$A$ beam of electrons is used in $Y.D.S.E.$ The slit width is $d$. When the velocity of the electrons is increased,then:

Two slits separated by $0.5 \,mm$ are illuminated by light of wavelength $500 \,nm$. The screen is at a distance of $120 \,cm$ from the slits. The phase difference between the interfering waves at a point $3 \,mm$ on the screen from the central bright fringe is ........... .

In Young's double slit experiment,if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled,then the fringe width becomes:

$A$ 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 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 a Young's double-slit experiment,the fringe pattern is observed on a screen placed at a distance $D$. The slits are separated by $d$ and are illuminated by light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to half the maximum intensity.