Two slits are separated by a distance of $0.5\, mm$ and illuminated with light of $\lambda = 6000\ \mathring A$. If the screen is placed $2.5\, m$ from the slits,the distance of the third bright fringe from the centre will be........$mm$.

$A$ double slit interference experiment performed with a light of wavelength $600 \ nm$ forms an interference fringe pattern on a screen with $10^{\text{th}}$ bright fringe having its centre at a distance of $10 \ mm$ from the central maximum. The distance of the centre of the same $10^{\text{th}}$ bright fringe from the central maximum when the source of light is replaced by another source of wavelength $660 \ nm$ would be . . . . . . $mm$.

In a Young's double-slit experiment, the angular width of a fringe is $1^\circ$ for light of wavelength $6000 \, \mathring{A}$. What is the distance between the slits in $mm$?

$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 \,mm$, and the distance between the slit and the screen is $1.33 \,m$. The slits are illuminated by light of wavelength $6300 \,Å$. The fringe width is:

In a Young's double slit experiment,a laser light of $560\,nm$ produces an interference pattern with consecutive bright fringes' separation of $7.2\,mm$. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of $8.1\,mm$. The wavelength of the second light is $......nm$.

In a Young's double slit experiment,the angular width of a fringe is $0.35^{\circ}$ on a screen placed at $2\,m$ away for a particular wavelength of $450\,nm$. The angular width of the fringe,when the whole system is immersed in a medium of refractive index $7/5$,is $\frac{1}{\alpha}$. The value of $\alpha$ is ..............