Light of wavelength $520\, nm$ passing through a double slit produces an interference pattern of relative intensity versus deflection angle $\theta$ as shown in the figure. The separation $d$ between the slits is

In a Young's double slit experiment,$15$ fringes are observed on a small portion of the screen when light of wavelength $500 \; nm$ is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is (in $nm$):

In Young's double slit experiment,the slits are $0.5\, mm$ apart and interference pattern is observed on a screen placed at a distance of $1.0\, m$ from the plane containing the slits. If the wavelength of the incident light is $6000\ \mathring A$,then the separation between the third bright fringe and the central maxima is......$mm$.

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 a Young's double-slit experiment, interference fringes are obtained on a screen at a distance of $1 \, m$ using light of wavelength $6000 \, \mathring{A}$. The distance between the slits is $1 \, mm$. 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.