In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,$I/I_0$ is equal to

Two slits separated by a distance of $1 \,mm$ are illuminated with light of wavelength $6.5 \times 10^{-7} \,m$. The interference fringes are observed on a screen placed at $1 \,m$ from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (in $\,mm$)

$A$ 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,the wavelength of light used is $6000 \, \mathring A$. The phase difference between the central bright fringe and the third bright fringe is .......

If fringe width is $0.4 \,mm$, the distance between the fifth bright and third dark band on the same side is: (in $\,mm$)

In Young's double slit experiment shown in the figure,$S_1$ and $S_2$ are coherent sources and $S$ is the screen having a hole at a point $1.0 \, mm$ away from the central line. White light ($400 \, nm$ to $700 \, nm$) is sent through the slits. Which wavelength passing through the hole has strong intensity (in $, nm$)? (Given: $d = 0.5 \, mm$,$D = 50 \, cm$)