A) 2 mm
B) 2 cm
C) 2 m
D) 2 km
Correct Answer: A
Solution :
For vacuum \[t=n\,{{\lambda }_{o}}\] .....(i) For air \[t=(n+1)\,{{\lambda }_{a}}\] .....(ii) From equation (i) and (ii) \[t=\frac{\lambda }{\mu -1}=\frac{6\times {{10}^{-7}}}{1.0003-1}\] \[\left( \mu =\frac{{{\lambda }_{o}}}{{{\lambda }_{a}}} \right)\] \[=2\times {{10}^{-3}}m\] = 2mm.
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