# A beam of light passes from zircon to quartz at an incident angle of 8°. What is the angle of refraction?

Jan 4, 2017

${9.4}^{\circ}$, rounded to one decimal place.

#### Explanation:

Refractive index of Zircon and Quartz varies as below

Zircon: $1.81 - 2.024$
Quartz: $1.544 - 1.553$

For this problem the refractive indices of samples are assumed as lowest.

From Snell's law we know that when a beam passes from a medium $1$ having refractive index ${n}_{1}$ making and angle ${\theta}_{1}$ to medium $2$ having refractive index ${n}_{2}$, angle of refraction ${\theta}_{2}$ is given by the equation

${n}_{1} \sin {\theta}_{1} = {n}_{2} \sin {\theta}_{2}$

Inserting given values we get

$1.81 \sin {8}^{\circ} = 1.544 \sin {\theta}_{2}$
$\implies \sin {\theta}_{2} = \frac{1.81 \sin {8}^{\circ}}{1.544}$
$\implies {\theta}_{2} = {\sin}^{-} 1 \left(\frac{1.81 \sin {8}^{\circ}}{1.544}\right)$
$= {9.4}^{\circ}$, rounded to one decimal place.