The Brewster's angle $i_b$ for an interface should be

For a given medium, the polarizing angle is $60^o$. What will be the critical angle for this medium?

Given below are two statements:
Statement $I$: If the Brewster's angle for the light propagating from air to glass is $\theta_B$,then Brewster's angle for the light propagating from glass to air is $\frac{\pi}{2} - \theta_B$.
Statement $II$: The Brewster's angle for the light propagating from glass to air is $\tan^{-1}(\mu_g)$ where $\mu_g$ is the refractive index of glass.
In the light of the above statements,choose the correct answer from the options given below:

The velocity of light in air is $3 \times 10^8 \, ms^{-1}$ and that in water is $2.2 \times 10^8 \, ms^{-1}$. The polarising angle of incidence is.......$^o$

Explain the polarisation of light by reflection,state Brewster's law,and derive its formula.

When unpolarized light is incident from air onto glass $(\mu = 1.5)$,then: