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Chapter 1: Problem 72

It is found that when light traveling in water falls on a plastic block, Brewster's angle is \(50.0^{\circ} .\) What is the refractive index of the plastic?

Short Answer

Expert verified
The refractive index of the plastic block is approximately 1.58, determined using Brewster's Law and the given Brewster's angle of \(50.0^{\circ}\) between water and the plastic block.

Step by step solution

01

Understand Brewster's Law

The Brewster's angle is the angle of incidence at which the light rays reflected on the boundary surface between two materials are entirely polarized. Brewster's Law relates the refractive indices of two media (n1 and n2) to the Brewster's angle, θ_B, as follows: \(tan(\theta_{B}) = \frac{n_{2}}{n_{1}}\) In our case, the angle of incidence, θ_B is given as 50.0 degrees.
02

Calculate the refractive index of water

Before finding the refractive index of the plastic block, we need to know the refractive index of water. The refractive index of water at room temperature is typically approximately 1.33.
03

Determine the refractive index of the plastic

We can now calculate the refractive index of the plastic block using Brewster's Law. We know the Brewster's angle and the refractive index of water. \(tan(\theta_{B}) = \frac{n_{2}}{n_{1}}\) Substitute the given values: \(tan(50.0^{\circ}) = \frac{n_{2}}{1.33}\) Now solve for n2: \(n_{2} = 1.33 * tan(50.0^{\circ})\) \(\Rightarrow n_{2} \approx 1.33 * 1.19\) \(\Rightarrow n_{2} \approx 1.58\)
04

State the Result

The refractive index of the plastic block is approximately 1.58. This means that the speed of light in the plastic block is about 1.58 times slower than in a vacuum.

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Most popular questions from this chapter

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