Question

A ray of light is incident at angle thetaθ = 59° on the side of two stacked transparent blocks. The bottom block has index of refraction 1n1 = 1.41. The ray enters the top block at angle phiϕ = 40.6°. What is the index of refraction 2n2 of the top block? Note: Angles may not be drawn to scale.

Answer 1

The index of refraction of the top block can be calculated using Snell's law, which relates the angles of incidence and refraction to the indices of refraction of the two media. In this case, the index of refraction of the bottom block is known as 1.41, and by substituting the given values into Snell's law, we can determine that the index of refraction of the top block is approximately 1.79.

Step-by-step explanation:

When light passes from one medium to another, it changes direction due to a change in its speed. The index of refraction, represented by the symbol n, is a measure of how much a medium slows down light compared to its speed in a vacuum. It is calculated using Snell's law:

n1 * sin(theta1) = n2 * sin(theta2)

For the given scenario, we can use Snell's law to find the index of refraction of the top block (n2). We know that the angle of incidence (theta1) is 59° and the angle of refraction (theta2) is 40.6°. The index of refraction of the bottom block (n1) is given as 1.41.

Using the values in Snell's law:

1.41 * sin(59°) = n2 * sin(40.6°)

Solving for n2, we find that the index of refraction of the top block is approximately 1.79.