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 = \[2Y=\frac{2\lambda D}{a}\]You need to login to perform this action.
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