In a single slit diffraction experiment first minimum for red light `(660nm)` coincides with first maximum of some other wavelength `lambda`'. The value of `lambda`' is

Options

A. `4400 A^(@)`
B. `6600 A^(@)`
C. `2000 A^(@)`
D. `3500 A^(@)`

Answer

A

Explanation

In a single slit diffraction experiment, position of minima is given by `dsin theta = nlambda`.
So, for first minima of red, `sintheta = 1 xx (lambda_(R)/d)`.
And as first maxima is midway between first and second minima, for wavelength A., its position will be
`dsintheta^(.) = (lambda. + 2lambda.)/2 rArr sintheta^(.) = (3lambda^(.))/(2d)`.
According to given condition `sin theta = sin theta.`
`rArr lambda^(.) = 2/3 lambda_(2)`
So, `lambda^(.) = 2/3 xx 6600 = 440 nm = 4400 A^(@)`.