A beam of light contains two wavelengths `6500Å` and `5200Å` They form interference fringes in Young's double slit experiment. What is the least distance (approx) from central maximum. Where the bright fringes due to both wavelength coincide? The distance between the slits is 2 mm and the distance of screen is 120 cm.

Explanation

The position of nth bright fringe on the screen is
`y_(n)=(n lambda D)/(d)`
Let the nth bright fringe of 6500 Å and the nth bright fringe of 5200 Å coicide, then
`(m xx 6500 xx D)/(d)=(n xx 5200 xx D)/(d)`
`(m)/(n)=(5200)/(6500)=(4)/(5)`
Thus the minimum values of m and n are 4 and 5 respectively.
Hence `y=(4xx6500xx120)/(0.2)=0.156 cm = 1.56 mm`