91Ó°ÊÓ

The light could be polarized by total reflection. This is Brewster’s law which states that if the incident angle equals the polarized angle, the light will be polarized. The polarization occurs due to reflection. Brewster’s law is given by

$\tan {\mathrm{θ}}_p=\frac{n_b}{n_a}$

Where $n_b$is the refractive index of the plastic wall while$n_a$ is the refractive index of air and it equals 1 Now, substitute the values for$n_a$ and$n_b$ into equation to get${\mathrm{θ}}_p$

${\mathrm{θ}}_b={\tan }^{-1}\left(\frac{n_b}{n_a}\right)={\tan }^{-1}\left(\frac{1.61}{1}\right)=58.15°$

As we mentioned above, the incident angle at the wall equals the polarized angle.

So, we can consider this angle to be the angle of the incident at the wall. Hence the incident angle on the water${\mathrm{θ}}_a$ will be

${\mathrm{θ}}_a=90°-58.15°=31.85°$

When the light is reflected and travels to the water, the refraction occurs where the refractive index of water is $n_W$. The refractive index of an optical material is expressed by n and the speed of light in the vacuum divided by the speed of light in the material. Snell’s law is given in the form

$n_a\sin {\mathrm{θ}}_a=n_w\sin {\mathrm{θ}}_w$

Now substitute the values$n_w,n_a$ and${\mathrm{θ}}_a$ .

${\mathrm{θ}}_b={\sin }^{-1}\left(\frac{n_a\sin {\mathrm{θ}}_a}{n_w}\right)={\sin }^{-1}\left(\frac{\left(1\right)\sin 31.85°}{1.333}\right)=23.3°$

Hence, the angle this beam makes with the normal inside the water is$23.3°$ .