A) \[\theta <{{\sin }^{-1}}\frac{{{\mu }_{1}}}{{{\mu }_{2}}}\]
B) \[\theta <{{\sin }^{-1}}\sqrt{\frac{\mu _{2}^{2}}{\mu _{1}^{2}}-1}\]
C) \[\theta >{{\sin }^{-1}}\frac{{{\mu }_{1}}}{{{\mu }_{2}}}\]
D) \[\theta >{{\sin }^{-1}}\sqrt{\frac{\mu _{2}^{2}}{\mu _{1}^{2}}-1}\]
Correct Answer: B
Solution :
\[\sin c=\frac{{{\mu }_{1}}}{{{\mu }_{2}}}\] \[{{\mu }_{1}}\sin \theta ={{\mu }_{2}}\sin ({{90}^{o}}-C)\] \[\sin \theta =\frac{{{\mu }_{2}}\sqrt{1-\frac{\mu _{1}^{2}}{\mu _{2}^{2}}}}{{{\mu }_{1}}}\] \[\theta <{{\sin }^{-1}}\sqrt{\frac{\mu _{2}^{2}-\mu _{1}^{2}}{\mu _{1}^{2}}}\] For TIR \[\theta <{{\sin }^{-1}}\sqrt{\frac{\mu _{2}^{2}}{\mu _{1}^{2}}-1}\]You need to login to perform this action.
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