Home exercise 6.4.1: Why a diamond sparkles

${\theta }_c=\arcsin (1/2.42)=24.4^{\circ }$.

Air to diamond: ${\theta }_B=\arctan 2.42=67.5^{\circ }$. Diamond to air: ${\theta }_B^{\prime }=\arctan (1/2.42)=22.5^{\circ }$. Since $\tan {\theta }_B^{\prime }=1/\tan {\theta }_B=\cot {\theta }_B$, the two angles are complementary: $67.5^{\circ }+22.5^{\circ }=90^{\circ }$. They are also the two angles ${\theta }_i$, ${\theta }_t$ of one and the same ray, crossed in opposite directions.

$R=(\frac{2.42-1}{2.42+1}{)}^2=0.172=17.2\%$.

Any internal ray meeting a facet at more than $24.4^{\circ }$ is totally reflected. In a brilliant cut the lower (pavilion) facets make about $41^{\circ }$ with the plane of the girdle. A ray entering the top (table) at normal incidence meets the first pavilion facet at $41^{\circ }$ and, after reflection, the opposite facet at $180^{\circ }-3\times 41^{\circ }=57^{\circ }$: both beyond $24.4^{\circ }$, both total. It then travels upwards at $180^{\circ }-4\times 41^{\circ }=16^{\circ }$ to the vertical, below ${\theta }_c$, and leaves through the top.

${\theta }_B^{\prime }=22.5^{\circ }<{\theta }_c=24.4^{\circ }$, as it must be for any pair of media (Subsection 6.4.10: $\sin {\theta }_B<\tan {\theta }_B=\sin {\theta }_c$). The ray in the figure is drawn to scale: it leaves the table at $16^{\circ }$ inside the stone and, by Snell's law, at $\arcsin (2.42\sin 16^{\circ })=41.8^{\circ }$ in the air.