In a Young's double-slit experiment, the angular width of a fringe is $1^\circ$ for light of wavelength $6000 \, \mathring{A}$. What is the distance between the slits in $mm$?

Two wavelengths of light $\lambda_1$ and $\lambda_2$ are sent through a Young's double-slit experiment simultaneously. If the third-order bright fringe of $\lambda_1$ coincides with the fourth-order bright fringe of $\lambda_2$,then

$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 \,mm$, and the distance between the slit and the screen is $1.33 \,m$. The slits are illuminated by light of wavelength $6300 \,Å$. The fringe width is:

Statement-$1$: If white light is used in $YDSE$,then the central bright fringe will be white.
Statement-$2$: In the case of white light used in $YDSE$,all the wavelengths produce their zero-order maxima at the same position.

In Young's double slit experiment,the phase difference between the light waves reaching the third bright fringe from the central fringe is: $(\lambda = 6000 \ \mathring{A})$

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: