question_answer

A light beam is incident on the boundary between two transparent media. At a particular angle of incidence the reflected ray is perpendicular to the refracted ray. Obtain an expression for this angle of incidence. Does this angle depend on the wavelength of light used?

Answer:

                                Refer to Fig. The angle of incidence is equal to the polarising or Brewster angle \[({{i}_{p}})\]. If \[{{r}_{p}}\] is the angle of refraction, then \[{{i}_{p}}+{{r}_{p}}={{90}^{\circ }}\] or \[{{r}_{p}}={{90}^{\circ }}-{{i}_{p}}\] \[\therefore \]  \[\mu =\frac{\sin {{i}_{p}}}{\sin {{r}_{p}}}=\frac{\sin {{i}_{p}}}{\sin ({{90}^{\circ }}-{{i}_{p}})}\] \[=\frac{\sin {{i}_{p}}}{\cos {{i}_{p}}}=\tan {{i}_{p}}\] \[\therefore \]  \[{{i}_{p}}={{\tan }^{-1}}(\mu )\] Yes, \[{{i}_{p}}\] depends on wavelength of light used because \[\mu \propto \frac{1}{\lambda }\]