In an interference experiment,the third bright fringe is obtained using light of wavelength $700 \, nm$. What should be the wavelength of light to obtain the fifth bright fringe at the same point?

$A$ light of wavelength $500 \, nm$ is incident on a Young's double-slit. The distance between the slits and the screen is $D = 1.8 \, m$ and the distance between the slits is $d = 0.4 \, mm$. If the screen moves with a speed of $4 \, m/s$,with what speed will the first maxima move? (in $mm/s$)

In Young's double-slit experiment,the angular width of the fringes formed on the screen is $\pi / 200$. If light of wavelength $4800 \, \mathring{A}$ is used,find the distance between the slits.

In Young's double slit experiment,if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled,then the fringe width becomes:

In a Young's double-slit experiment,the slits are separated by $0.28 \; mm$ and the screen is placed $1.4 \; m$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2 \; cm$. Determine the wavelength of light used in the experiment (in $nm$).

In a certain region of the screen, the number of fringes formed is observed to be $8$ in Young's double slit experiment when light of wavelength $6600 \text{ Å}$ is used. If light of wavelength $4400 \text{ Å}$ is used, the number of fringes observed in the same region of the screen will be: