Question

The angle of polarization of glass is 58° and that for water is 53°. The angle of polarization for glass in water is

• A. $55.5°$
• B. ${\tan }^{-1}x\left[\frac{\tan -1\left({53}^{°}\right)}{\tan -1\left({58}^{°}\right)}\right]$
• C. ${\tan }^{-1}x\left[\frac{\tan -1\left({58}^{°}\right)}{\tan -1\left({53}^{°}\right)}\right]$
• D. $ta{n}^{-1}(55.5°)$

Answer 1

The Correct Option is C

Explanation:
Given:The angle of polarization for glass, ig = 58°The angle of polarization for water, iw = 53°According to Brewster's law, the angle of polarization is related to the refractive index of the transparent medium:$\tan \left(i_p\right)={\mu }_g$ .....(i)and when the medium is water:$\tan \left(i_w\right)=\mu w$ .......(ii)Now, when the glass is in water, the relative refractive index of the system is$\frac{{\mu }_g}{{\mu }_w}=\frac{\tan \left(i_g\right)}{\tan \left(i_w\right)}$$\Rightarrow \frac{{\mu }_g}{{\mu }_w}={\mu }_{gw}=\frac{\tan \left({58}^{\circ }\right)}{\tan \left({53}^{\circ }\right)}$Let the angle of polarization be ip when a glass is in water. Then we get$\tan \left(i_p\right)={\mu }_{gw}=\frac{\tan \left({58}^{\circ }\right)}{\tan \left({53}^{\circ }\right)}$or, $i_p={\tan }^{-1}\left[\frac{\tan \left({58}^{\circ }\right)}{\tan \left({53}^{\circ }\right)}\right]$Hence, the correct option is (C).