Light of wavelength $6000 \,\mathring A$ is incident on two slits. The distance between the slits is $0.1 \, cm$ and the screen is placed at a distance of $1 \, m$ from them. What is the distance between two consecutive minima in $mm$?

For sustained interference fringes in a double-slit experiment,the essential condition$(s)$ is/are:
$1$. Sources must be coherent.
$2$. The intensities of the two sources must be equal.
Here,the correct option$(s)$ is/are:

Imagine a Young's double-slit interference experiment performed with waves associated with fast-moving electrons produced from an electron gun. The distance between successive maxima (fringe width) will decrease if:

If the separation between slits in Young's double slit experiment is reduced to $\frac{1}{3}rd,$ the fringe width becomes $n$ times. The value of $n$ is

$A$ source simultaneously emitting light at two wavelengths $400 \,nm$ and $800 \,nm$ is used in the Young's double slit experiment. If the intensity of light at the slit for each wavelength is $I_{0}$,then the maximum intensity that can be observed at any point on the screen is

In Young's double slit experiment,the distance between slits and the screen is $1.0\,m$ and monochromatic light of $600\,nm$ is being used. $A$ person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied,the interference pattern disappears for a particular distance $d_0$ between the slits. If the angular resolution of the eye is $\frac{1}{60}^o,$ the value of $d_0$ is close to......$mm$