Two slits at a distance of $1\, mm$ are illuminated by light of wavelength $6.5 \times 10^{-7}\, m$. The interference fringes are observed on a screen placed at a distance of $1\, m$. The distance between the third dark fringe and the fifth bright fringe will be......$mm$.

In Young's double slit experiment,if the distance between the two slits is halved,then the fringe width will become

White light is passed through a double slit and interference is observed on a screen $1.5 \, m$ away. The separation between the slits is $0.3 \, mm$. The first violet and red fringes are formed $2.0 \, mm$ and $3.5 \, mm$ away from the central white fringes. The difference in wavelengths of red and violet light is $.... \, nm$.

In Young's double slit experiment,the phase difference between the light waves reaching the third bright fringe from the central fringe is: $(\lambda = 6000 \ \mathring{A})$

Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation (in $mm$) when blue-green light of wavelength $500 \; nm$ is used?

In Young's double-slit experiment,the distance between the two slits is $1 \, mm$ and the distance between the slits and the screen is $1 \, m$. If the wavelength of the light used is $500 \, nm$,then the distance of the third minimum from the central maximum is ... $mm$.