In a Young's double-slit experiment with light of wavelength $\lambda$,the separation of slits is $d$ and the distance of the screen is $D$ such that $D >> d >> \lambda$. If the fringe width is $\beta$,the distance from the point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

In Young's double-slit experiment,$\frac{d}{D} = 10^{-4}$. The intensity at point $P$ on the screen is equal to the intensity of one of the sources. If the wavelength of the light used is $\lambda = 6000 \, \mathring{A}$,what is the distance of point $P$ from the central bright fringe in $mm$?

The angular width of the central maximum in a single slit diffraction pattern is $60^o$. The width of the slit is $1 \mu m$. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it,Young's fringes can be observed on a screen placed at a distance $50 \ cm$ from the slits. If the observed fringe width is $1 \ cm$,what is the slit separation distance in $\mu m$ (i.e.,distance between the centres of each slit)?

In a double-slit experiment using light of wavelength $600 \; nm$,the angular width of a fringe formed on a distant screen is $0.1^{\circ}$. What is the spacing between the two slits?

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 $YDSE$, the distance of the slits from the screen is increased by $25 \%$ and the separation between the slits is halved. If $W$ represents the original fringe width, the new fringe width is (in $\,W$)