In Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength `lamda_0` = 750 nm and `lamda` = 900 nm. What is the minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other?

Explanation

Given data: The fringe width `(beta_1) " for " lamda_1 = 900 nm `
` = 900 xx 10^(-9) m`
The fringe width ` (beta_2) " for " lamda_2 = 750 nm`
` = 750 xx 10^(-9) m`
(i) This means that the central maxima of two coincide, but first maximum for `lamda_1 = 900 xx 10^(-9)` m will be further away from the first maxima for `lamda_2` = 750 nm and so on. (ii) A stage may come when this mismatch equals to `beta^2` then again maxima of `lamda_1 = 900 nm` will coincide with a maxima of `lamda_2` = 750 nm let this correspond to ` n^(th)` order fringe for 11. Then it will correspond to `(n+1)^(th)` order fringe for 12.
` therefore (n lamda_1 D)/(d) = ((n+1) lamda_2 D)/(d)`
`n lamda_1 = (n+1) lamda_2`
`lamda_1/lamda_2 = (n+1)/(n) rArr (1+ 1/n) = lamda_1/lamda_2`
` 1+ 1/n = (900 xx 10^(-9))/(750 xx 10^(-9)) = 6/5`
`1+ 1/n = 1 + 1/ 5`
` rArr n = 5`
Minimum distance from central maxima
` = therefore (nlamda_1D)/(d)`
` = ( 5 xx (900 xx 10^(-9) ) xx 2)/(2 xx 10^(-3))`
` = 45 xx 10^(-4)m = 4.5 mm`