Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

In Young's double slit experiment,white light is used. The separation between the slits is $b$. The screen is at a distance $d$ $(d >> b)$ from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are

If yellow light emitted by a sodium lamp in Young's double-slit experiment is replaced by monochromatic blue light of the same intensity,then:

In Young's double slit experiment,when the wavelength used is $6000 \ \mathring{A}$ and the screen is $40 \ cm$ from the slits,the fringes are $0.012 \ cm$ wide. What is the distance between the slits in $cm$?

In Young's double-slit experiment,if the amplitudes of the interfering waves are not equal,then . . . . .

Two light waves of wavelengths $800 \, nm$ and $600 \, nm$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 \, m$ away from the plane of the slits. If the two slits are separated by $0.35 \, mm$,then the shortest distance from the central bright maximum to the point where the bright fringes of the two wavelengths coincide will be $............. \, mm$.