Problem 6.90: In Young’s double slit experiment two narrow slits $0.8\:mm$ apart are illuminated by the same source of yellow light of wavelength $600\:nm$. How far apart are the adjacent bright bands in the interference pattern observed on a screen $2\:m$ away?

Solution:

29-11-2015A

The problem demands the calculation of distance between adjacent bright bands which is known as fringe width or bandwidth. In Young’s double slit experiment, this is calculated using the equation $\beta\:=\:\frac{\lambda\:D}{d}$ where $\lambda$ is the wavelength of light used, $D$ is the distance to the screen and $d$ is the distance between the slits.

It is given in the problem that  wavelength of light used $\lambda\:=\:600\:nm\:=\:600\times 10^{-9}\:=\:6\times 10^{-7}\:m$

The distance to the screen $D\:=\:2\:m$

The distance between the slits $d\:=\:0.8\:mm\:=\:0.8\times 10^{-3}\:=\:8\times 10^{-4}\:m$.

Using these given values, fringe width $\beta\:=\:\frac{6\times 10^{-7}\times 2}{8\times 10^{-4}}\:=\:0.0015\:m\:=\:1.5\times 10^{-3}\:mm$.

So the bright bands formed on the screen are separated by  a distance $1.5\times 10^{-3}\:mm$.