In a Young's double-slit experiment,$D$ is the distance of the screen from the slits and $d$ is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is the same as that due to a single slit is equal to:

In $Y.D.S.E$, two slits are made $1 \, mm$ apart and the screen is placed $1 \, m$ away. The fringe separation when light of wavelength $500 \, nm$ is used is:

The maximum intensity in Young's double slit experiment is $I_0$. The distance between the slits is $d = 5\lambda$,where $\lambda$ is the wavelength of the monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D = 10d$?

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 a Young's double slit experiment, the slits are $2 \,mm$ apart and are illuminated with a mixture of two wavelengths ${\lambda _1} = 750 \,nm$ and ${\lambda _2} = 900 \,nm$. The minimum distance from the common central bright fringe on a screen $2 \,m$ from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is.....$mm$

The distance between two coherent sources is $0.1 \, mm$. The fringe-width on a screen $1.2 \, m$ away from the source is $6.0 \, mm$. The wavelength of light used is ......... $\mathring{A}$.