In Young's double slit experiment,if the distance between the two slits is halved,then the fringe width will become

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 a Young's double-slit experiment,the light beam consists of two wavelengths $6500 \, \mathring A$ and $5200 \, \mathring A$. The distance between the slits is $2 \, mm$ and the distance between the plane of the slits and the screen is $120 \, cm$. Find the distance of the third bright fringe from the central maximum for the wavelength $6500 \, \mathring A$ in $mm$.

In a double slit experiment shown in the figure,when light of wavelength $400 \ nm$ is used,a dark fringe is observed at $P$. If $D=0.2 \ m$,the minimum distance between the slits $S_1$ and $S_2$ is . . . . . . $mm$.

In Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $k$ units; $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\lambda/4$ will be:

$A$ double slit experiment is performed by using light of wavelength $6000 \, Å$. If the distance of the screen is $1 \, m$ and the slits are $0.1 \, cm$ apart, calculate the angular position of the $10^{th}$ bright fringe.