question_answer

The critical angle between a given transparent medium and air is denoted by \[{{i}_{c}}\]. A ray of light in air medium enters this transparent medium at an angle of incidence equal to the polarizing angle \[({{i}_{p}}).\] Deduce a relation for the angle of refraction \[({{r}_{p}})\] in terms of \[{{i}_{c}}\].

Answer:

                                                      According to Brewster law, when a ray of light is incident on a transparent refracting medium at polarising angle, \[\mu =\tan {{i}_{p}}\]                 But         \[{{I}_{p}}+{{r}_{p}}={{90}^{\circ }}\] or \[{{i}_{p}}={{90}^{\circ }}-{{r}_{p}}\]                 \[\therefore \]  \[\mu =\tan ({{90}^{\circ }}-{{r}_{p}})=\cot {{r}_{p}}=\frac{1}{\tan {{r}_{p}}}\]?.(i)                 As \[{{i}_{c}}\]is the critical angle for the transparent medium, so                 \[\mu =\frac{1}{\sin {{i}_{c}}}\]                                                 ? (ii)                 On comparing (i) and (ii) , we get \[\tan {{r}_{p}}=\sin {{i}_{c}}\] or \[{{r}_{p}}={{\tan }^{-1}}(\sin {{i}_{c}})\].