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, ` d sin theta =n lamda`
For first minima of `lamda_1` , we have
`d sin theta_1 = (1) lamda_1 " or " sin theta_1 =(lamda_1)/(d)` .....(i)
The first maxima approximately lies between first and second minima. For wavelength `lamda_2` its position will be ,
`d sin theta_2 = 3/2 lamda_2 therefore sin theta_2 = (3 lamda_2)/(2d)` ...(ii)
The two will coincide if ,
`theta_1 = theta_2 " or " sin theta_1 = sin theta_2 therefore (lamda_1)/(d) =(3lamda_2)/(2d)`
`lamda_2 = 2/3 lamda_1 = 2/3 x 660 nm = 440 nm`