In Young's double-slit experiment,the angular width of the fringes formed on the screen is $\pi / 200$. If light of wavelength $4800 \, \mathring{A}$ is used,find the distance between the slits.

Explain Young's double-slit experiment,its experimental arrangement,and how it produces an interference pattern.

In Young's double-slit experiment, if the wavelength of light is decreased, the fringe width will .....

Interference fringes are observed on a screen by illuminating two thin slits $1 \, mm$ apart with a light source $(\lambda = 632.8 \, nm)$. The distance between the screen and the slits is $100 \, cm$. If a bright fringe is observed on a screen at a distance of $1.27 \, mm$ from the central bright fringe,then the path difference between the waves,which are reaching this point from the slits,is close to $.... \mu m$.

Two sources produce an interference pattern which is observed on a screen,at a distance $D$ from the sources. The fringe width is $2w$. If the distance $D$ is now doubled,the fringe width will be:

In $YDSE$ experiment,the $4^{\text{th}}$ dark band is formed opposite to one of the slits. The wavelength of light used is ($d=$ distance between the slits,$D=$ distance between source and the screen)