A diver, below the surface of a lake, notices that no light appears to be coming toward her greater than the angle è, measured from the vertical. If the index of refraction of water is 1.33, what is the angle è (which is the critical angle of total internal reflection, in degrees)?

48.6

To solve this problem, we need to use the relationship between the incident angle and the critical angle of total internal reflection.

The critical angle (θc) is the angle of incidence at which light entering a medium with higher refractive index is refracted along the boundary instead of being transmitted through it. Beyond the critical angle, total internal reflection occurs.

In this case, the medium is water, and the refractive index of water (n) is given as 1.33.

We are looking for the critical angle (θc) in degrees.

To find the critical angle, we can use Snell's law:

n₁ * sin(θ₁) = n₂ * sin(θ₂),

where
n₁ = refractive index of the initial medium,
θ₁ = angle of incidence,
n₂ = refractive index of the final medium,
θ₂ = angle of refraction.

In our scenario, the initial medium is air (which has a refractive index close to 1), and the final medium is water (with a refractive index of 1.33).

So, we can rewrite Snell's law as:

sin(θ₁) = n₂ / n₁ * sin(θ₂).

Since the diver notices no light appearing at angles greater than the critical angle, that means the light is completely internally reflected. Therefore, θ₂ is 90 degrees.

Plugging in the values, we have:

sin(θ₁) = 1.33 / 1 * sin(90°).

Since sin(90°) = 1, the equation simplifies to:

sin(θ₁) = 1.33.

To find θ₁, we need to take the inverse sine (sin⁻¹) of both sides:

θ₁ = sin⁻¹(1.33).

Calculating this value using a scientific calculator, we find that:

θ₁ ≈ 51.44 degrees.

Hence, the critical angle (è) is approximately 51.44 degrees.