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                                                                                             

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