A) 0.2
B) 0.4
C) 0.8
D) 0.9
Correct Answer: C
Solution :
Here, \[\sin {{\theta }_{ic}}/\sin {{\theta }_{iB}}=1.28\] As we know, \[\mu =\frac{\sin {{\theta }_{iB}}}{\sin \left( \frac{\pi }{2}-{{\theta }_{iB}} \right)}\]where, \[{{\theta }_{iB}}\]is Brewster?s angle of incidence, And, \[\mu =\frac{1}{\sin {{\theta }_{ic}}}\] On solving we get, relative refractive index of the two media.You need to login to perform this action.
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