A) \[30{}^\circ \]
B) \[60{}^\circ \]
C) \[90{}^\circ \]
D) \[0{}^\circ \]
Correct Answer: B
Solution :
Angular width of central maximum\[=2\theta \] Now, \[\sin \theta =\frac{\lambda }{a}\] \[\sin \theta =\frac{6000\times {{10}^{-10}}}{12\times {{10}^{-7}}}\] \[\sin \theta =0.5\] \[\theta =30{}^\circ \] Hence, \[2\theta =60{}^\circ \]You need to login to perform this action.
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