A) \[\sin \,\,i\,\,=\,\,\frac{{{\mu }_{2}}}{{{\left( \mu _{1}^{2}+\mu _{2}^{2} \right)}^{1/2}}}\]
B) \[\tan \,\,i\,\,=\,\,\frac{{{\mu }_{1}}}{{{\mu }_{2}}}\]
C) \[\sin \,\,i\,\,=\,\,{{\mu }_{1}}{{\mu }_{2}}\]
D) \[\sin \,\,i\,\,=\,\,\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\]
Correct Answer: A
Solution :
By theory Brewster angle \[\mu =\,\,\tan \,\,{{i}_{p}}\] \[\frac{{{\mu }_{1}}}{{{\mu }_{2}}}\,\,=\,\,\tan \,\,{{i}_{p}}\] \[\therefore \,\,\,\,\sin \,\,i\,\,=\,\,\frac{{{\mu }_{2}}}{{{(\mu _{1}^{2}\,+\,\,\mu _{2}^{2})}^{1/2}}}\]
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