question_answer

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 a the of the central maxima will be                                                                                                      [NEET  2015]

A)  \[\frac{2\lambda D}{a}\]           

B)  \[\frac{D\lambda }{a}\]

C)  \[\frac{Da}{\lambda }\]            

D)       \[\frac{2Da}{\lambda }\]

Correct Answer: A

Solution :

For the condition of maxima

\[\sin \theta =\frac{\lambda }{a}\]

From the geometry,

\[\sin \theta =\theta =\frac{Y}{D}\]         (for small angle)

So,       \[\frac{Y}{D}=\frac{\lambda }{a}\]

\[\Rightarrow \]   \[\,Y=\frac{\lambda D}{a}\]

Hence, width of central maxima \[=2Y=\frac{2\lambda D}{a}\]