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

In a Young's double-slit experiment,light of two wavelengths $6500 \, \mathring A$ and $5200 \, \mathring A$ is used. Find the distance of the third bright fringe from the central maximum for the wavelength $6500 \, \mathring A$. The distance between the two slits is $2 \, mm$ and the distance between the plane of the slits and the screen is $120 \, cm$.

In Young's double-slit experiment,if the distance between the slits is twice the wavelength,the number of possible interference maxima is:

In a double-slit experiment,the angular width of the fringes is $0.20^o$ for sodium light $(\lambda = 5890 \ \mathring{A})$. In order to increase the angular width of the fringes by $10\%$,the necessary change in the wavelength is:

In an experiment, light passing through two slits separated by a distance of $0.3 \,mm$ is projected onto a screen placed at $1 \,m$ from the plane of the slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 \,mm$. The wavelength of light in $nm$ is

In two different Young's double-slit experiments,the fringe width is the same when the ratio of wavelengths is $1:2$. If the ratio of the distance between the slits in the two cases is $2:1$,then the ratio of the distance between the slits and the screen in the two experiments is: