In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.

In Young's double-slit experiment,the intensity at a point is $1/4^{th}$ of the maximum intensity. The angular position of this point is

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 $\lambda = 6000 \, \mathring{A}$.

In a Young's double slit experiment,the fringe width will remain the same if ($D =$ distance between screen and plane of slits,$d =$ separation between two slits,and $\lambda =$ wavelength of light used).

In Young's double-slit experiment,an interference pattern is obtained on a screen by light of wavelength $6000 \ \mathring A$,coming from coherent sources $S_1$ and $S_2$. At a certain point $P$ on the screen,the third dark fringe is formed. Then the path difference $S_1P - S_2P$ in microns is:

In a Young's double-slit experiment,the fringe pattern is observed on a screen placed at a distance $D$. The slits are separated by $d$ and are illuminated by light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to half the maximum intensity.