In an interference experiment,the spacing between successive maxima or minima is

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ was $1.2 \ \text{mm}$ for parallel narrow slits placed $2 \ \text{mm}$ apart. In this arrangement, if the slit separation is increased by one-and-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

In Young's double-slit experiment,the distance between the two slits is $3 \, cm$,the distance from the slits to the screen is $7 \, cm$,and the wavelength of light used is $1000 \, \mathring{A}$. Calculate the fringe width.

In Young's double slit experiment,if one of the slits is closed fully,then in the interference pattern:

In a $YDSE$ apparatus,if we use white light,then:

$A$ mixture of light of wavelength $590 \, nm$ and an unknown wavelength is incident on the two slits of a Young's double-slit experiment. The central bright fringes of both lights coincide. The $3^{rd}$ bright fringe of the known wavelength coincides with the $4^{th}$ bright fringe of the unknown wavelength. Find the unknown wavelength in $nm$.