In Young's double-slit experiment, light of wavelength $4000 \, Å$ is used to produce bright fringes of width $0.6 \, mm$ at a distance of $2 \, m$ from the slits. If the whole apparatus is immersed in a liquid of refractive index $1.5$, what will be the new fringe width 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

In a Young's double-slit experiment,the slits are placed $0.320 \, mm$ apart. Light of wavelength $\lambda = 500 \, nm$ is incident on the slits. The total number of bright fringes that are observed in the angular range $-30^\circ \le \theta \le 30^\circ$ is:

In Young's double-slit experiment with a source of light of wavelength $6320 \ \mathring{A}$,the first maxima will occur when:

In Young's double slit interference experiment,the slit separation is made $3$ folds. The fringe width becomes ......... times.

$A$ Young's double-slit experimental setup is immersed in water of refractive index $1.33$. It has a slit separation of $1 \ mm$ and the distance between the slits and the screen is $1.33 \ m$. If the wavelength of incident light on the slits is $6300 \ \mathring{A}$,then the fringe width on the screen is: