The path difference between two identical light waves at a point $Q$ on the screen is $3 \mu m$. If the wavelength of the waves is $5000 \text{ Å}$,then at point $Q$ there is:

$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{87}{2}\right) \lambda$. The band obtained at that point is

In the figure, Young's double-slit experiment is shown. $Q$ is the position of the first bright fringe on the right side of $O$. $P$ is the $11^{th}$ fringe on the other side, as measured from $Q$. If the wavelength of the light used is $6000 \times 10^{-10} \text{ m}$, then $S_1B$ will be equal to:

In Young's double slit experiment,the slits are $0.5\, mm$ apart and interference pattern is observed on a screen placed at a distance of $1.0\, m$ from the plane containing the slits. If the wavelength of the incident light is $6000\ \mathring A$,then the separation between the third bright fringe and the central maxima is......$mm$.

In Young's double-slit experiment,the distance between the two slits is $3 \, cm$,the distance from the slits to the screen is $7 \, cm$,and the wavelength of light used is $1000 \, \mathring{A}$. Calculate the fringe width.