In a single slit diffraction expriment, first minimum for red light `(660 nm)` coincide with first maximum of some other wavelength `lambda'`. Calculate `lambda'`.

Answer

`440.00`

Explanation

Secondary minimum
`sin theta = (n lambda)/(alpha)`
`sin theta = (nlambda_(r))/(d)`
`n = 1`
`sin theta = (lambda_r)/(alpha)` ……….(i)
Secondary maximum
`sin theta = ((2n+1)lambda)/(2alpha)`
`n =1`
`sin theta = (3 lambda)/(2d)` ........(ii)
(i) = (ii)
`(lambda_r)/(d) = (3lambda)/(2d)`
`lambda = (2lambda_r)/(3)`
`=(2xx660xx10^(-9))/3`.