A) 1.62 mm
B) 2.62 mm
C) 5.62 mm
D) 3.62 mm
Correct Answer: A
Solution :
Distance of third dark fringe \[{{x}_{1}}=\left( 3-\frac{1}{2} \right)\,\frac{D\lambda }{d}=\frac{5}{2}\times \frac{1\times 650\times {{10}^{-9}}}{1\times {{10}^{-3}}}\] = 1.62 mm Distance of fifth bright fringe \[{{x}_{2}}=5\frac{D\lambda }{d}=3.24mm\] \[\therefore \] Required distance \[={{x}_{2}}-{{x}_{1}}\] \[=3.24=1.62\,\,mm\]
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