The speed of light in a material is 0.50 c. what is the critical angle of a light ray at the interface between the material and a vacuum?

Explanation:

This problem indicates that the speed of light in a material medium is 0.5 10⁸ m / s, they ask to find the critical angle between the material and the vacuum

Let's find the refractive index of the material

            n = c / v

           n = 3 10⁸ / 0.5 10⁸

           n = 6

When the material passes from one medium to another, it must comply with the law of refraction

           n₁ sin θ = n₂ sin θ₂

for the angle criticize the angles tea2 = 90

          tea = sin⁻¹n₂/ n₁

The vacuum replacement index is n₂ = 1

         tea = sin⁻¹ (1/6)

         tea = 9.59º

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