The speed of light in clear plastic is 1.866 ×10^8 m/s. A ray of light enters the plastic at an angle of 29.3 degrees.

At what angle is the ray refracted?

To determine the angle at which the light ray is refracted, we can use Snell's law, which relates the angle of incidence to the angle of refraction when light passes through a boundary between two mediums.

Snell's law is given by:
n₁ * sin(θ₁) = n₂ * sin(θ₂)

Where:
n₁ = refractive index of the first medium (in this case, the refractive index of air/vacuum is approximately 1)
θ₁ = angle of incidence
n₂ = refractive index of the second medium (in this case, the refractive index of clear plastic is not provided)
θ₂ = angle of refraction (what we need to find)

Since the refractive index of the plastic is not provided, we can't directly calculate the angle of refraction. However, we can calculate the unknown refractive index using the given information and then determine the angle of refraction.

First, we need to find the refractive index of the plastic using the speed of light in the plastic compared to the speed of light in vacuum:

Refractive index (n) = speed of light in vacuum / speed of light in the material

Given:
Speed of light in vacuum (c) = 3.0 × 10^8 m/s
Speed of light in clear plastic = 1.866 × 10^8 m/s

Refractive index (n) = (3.0 × 10^8 m/s) / (1.866 × 10^8 m/s)
Refractive index (n) = 1.609

Now that we have the refractive index of the clear plastic, we can substitute the values into Snell's law:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

Substituting the known values:
(1) * sin(29.3°) = 1.609 * sin(θ₂)

Now we can solve for θ₂:
sin(θ₂) = (1 * sin(29.3°)) / 1.609
sin(θ₂) ≈ 0.4629

To find the angle θ₂, we can take the inverse sine (sin⁻¹) of 0.4629:
θ₂ ≈ sin⁻¹(0.4629)
θ₂ ≈ 27.4°

Therefore, the ray of light entering the plastic at an angle of 29.3 degrees will be refracted at an angle of approximately 27.4 degrees.