Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)

$A$ student is asked to measure the wavelength of monochromatic light. He sets up the apparatus as shown in the figure. $S_1, S_2, S_3$ are narrow parallel slits,$L$ is a sodium lamp,and $M$ is a microscope eyepiece. The student fails to observe interference fringes. Your first advice to him will be:

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 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 a certain double-slit experimental arrangement,interference fringes of width $1.0 \ mm$ each are observed when light of wavelength $5000 \ \mathring{A}$ is used. Keeping the setup unaltered,if the source is replaced by another source of wavelength $6000 \ \mathring{A}$,the fringe width will be $...... \ mm$.

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 $\lambda = 6000 \, \mathring{A}$.