Question

A ray of light is traveling through a mineral sample is submerged inwater. The ray refracts as it enters the water, as shown in the diagrambelow.NormalWater41°149°63°27°MineralCalculate the absolute index of refraction of the mineral.

Answer 1

We are asked to determine the absolute index of refraction of a mineral submerged in water. To do that we will use Snell's law:

$n_1\sin \theta_1=n_2\sin \theta_2$

Where n1 and n2 are the refraction indices of water and mineral respectively and the angles "theta 1" and "theta 2" are the incidence and refraction angles. We will solve for n1:

$n_1=\frac{n_2\sin \theta_2}{\sin \theta_1}$

Replacing the values:

$n_1=\frac{(589.29nm)\sin 41}{\sin 27}$

Solving the operations:

$n_1=851.58nm$

Therefore, the index of refraction of the mineral is 851.58 nm.