In Young's double-slit experiment,the distance between the two slits is $1 \, mm$ and the distance between the slits and the screen is $1 \, m$. If the wavelength of the light used is $500 \, nm$,then the distance of the third minimum from the central maximum is ... $mm$.

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?

$A$ beam of light consisting of wavelengths $650 \ nm$ and $550 \ nm$ illuminates the Young's double slits with separation of $2 \ mm$ such that the interference fringes are formed on a screen, placed at a distance of $1.2 \ m$ from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is . . . . . . $\times 10^{-5} \ m$.

Interference was observed in an interference chamber when air was present. Now,the chamber is evacuated and if the same light is used,a careful observer will see:

In $Y.D.S.E.$ using red and blue lights of wavelengths $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$,the $n^{th}$ red fringe coincides with the $(n + 1)^{th}$ blue fringe. The value of $n$ is:

In an interference experiment,the spacing between successive maxima or minima is (where the symbols have their usual meanings):