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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:

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 $1.5\times 10^{-3}\:mm$.