Two slits,$4 \, mm$ apart,are illuminated by light of wavelength $6000 \, \mathring{A}$. What will be the fringe width on a screen placed $2 \, m$ from the slits in $mm$?

In Young's double slit experiment,the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to

In a Young's double-slit experiment,interference is obtained using electron waves. The electrons are produced by an electron gun. To increase the fringe width,one must:

In a Young's double-slit experiment,two slits act as coherent sources of waves of equal wavelength $\lambda$ and amplitude $A$. In another experiment,the two slits are arranged to act as incoherent sources of waves of the same amplitude and wavelength. If the intensity at the central point of the screen in the first case is $I_1$ and in the second case is $I_2$,then the ratio $I_1/I_2$ is:

Two ideal slits $S_1$ and $S_2$ are at a distance $d$ apart and are illuminated by light of wavelength $\lambda$ passing through an ideal source slit $S$ placed on the line through $S_2$ as shown. The distance between the plane of the slits and the source slit is $D$. $A$ screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is

In Young's double slit experiment, carried out with light of wavelength $\lambda = 5000\;\mathring{A}$, the distance between the slits is $d = 0.2\;mm$ and the screen is at $D = 200\;cm$ from the slits. The central maximum is at $x = 0$. The third maximum (taking the central maximum as the zeroth maximum) will be at $x$ equal to......$cm$.