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

Answer

C

Explanation

Given situation is shown in the figure.

For central maxima,
`sintheta=(lambda)/(a)`
Also, `theta` is very-very small so `sintheta~~tantheta=(y)/(D)`
`:.(y)/(D)=(lambda)/(a),y=(lambdaD)/(a)`
Width of central maxima `=2y=(2lambdaD)/(a)`.