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 \[\mu \propto \frac{1}{\lambda }\]
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