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Show that when a ray of light is incident on the surface of a transparent medium at the polarizing angle, the reflected and transmitted rays are perpendicular to each other.                               

Answer:

                From Snell?s law, \[\frac{\sin {{i}_{p}}}{\sin {{r}_{p}}}=\mu \] From Brewster law, \[\tan {{i}_{p}}=\frac{\sin {{i}_{p}}}{\cos {{i}_{p}}}=\mu \] \[\therefore \]  \[\frac{\sin {{i}_{p}}}{\sin {{r}_{p}}}=\frac{\sin {{i}_{p}}}{\cos {{i}_{p}}}\]or \[\sin {{r}_{p}}=\cos {{i}_{p}}\] or            \[\sin {{r}_{p}}=\sin ({{90}^{\circ }}-{{i}_{p}})\] \[\therefore \]  \[{{r}_{p}}={{90}^{\circ }}-{{i}_{p}}\] or \[{{i}_{p}}+{{r}_{p}}={{90}^{\circ }}\] Hence the reflected and transmitted rays are perpendicular to each other.