Question

In: Physics

The angle of incidence of a light beam in air onto a reflecting surface is continuously...

The angle of incidence of a light beam in air onto a reflecting surface is continuously variable. The reflected ray is found to be completely polarized when the angle of incidence is 48.0. (a) What is the index of refraction of the reflecting material? (b) If some of the incident light (at an angle of 48.0) passes into the material below the surface, what is the angle of refraction?

Solutions

Expert Solution

Concepts and reason

The required concept to solve the problem is Brewster's law and polarization. First, using Brewster's law, find the refractive index of the reflecting material. Then, using Brewster's law, find the angle of refraction.

Fundamentals

According to Brewster's law, the maximum polarization occurs when the angle of reflection and angle of refraction make an angle of \(90^{\circ}\) Another statement for Brewster's law is that the tangent of the polarization angle is equal to the ratio of the refractive indices of the contacting mediums. \(\tan \theta_{p}=\frac{n_{2}}{n_{1}}\)

Here, \(\theta_{p}\) is the angle of polarization, \(n_{2}\) is the refracted medium and \(n_{1}\) is the incident medium.

(a) According to Brewster's law as given below:

\(\tan \theta_{p}=\frac{n_{2}}{n_{1}}\)

Rewrite the above equation in terms of \(n_{2}\).

\(n_{2}=n_{1} \tan \theta_{p}\)

Substitute 1 for \(n_{1}\) and \(48^{\circ}\) for \(\theta_{p}\) \(n_{2}=(1) \tan 48^{\circ}\)

\(=1.11\)

Part a

Thus, the refractive index of the reflecting material is 1.11.

Brewster's angle is the angle where the unpolarized light after complete polarization is reflected.The first medium is air and the refractive index of air is 1 .

(b) As the reflected ray is completely polarized, the condition is as given below:

\(\theta_{p}+\theta_{r e f}=90^{\circ}\)

Here, \(\boldsymbol{\theta}_{\text {ref }}\) is the angle of refraction. Rewrite the above relation for the angle of refraction. \(\theta_{r e f}=90^{\circ}-\theta_{p}\)

Substitute \(48^{\circ}\) for \(\theta_{p}\)

\(\theta_{r e f}=90^{\circ}-48^{\circ}\)

\(=42^{\circ}\)

Part \(b\) Thus, when the reflected light is completely polarized, the angle of refraction is \(42^{\circ}\).

According to Brewster's law, the maximum polarization occurs when the angle of reflection and angle of refraction make an angle of \(90^{\circ}\). So, the angle of refraction is the difference between \(90^{\circ}\) and the angle of polarization.


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