# What is the index of refraction of a refractive medium if the angle of incidence in air is 30° and the angle of refraction is 15°?

May 21, 2017

Okay. Let's go ahead and write out Snell's law for our first step which has been altered to fit this probelm

color(white)(------)color(magenta)(n_(air)sin(30^(@))= n_(2)sin(15^(@))

Where
${n}_{a i r} = \text{index of refraction of air (1.00)}$
$\sin \left({30}^{\circ}\right) = \text{angle of incidence in air}$
${n}_{2} = \text{index of refraction of medium}$
$\sin \left({15}^{\circ}\right) = \text{angle of refraction in medium}$

Since we want to figure out the index of refraction, we will rearrange to get ${n}_{2}$ on one side and solve.

• color(magenta)(n_(air)sin(30^(@))= n_(2)sin(15^(@))

• $\frac{\left(1\right) \left(0.5\right)}{\left(0.26\right)} = {n}_{2}$

• $1.92 = {n}_{2}$

$\textcolor{\mathmr{and} a n \ge}{\text{Answer} : 1.92 = {n}_{2}}$

This answer intuitively makes sense because when light travels from one medium to the next and the angle of refraction is smaller than the angle of incident, than that tells us that the light rays bend toward the Normal, indicating a index of refraction greater than the other medium, in this case, air.