Answer:
Using Smell?s law for refraction from glass to air, \[\frac{\sin i}{\sin r}{{=}^{g}}{{\mu }_{a}}=\frac{\upsilon }{c}\] where c is the speed of light in air and v is the speed of light in glass. In the condition of critical incidence, we have \[i={{i}_{c}}\] and \[r={{90}^{\circ }}\] \[\therefore \] \[\frac{\sin {{i}_{c}}}{\sin {{90}^{\circ }}}=\frac{\upsilon }{c}\] or \[\sin {{i}_{c}}=\frac{\upsilon }{c}\] or \[{{i}_{c}}={{\sin }^{-1}}\left( \frac{\upsilon }{c} \right).\]
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