Question

in the figure below, a light ray travels in material 1, which has an index of refraction ( n_1 = 1.33 ). the ray is incident on a flat layer of material 2 that has an index of refraction ( n_2 = 1.58 ). beneath material 2 is material 3 with an index of refraction ( n_3 ). the ray is incident on the material 1 - material 2 interface at the brewster angle for that interface. the ray of light refracted into material 3 happens to be incident on the material 2 - material 3 interface at the brewster angle for that interface as well. all of the interfaces are parallel to each other. what is the value of ( n_3 )?

Explanation:

Step1: Use Brewster's law for material 1 - material 2 interface

Brewster's law is \( \tan\theta_{B}=\frac{n_{2}}{n_{1}}\). Also, from Snell's law \(n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}\). When \(\theta_{1}\) is the Brewster angle \(\theta_{B1}\), \(\theta_{1}+\theta_{2} = 90^{\circ}\), so \(\sin\theta_{1}=\cos\theta_{2}\). Then \(n_{1}\cos\theta_{2}=n_{2}\sin\theta_{2}\), or \(\tan\theta_{2}=\frac{n_{1}}{n_{2}}\)

Step2: Use Brewster's law for material 2 - material 3 interface

For the material 2 - material 3 interface, the Brewster angle condition gives \(\tan\theta_{3}=\frac{n_{3}}{n_{2}}\). But since the interfaces are parallel, the angle of refraction at the first interface is related to the angle of incidence at the second interface. Also, using the fact that for two - step Brewster angle incidence (parallel interfaces) \(n_{3}=\frac{n_{2}^{2}}{n_{1}}\)

Step3: Substitute the given values

Given \(n_{1} = 1.33\) and \(n_{2}=1.58\). Then \(n_{3}=\frac{1.58^{2}}{1.33}=\frac{2.4964}{1.33}\approx1.88\)

Answer:

1.88