In Young's double-slit experiment,which of the following statements is correct?

In Young's double-slit experiment,the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

In Young's double slit experiment,if the wavelength of the light used is doubled and the distance between the two slits is halved,the resultant fringe width becomes $n$ times the initial fringe width. Find the value of $n$.

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,$I/I_0$ is equal to

$A$ 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 . . . . . . $\times 10^{-5} \ m$.

Two parallel slits $0.6 \,mm$ apart are illuminated by a light source of wavelength $6000 \,\mathring{A}$. The distance between two consecutive dark fringes on a screen $1 \,m$ away from the slits is: