Chapter 33: Q70P (page 1006)

Question: In Fig. 33-64, a light ray in air is incident on a flat layer of material 2 that has an index of refraction $n_{2} = 1.5$ . Beneath material 2is material 3 with an index of refraction $n_{3}$ .The ray is incident on the air–material 2 interface at the Brewster angle for that interface. The ray of light refracted into material 3happens to be incident on the material 2-material 3 interface at the Brewster angle for that interface. What is the value of $n_{3}$ ?

Short Answer

Expert verified

The value of $n_{3}$ is 1.

Step by step solution

Given

The refractive index of the material 2 is $n_{2} = 1.5$

The refractive index of the first medium i.e. air is $n_{1} = 1$

Understanding the concept

By using the formula of Brewster’s angle and as the figure say that the layers are parallel so that the refraction regarding the first is the same as the angle of incidence regarding the second surface.

Formula:

Brewster’s angle

$θ_{B} = \tan (\frac{n_{2}}{n_{1}})$

Step 3: Calculate the value of n3.

Since the layers are parallel, the angle of refraction of the first surface is the same as the angle of incidence of the second surface. Thus,

$θ_{2} = {(θ_{1})}_{c} = 90^{\circ} - θ$

Brewster’s angle is given by

$θ_{B} = \tan^{- 1} (\frac{n_{2}}{n_{1}}) θ_{1} = \tan^{- 1} (\frac{1.5}{1}) = 56 . 31^{\circ}$

Therefore,

$θ_{2} = 90^{\circ} - 56 . 31^{o} = 33 . 69^{\circ}$

We can apply Brewster’s law to both refractions. Therefore,

$(\frac{n_{2}}{n_{1}}) (\frac{n_{3}}{n_{2}}) = ({tanθ}_{1 \to 2}) ({tanθ}_{2 \to 3}) \frac{n_{3}}{n_{1}} = ({tanθ}_{1}) ({tanθ}_{2}) \frac{n_{3}}{1} = (\tan 56 . 31^{\circ}) (\tan 33 . 69^{\circ}) n_{3} = 1$