A light ray propagates in Material 1 with index of refraction n 1 = 1.19 , strikes an interface, then passes into Material 2 with index of refraction n 2 = 1.37 . The angle of incidence at the interface is θ 1 = 31.9 °. Find the angle of refraction θ 2 .

    θ₂ = 27.3º

Explanation:

For this exercise we will use the law of refraction

         n₁ sin θ₁ = n₂ sin  θ₂

Where we use index 1 for material 1 and index 2 for material 2

Let's clear

           sin θ₂ = n₁ / n₂  sin θ₁

Let's calculate

          sin θ₂ = 1.19 / 1.37 sin 31.9

          θ₂ = sin⁻¹ (0.4590)

         θ₂ = 27.3º

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

2b2t

explanation:

2b2t