For a parallel beam of monochromatic light of wavelength `'lambda'` diffraction is produced by a single slit whose width `'a'` is of the order of the wavelength of the light. If `'D'` is the distance of the screen from the slit, the width of the central maxima will be

Options

A. `(2Dlambda)a`
B. `(Dlambda)/a`
C. `(Da)/lambda`
D. `(2Da)/lambda`

Answer

A

Explanation

For the condition of maxima `sin theta = lambda/a`

From the geometry
`sin theta=theta=y/D`
So, `y/D=lambda/a rArry=(lambdaD)/a`
Hence, width of central maxima `=2y=(2lambdaD)/a`