Question

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?

Answer 1

Radius = 2.395m

Step-by-step 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

Answer 2

The radius of the circle from which light escapes from the liquid into the air above the surface = 2.395 m = 2.40 m

Step-by-step explanation:

The maximum radius of this circle can be obtained by examining the critical angle for refraction.

The rays in the circle can only be refracted into air if the refracted angle made with the vertical is less than the critical angle.

Any rays that makes an angle greater than the critical angle is totally internally reflected back into the liquid.

The critical angle is evidently 90°.

Using Snell's law

n₁ sin θ₁ = n₂ sin θ₂

where n₁ = refractive index of the liquid = 1.33

θ₁ = angle of incidence = ?

n₂ = refractive index of air = 1.00

θ₂ = refracted angle = 90°

1.33 sin θ₁ = 1.00 × sin 90°

Sin θ₁ = (1/1.33)

θ₁ = sin⁻¹ (0.752) = 48.75°

From the conceptual diagram of the situation described by the question, attached to this solution,

R = 2.1 tan θ₁

R = 2.1 × tan 48.75° = 2.395 m

Hope this Helps!!!