Question

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 he screen from the slit, the width of the central maxima will be

• A. $\frac{Da}{\lambda}$
• B. $\frac{2Da}{\lambda}$
• C. $\frac{ 2D \lambda}{a}$
• D. $\frac{ D\lambda }{ a}$

Answer 1

Answer: (C)

$\frac{ 2D \lambda}{a}$

Answer 2

Given situation is shown in the figure.
For central maxima , $\sin \theta = \frac{\lambda}{ a}$
Also, $\theta$ is very-very small so
$sin \theta = \tan \theta = \frac{y}{D}$
$\frac{y}{D} = \frac{\lambda}{ a} , y= \frac{ \lambda D}{a}$
Width of central maxima $= 2y = \frac{ 2 \lambda D }{ a}$