Solution:
Given,
Refractive index of the water (aμw) = 1.33,
The refractive index of the glass (aμg) = 1.53,
Now, the refractive index of glass with respect to water is
\({}_w\mu_g={}_w\mu_a\times{}_a\mu_g=\frac{{}_a\mu_g}{{}_a\mu_w}\)
Now, using Brewster's law, the polarizing angle is
\(\tan\left(\theta_p\right)=\mu=\frac{1.53}{1.33}\)
Or, \(\theta_p=\tan^{-1}\left(\frac{1.53}{1.33}\right)=49^\circ\)
Hence, the required polarizing angle is 49°