A ray of light falls at an angle pi / 3 at the air-liquid interface. Reflected and refracted

A ray of light falls at an angle pi / 3 at the air-liquid interface. Reflected and refracted rays are perpendicular to each other. Find the refractive index of the liquid.

Let’s apply Snell’s law:

nв * sina = nж * sinβ,

where nw is the optical density of air, nl is the optical density of the liquid, a is the angle of incidence, β is the angle of refraction.

Air optical density nв ≈ 1:

sina = nж * sinβ (1).

Find the angle β. The angle of reflection γ is equal to the angle of incidence a:

<γ = <a = π / 3.

Consider the unfolded corner FOJ:

<FOJ = <γ + <δ + <β;

π = π / 3 + π / 2 + <β;

<β = π – π / 3 – π / 2 = 6π / 6 – 2π / 6 – 3π / 6 = π / 6.

Let’s rewrite (1):

sin (π / 3) = nzh * sin (π / 6);

√3 / 2 = nzh * (1/2);

nzh = √3.

Answer. The density of the liquid nzh = √3 ≈ 1.732.