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 give by, `d sin theta = n lambda`. For first minima of `lambda_(1)`, we have
`d sin theta_(1)= (1)lambda_(1)" or "sin theta_(1)= (lambda_1)/(d)" ""……….."(i)`
The first manima approximately lies between first ans second minima. For wavelength `lambda_(2)` its position will be,
`d sin theta_(2)= (3)/(2)lambda_(2)" "therefore sin theta_(2)= (3 lambda_2)/(2d)" ""..........."(ii)`
The two will coincide if, `theta_(1)=theta_(2)" or "sin theta_(1)= sin theta_(2)`
`therefore (lambda_1)/(d)= (3lambda_(2))/(2d)" or "lambda_(2)= (2)/(3)lambda_(1)= (2)/(3)xx 660 nm= 440nm`.