A) zero
B) \[\pi /2\]
C) \[\pi \]
D) \[2\pi \]
Correct Answer: C
Solution :
The phase difference \[(\phi )\] between the wavelets from the top edge and bottom edge of the slit is \[\phi =\frac{2\pi }{\lambda }(d\sin \theta )\]where d is the 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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