light,the intensity reaching the screen from each slit is $I$. What is the resultant intensity at a point where the phase difference is $60^{\circ}$?