In a Young's double slit experiment,the fringe width is found to be $0.4 \, mm$. If the whole apparatus is immersed in water of refractive index $4/3$ without disturbing the geometrical arrangement,the new fringe width will be

$A$ beam of light consisting of two wavelengths $6500 \mathring{A}$ and $5200 \mathring{A}$ is used to obtain interference fringes in Young's double-slit experiment. The distance between slits is $2 \text{ mm}$ and the distance of the screen from the slits is $120 \text{ cm}$. What is the least distance from the central maximum where the bright fringes due to both wavelengths coincide (in $\text{ cm}$)?

In an interference experiment,let $D$ be the distance between the slit and the eyepiece. When the distance between two virtual sources is changed from $d_{A}$ to $d_{B}$,the fringe width changes from $Z_{A}$ to $Z_{B}$. The ratio $Z_{A} / Z_{B}$ is:

In Young's double slit experiment,the wavelength of red light is $7800\,\mathring{A}$ and that of blue light is $5200\,\mathring{A}$. The value of $n$ for which the $n^{th}$ bright band due to red light coincides with the $(n + 1)^{th}$ bright band due to blue light is:

If the monochromatic source in Young's double slit experiment is replaced by white light,then

In Young's double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is $d$ and $D$ is the distance between the source and the screen. The possible wavelengths used are inversely proportional to