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`.

Options

A. 4400 Å
B. 6600 Å
C. 2000 Å
D. 3500 Å

Answer

A

Explanation

In a single slit diffraction experiment, position of minima is given by `d sin theta=n lambda`
So, for first minima of red `sin theta=1 xx ((lambda_(R))/(d))` and as first maxima is midway between first and second minima, for wavelength `lambda` its position will be `d sin theta'=(lambda+2lambda)/(2) rArr sin theta' =(3lambda')/(2d)`
According to given condition `sin theta=sin theta' rArr lambda'=2/3 lambdaR`
So, `lambda' =2/3 xx 6600=440nm =4400 Å`