A) 0.65 cm
B) 4.8 mm
C) 1.63 mm
D) 3.25 cm
Correct Answer: C
Solution :
Here the fringe width for nth bright fringe is \[X_{n}^{}=(2n-1)\frac{D\lambda }{2d}\] Now for fifth brigth fringe \[{{X}_{5}}=\frac{5D\lambda }{d}\] For nth dark fringe, the fringe width is given by \[X_{n}^{}=(2n-1)\frac{D\lambda }{2d}\] Now for third dark fringe \[X_{3}^{}=(6-1)\frac{D\lambda }{2d}=\frac{5D\lambda }{2d}\] \[{{X}_{5}}-{{X}_{3}}=\frac{5D\lambda }{d}-\frac{5}{2}\frac{D\lambda }{d}\] \[=\frac{5}{2}\frac{D\lambda }{d}=\frac{5\times 1\times 6.5\times {{10}^{-7}}}{2\times {{10}^{-3}}}\] \[=1.625\times {{10}^{-3}}m\] \[=1.625mm\] \[=1.63mm\]
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