A) \[0\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) \[2\pi \]
Correct Answer: D
Solution :
The phase difference \[(\phi )\] between the wavelets from the top edge and the bottom edge of the slit is \[\phi =\frac{2\pi }{\lambda }(d\sin \theta )\] where \[d\]is the slit width. The first minima of the diffraction pattern occurs at\[\sin \theta =\frac{\lambda }{d},\]so\[\phi =\frac{2\pi }{\lambda }\left( d\times \frac{\lambda }{d} \right)=2\pi \].You need to login to perform this action.
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