Answer:
As shown in Fig. when \[i={{i}_{c}},r={{90}^{\circ }}\] Using Snell?s law of refraction, \[{{\mu }_{1}}\sin {{i}_{c}}={{\mu }_{2}}\sin {{90}^{\circ }}\] or \[\sin {{i}_{c}}=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\] or \[{{i}_{c}}={{\sin }^{-1}}\left( \frac{{{\mu }_{2}}}{{{\mu }_{1}}} \right)\]
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