A beam of light is emitted 8.6 cm beneath the surface of a liquid and strikes the air surface 7.2 cm from the point directly above the source. Part A If total internal reflection occurs, what can you say about the minimum possible index of refraction of the liquid?

n = 1.56

Explanation:

The total reflection attempts occurs when a light beam passes from a medium with a higher index to a medium with a lower nest, at an angle where it occurs we can find them by the refractive relationship

             n₁ sin θ₁ = n₂ sin θ₂

             n1 = n2 / sin θ₁

For this relationship to be fulfilled, the liquid index must be greater than the air index divided by the sine of the critical angle

Let's use trigonometry to find angle

               tan θ = y / x

                θ = tan⁻¹ 7.2 / 8.6

                θ = 39.94º

                n₁ = 1 / sin 39.94

                n = 1.56

This is the refractive index of the liquid

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