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)` …..(i)
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)= (3lambda_(2))/(2a)` .....(ii)
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)= (2)/(3)lambda_(1)= (2)/(3)xx660 nm = 440 nm`