A) \[\frac{B}{4}\]
B) B
C) \[\frac{B}{2}\]
D) \[\frac{3B}{2}\]
Correct Answer: C
Solution :
Position of\[{{n}^{th}}\]bright fringe \[{{x}_{1}}=\frac{n\lambda D}{d}\] For first bright fringe\[n=1\] \[\therefore \] \[{{x}_{1}}=\frac{\lambda D}{d}\] Position of\[{{n}^{th}}\]dark fringe\[{{x}_{2}}=\frac{(2n-1)\lambda D}{2d}\] For first dark fringe \[n=1\] \[\therefore \] \[{{x}_{2}}=\frac{\lambda D}{2d}\] Now, \[{{x}_{1}}-{{x}_{2}}=\frac{\lambda D}{2d}\] If B is the band width, then \[{{x}_{1}}-{{x}_{2}}=\frac{B}{2}\]
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