In a young's double slit experiment, two wavelength of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas coincide again ? Take `D//d = 10^(3)` . Symbols have their usual meanings.

Answer

`3.5mm`

Explanation

Let `n^(th)` maxima of `500nm`coincides with `m^(th)` maxima of`700mm.`
`(nlambda_(1)D)/(d)=(mlambda_(2)D)/(d)rArr (n)/(m)=(lambda_(2))/(lambda_(1))rArr(700)/(400)rArr (n)/(m) =(7)/(4)`
`7^(th)` maxiam of `500nm` concides with`4^(th)`maxima of `700nm` next `14^(th)`maxima of `500nm`coinaides with `8^(th)`maxima of `700nm`=`x_(14)^(th)-x_(7)^(th)`
minimum supertion between two=`x_(14)^(th)-x_(7)^(th)`
`x_(14)^(th)-x_(7)^(th)=14beta-7beta rArr x_(14)^(th)-x_(7)^(th)=7beta`
`x_(14)^(th)-x_(7)^(th)=7(lambdaD)/(d) rArr x_(14)^(th)-x_(7)^(th)=7xx5xx10^(-7)xx10^(3)`
`x_(14)^(th)-x_(7)^(th)=3.5xx10^(-3)m rArr x_(14)^(th)-x_(7)^(th)=3.5mm`