(a) Light reflected at 62.5º from a gemstone in a ring is completely polarized. Can the gem be a diamond?

(b) At what angle would the light be completely polarized if the gem was in water?

Polarization is a phenomenon generated by the wave nature of electromagnetic radiation, according to physics.

(a) To determine if a gemstone is a diamond or not, we calculate its index of redfraction and compare it to that of a diamond.

\(\begin{aligned}{n_2} &= \tan \left( {{\theta _b}} \right){n_1}\\ &= \tan \left( {{{62.5}^\circ }} \right) \times 1.00\\ &= 1.92\end{aligned}\)

The index of red fraction that we obtained is for zircon, not diamond.

(b)

We utilise Brewster's law to calculate the Brewster angle for a diamond in water, where\({\rm{n1}}\)is the index of redfraction for water.

\(\begin{aligned}{\rm{tan}}\left( {{{\rm{\theta }}_{\rm{b}}}} \right){\rm{ = }}\dfrac{{{{\rm{n}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{1}}}}}\\{\rm{ = }}\dfrac{{{\rm{1}}{\rm{.92}}}}{{{\rm{1}}{\rm{.33}}}}\\{\rm{ = 1}}{\rm{.44}}\end{aligned}\)

TheBrewster's angle for the gem in water is,

\(\begin{aligned}{\theta _b} &= {\tan ^{ - 1}}(1.44)\\ &= {55.3^\circ }\end{aligned}\)

Therefore, the Brewster's angle for the gem in water is \({\rm{55}}{\rm{.}}{{\rm{3}}^{\rm{^\circ }}}{\rm{.}}\)