In a single slit diffraction experiment first minima for `lambda_1 = 660nm` coincides with first maxima for wavelength `lambda_2`. Calculate the value of `lambda_2`.

Explanation

Position of minima in diffraction pattern is given by ,
`a sin theta = n lambda`
For first minima of `lambda_(1)` , we have
`a sin theta_(1) = (1) lambda_(1) ` or sin `theta_(1) = (lambda_(1))/(a)` . . . (1)
The first maxima approximately lies between first and second minima . For wavelength `lambda_(2)` its position will be
`a sin theta_(2) = (3)/(2) lambda_(2) :. sin theta_(2) = (3 lambda_(2))/(2 a)` . . . (ii)
The two will coincide if .
`theta_(1) = theta_(2) "or" sin theta_(1) = sin theta_(2)`
`:. (lambda_(1))/(a) = (3 lambda_(2))/(2a) ` or
`lambda _(2) = (2)/(3) lambda_(1) = (2)/(3) xx 660 nm = 440 nm `