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 =nlambda`
For first minima of `lambda_(1)` we have
`a sin theta_(1)=(1)lambda_(1) or sin theta_(1)=lambda_(1)/a`
The first maxima approximately lies between first and second maxima. For wavelength `lambda_(2)` its position will be
`a sin theta_(2)=3/2 lambda_(2)`
`sin theta_(2)=(3lambda_(2))/(2a)`
The two will coincide, if
`theta_(1)=theta_(2) or sin theta_(1)=sin theta_(2)`
`:. lambda_(1)/a=(3lambda_(2))/(2a)`
or `lambda=2/3 lambda_(1)=2/3xx660nm=440nm`