In $YDSE$,the angular width of a fringe formed on a distant screen is $1^o$. The slit separation is $0.01\,mm$. The wavelength of light is:

The intensity of each source 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. What will be the intensity of light in front of one of the slits on a screen (where the slit and screen are at a distance $D = 10d$)?

In Young's double-slit experiment,the distance between the two slits is $1 \, mm$ and the distance between the slits and the screen is $1 \, m$. If the wavelength of the light used is $500 \, nm$,then the distance of the third minimum from the central maximum is ... $mm$.

In an experiment, light passing through two slits separated by a distance of $0.3 \,mm$ is projected onto a screen placed at $1 \,m$ from the plane of the slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 \,mm$. The wavelength of light in $nm$ is

The ratio of intensities at two points $P$ and $Q$ on the screen in a Young's double-slit experiment,where the phase differences between two waves of the same amplitude are $\pi / 3$ and $\pi / 2$ respectively,is:

Light of wavelength $6000 \,\mathring A$ is incident on two slits. The distance between the slits is $0.1 \, cm$ and the screen is placed at a distance of $1 \, m$ from them. What is the distance between two consecutive minima in $mm$?