Question

The critical angle of refraction for calcite is 68.4 when it forms a boundary with water. Use this information to determine the speed of light in Calcite. Assume the refractive index of water ,1.333, is less than the refractive index of calcite.

Answer 1

\[\color{green}{\dfrac{\sin \angle i_c}{\sin 90^{\circ}} = \dfrac{1.33}{n}}\] Solve for n.

Answer 2

what is the original equation that we use ?

Answer 3

Snell's Law:
\[\color{green}{\dfrac{\sin \angle i_c}{\sin \angle \text{R}} = \dfrac{\text{n}_2}{\text{n}_1}}\]

Answer 4

ok thanks :), How do you medal?

Answer 5

what do I substitute for sin i then ?

Answer 6

\[\color{green}{\dfrac{\sin 68.4^{\circ}}{\sin 90^{\circ}} = \dfrac{1.33}{n}}\]

Answer 7

thank you