In Young's double-slit experiment, light of wavelength $4000 \, Å$ is used to produce bright fringes of width $0.6 \, mm$ at a distance of $2 \, m$ from the slits. If the whole apparatus is immersed in a liquid of refractive index $1.5$, what will be the new fringe width in $mm$?

In a Young's double slit experiment,the fringe width will remain the same if ($D =$ distance between screen and plane of slits,$d =$ separation between two slits,and $\lambda =$ wavelength of light used).

In Young's double-slit experiment,the fringes are displaced by a distance $x$ when a glass plate of refractive index $1.5$ is introduced in the path of one of the beams. When this plate is replaced by another plate of the same thickness,the shift of fringes is $(3/2)x$. The refractive index of the second plate is

$A$ source simultaneously emitting light at two wavelengths $400 \,nm$ and $800 \,nm$ is used in the Young's double slit experiment. If the intensity of light at the slit for each wavelength is $I_{0}$,then the maximum intensity that can be observed at any point on the screen is

Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation (in $mm$) when blue-green light of wavelength $500 \; nm$ is used?

Explain Young's double-slit experiment,its experimental arrangement,and how it produces an interference pattern.