Answer:
Polarizing angle,
![\theta_(P) = tan^(- 1){2.2} = 65.56^(\circ)](https://img.qammunity.org/2020/formulas/physics/college/30mtqbdx09ekr9smzejpxvrmyaeyfxeaf7.png)
Given:
Critical angle,
![\theta_(cr) = 27^(\circ)](https://img.qammunity.org/2020/formulas/physics/college/2pidse7hpg0bu7yjol48a54zj36x21gro3.png)
Solution:
Now, in Total Internal Reflection (TIR), the critical angle for cubic zirconia is given by:
(1)
where
= refractive index of zirconia
From eqn (1):
![\mu_(Z) = (1)/(sin\theta_(cr))](https://img.qammunity.org/2020/formulas/physics/college/nkedigzja0fvibqk97hqyp1pgdjnsoyzus.png)
![\mu_(Z) = (1)/(sin(27^(\circ))) = 2.2](https://img.qammunity.org/2020/formulas/physics/college/8huhnblx1wy3zbte15lw0gisxlwllzz5hx.png)
Now, the angle of polarization is given by:
tan
![\theta_(P) = \mu_(Z)](https://img.qammunity.org/2020/formulas/physics/college/lq35d3vtuow0kiwwsm87io3xxfn0p30pfr.png)
Therefore,
![\theta_(P) = tan^(- 1){2.2} = 65.56^(\circ)](https://img.qammunity.org/2020/formulas/physics/college/30mtqbdx09ekr9smzejpxvrmyaeyfxeaf7.png)