Question

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?

Answer 1

Angle of refraction at the air-ice interface`=32.7^@`
Angle of refraction at the ice-water interface`=32.1^@`

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
`1.00xxsin 45 = 1.31xxsin theta_2`
`=>sin theta_2 =(sin 45/1.31)= 0.5398`, rounded to four decimal places
`=>theta_2 =sin^-1(0.5398)=32.7^@`, 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

`1.31xxsin 32.67 = 1.33 xxsin theta_2`
`theta_2 = sin^-1((1.31xx0.5398)/1.33)=32.1^@`, rounded to one decimal place