A) \[1.5\]
B) \[\sqrt{2}\]
C) \[1.33\]
D) \[\sqrt{3}\]
Correct Answer: B
Solution :
\[AB=BC\] \[\therefore \] \[\angle A=\angle C={{45}^{o}}\] MN is normal drawn at the point of incidence. By geometry \[\angle LMN=45{}^\circ \] The ray is suffering total internal reflection, \[\therefore \] \[C=45{}^\circ \] So, \[\mu =\frac{1}{\sin C}\] \[=\frac{1}{\sin {{45}^{o}}}=\frac{1}{1/\sqrt{2}}=\sqrt{2}\]You need to login to perform this action.
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