A green light is submerged 2.10 m beneath the surface of a liquid with an index of refraction 1.33. What is the radius of the circle from which light escapes from the liquid into the air above the surface?

Radius = 2.395m

Explanation:

The radius will be determined by the angle of incidence which is equal to the critical angle.

Furthermore, if the angle is larger than that critical angle, the light is totally internally reflected. Thus, we can figure out that angle from Snell’s law where the refracted angle is 90°, and then use the tangent function.

tan (θc) = R/d

Thus, R = dtan (θc)

Now, we can find θ by;

n_air x sin 90 = n_liquid x sin (θc)

Where,

n_air = refraction index of air

n_liquid = refraction index of liquid

Refraction index of air = 1

Thus,

1 x 1 = 1.33sin(θc)

sin(θc) = 1/1.33 = 0.7519

Thus, θc = sin^(-1)0.7519

θc = 48.756°

Now,we can find the radius.

So,

R = dtan (θc)

R = 2.1 tan48.756 = 2.1 x 1.1405 = 2.395m