When unpolarised light ray is incident at an angle such that the angle between reflected of refracted rays is 90°, then reflected ray is linearly polarised. In that case incident angle is called polarising angle or Brewster angle(iP or iB).
$\sin\text{i}_{\text{c}}=\Big(\frac{1}{\sin\text{i}_{\text{c}}}=\mu_{\text{DR}}\Big)$
$\sin\text{i}_{\text{c}}=\mu_{\text{wg}}$
$\sin\text{i}_{\text{c}}=\frac{4}{3}\div\frac{3}{2}$
$=\frac{4}{3}\times\frac{2}{3}$
$\sin\text{i}_{\text{c}}=\frac{8}{9}=0.88$
Now, in this case
$\sin\text{i}=\sin60^\circ=\frac{\sqrt{3}}{2}=0.867$
$\because\ \sin\text{i}<\sin\text{i}_{\text{c}}$
$\text{So,}\ \text{i}<\text{i}_{\text{c}}$
So, ray will not suffer Total internal reflection.
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