# A ray of sunlight hits a frozen lake at a 45° angle of incidence. At what angle of refraction does the ray penetrate the ice? At what angle does it penetrate the water beneath the ice?

Dec 19, 2016

Angle of refraction at the air-ice interface$= {32.7}^{\circ}$
Angle of refraction at the ice-water interface$= {32.1}^{\circ}$

#### Explanation:

Using the following refractive indices
Air $= 1.00$
Water$= 1.33$
Ice $= 1.31$

Using Snell's Law of refraction applicable for a ray crossing the interface between two media which have refractive indices ${n}_{1}$ and ${n}_{2}$ as shown in the figure. n_1 sinθ_1 = n_2 sinθ_2
where θ_1 and θ_2 are the angles of incidence and refraction respectively.

For first interface when ray enters from air to ice
$1.00 \times \sin 45 = 1.31 \times \sin {\theta}_{2}$
$\implies \sin {\theta}_{2} = \left(\sin \frac{45}{1.31}\right) = 0.5398$, rounded to four decimal places
$\implies {\theta}_{2} = {\sin}^{-} 1 \left(0.5398\right) = {32.7}^{\circ}$, rounded to one decimal place.

Due to geometry this becomes angle of incidence for the second interface as ray enters from ice to water. Using Snell's law again we have

$1.31 \times \sin 32.67 = 1.33 \times \sin {\theta}_{2}$
${\theta}_{2} = {\sin}^{-} 1 \left(\frac{1.31 \times 0.5398}{1.33}\right) = {32.1}^{\circ}$, rounded to one decimal place