A beam of light consisting of two wavelength, `6500 Å` and `5200 Å` is used to obtain interference fringes in a Young's double slit experiment. Find the distance of the third fringe on the screen from the central maximum for the wavelength `6500Å`. What is the least distance from the central maximum at whixh the bright fringes due to both wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.

Explanation

`lambda_1 = 6500 Å`
`lambda_2 = 5200 Å`
`d = 2 mm = 2 xx 10^(-3) m, D = 120 cm = 120 xx 10^(-2)m`
For third fringe (for wavelength `lambda_1`)
`beta = (3 lambda_1 D)/d = (3 xx 6500 xx 10^(-10) xx 120 xx 10^(-2))/(2 xx 10^(-3))`
`= 1170000 xx 10^(-9) = 1.17 xx 10^(-3) m`.
Let x be the distance from central maxima at which bright fringes overlap.
`:. x = (n lambda_1 D)/d = (n + 1) (lambda_2 D)/(d)`
`n/(n + 1) = (lambda_2)/(lambda_1) = (5200 xx 10^(-10))/(6500 xx 10^(-10))`
Therefore, `x = (n lambda_1 D)/d = (4 xx 6500 xx 10^(-10) xx 120 xx 10^(-2))/(2 xx 10^(-3))`
`= 1.56 xx 10^(-3) m`.