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........$mm$.

In $Y.D.S.E.,$ the slits are $0.2 \, cm$ apart and are illuminated by yellow light $(\lambda = 600 \, nm).$ What would be the fringe width on a screen placed $1 \, m$ from the plane of the slits if the whole system is immersed in water of refractive index $\mu = \frac{4}{3}$ (in $, mm$)?

In Young's double-slit experiment,the distance between the two slits is $3 \, cm$,the distance from the slits to the screen is $7 \, cm$,and the wavelength of light used is $1000 \, \mathring{A}$. Calculate the fringe width.

In Young's double-slit experiment,the maximum intensity is $I_0$. If one slit is closed,the new maximum intensity is:

$A$ double slit of separation $1.5\, mm$ is illuminated by white light (between $4500\, \mathring{A} - 7000\, \mathring{A}$). On a screen $120\, cm$ away,a coloured interference pattern is formed. If a pinhole is made on this screen at a distance $3.0\, mm$ from the central white fringe,which of the following wavelengths will be absent in the transmitted light? (in $\mathring{A}$)

In Young's double slit experiment,if one of the slits is closed fully,then in the interference pattern: