In Young's double-slit experiment,when two light waves form the third minimum,the path difference between them is:

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 $YDSE$,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. Then,the fringe width:

In an experiment, light passing through two slits separated by a distance of $0.3 \,mm$ is projected onto a screen placed at $1 \,m$ from the plane of the slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 \,mm$. The wavelength of light in $nm$ is

Laser light $(630 \, nm)$ incident on a pair of slits produces an interference pattern in which the bright fringes are separated by $8.3 \, mm$. $A$ second light produces an interference pattern in which the bright fringes are separated by $7.6 \, mm$. What is the wavelength of this second light in $nm$?

In a $Young's$ double-slit experiment,light of wavelength $6000 \, \mathring{A}$ is used. If the third dark fringe is formed at point $P$ on the screen,then the path difference $(S_1P - S_2P)$ is ......... $\mu m$.