A) Fringe width has no relation with the diameter of wire
B) Increases
C) Decreases
D) Fringe width changes with change of wavelength only
Correct Answer: C
Solution :
Option [c] is correct. Explanation:\[\beta =\lambda \operatorname{D}/d\], where d is the diameter of the wire. So, if the diameter increases, fringe with decreases.You need to login to perform this action.
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