In Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ bright fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

Light of wavelength $520\, nm$ passing through a double slit produces an interference pattern of relative intensity versus deflection angle $\theta$ as shown in the figure. The separation $d$ between the slits is

Young's double slit experiment is first performed in air and then in a medium other than air. It is found that the $8^{th}$ bright fringe in the medium lies where the $5^{th}$ dark fringe lies in air. The refractive index of the medium is nearly:

In a Young's double slit experiment,the ratio of the amplitude of light coming from the slits is $2:1$. The ratio of the maximum to minimum intensity in the interference pattern is:

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 $, mm$)?

The distance between two slits in a double slit experiment is $1 \ mm$. The distance between the slits and the screen is $1 \ m$. If the distance of the $10^{th}$ fringe from the central fringe is $5 \ mm$, then the wavelength of light is $... \mathring{A}$.