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 ........ $mm$.

Two slits are made $1\, mm$ apart and the screen is placed $1\, m$ away. The fringe separation when monochromatic light of wavelength $500\, nm$ is used is

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:

On using red light $(\lambda = 6600 \ \text{Å})$ in Young's double slit experiment, $60$ fringes are observed in the field of view. If violet light $(\lambda = 4400 \ \text{Å})$ is used, the number of fringes observed will be

The graph shows the variation of fringe width $(X)$ versus the distance of the screen from the plane of the slits $(D)$ in Young's double-slit experiment (keeping other parameters constant,where $d$ is the distance between the slits). The wavelength of light used can be calculated as:

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\theta$ is the angle $S_1CS_2$ and $\lambda$ is the wavelength,the fringe width will be