In an experiment, light passing through two slits separated by a distance of $0.3 \,mm$ is projected onto a screen placed at $1 \,m$ from the plane of the slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 \,mm$. The wavelength of light in $nm$ is

In an interference pattern,the $(n + 4)^{th}$ blue bright fringe and $n^{th}$ red bright fringe are formed at the same spot. If red and blue light have wavelengths of $7800\,\mathring{A}$ and $5200\,\mathring{A}$ respectively,then the value of $n$ is:

In Young's double-slit experiment,when two light waves form the third minimum,the path difference between them is:

In Young's double-slit experiment,the value of $\lambda = 500\, nm$. The value of $d = 1\, mm$ and $D = 1\, m$. Then the minimum distance from the central maximum for which the intensity is half the maximum intensity will be:

$A$ beam of light consisting of two wavelengths $6300 \, \mathring{A}$ and $\lambda \, \mathring{A}$ is used to obtain interference fringes in a Young's double-slit experiment. If the $4^{th}$ bright fringe of $6300 \, \mathring{A}$ coincides with the $5^{th}$ dark fringe of $\lambda \, \mathring{A}$,the value of $\lambda$ (in $\mathring{A}$) is:

The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width,the ratio of minimum to maximum intensity in the interference pattern is $x: 4$ where $x$ is ..... .