In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

$A$ 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

In the Young's double slit experiment,the ratio of intensities of bright and dark fringes is $9$. This means that

In Young's double-slit experiment,the two slits are $2 \, mm$ apart and are illuminated by a mixture of two wavelengths $\lambda_1 = 12000 \, \mathring{A}$ and $\lambda_2 = 10000 \, \mathring{A}$. At what minimum distance from the common central bright fringe on the screen will a bright fringe from one interference pattern coincide with a bright fringe from the other? The distance between the slits and the screen is $2 \, m$.

In Young's double-slit experiment, light of wavelength $4000 \, Å$ is used to produce bright fringes of width $0.6 \, mm$ at a distance of $2 \, m$ from the slits. If the whole apparatus is immersed in a liquid of refractive index $1.5$, what will be the new fringe width in $mm$?

In Young's double-slit experiment,both slits are illuminated by a laser beam and the interference pattern was observed on a screen. If the viewing screen is moved farther from the slit,what happens to the interference pattern?