Polarization

Polarization	A property of light is that it can be polarized. Unpolarized light points in all sorts of direction, while polarized light has its E vectors going in the same direction.
Malus' Law		Where I₀ is the intensity of unpolarized light, and I is the intensity after polarization: For polarized light going at angle θ to the polarizer: I = I₀ cos² θ For unpolarized light, since Since the average value of cos² θ is ½ then: I = ½ I₀
Brewster's Angle (θ_P)		When a ray travelling through a medium with index of refraction n₁ strikes a second medium with n₂, then the angle θ_P is when the reflected ray and refracted ray have an angle of 90° between them. The angle θ_P is formed between the incident ray and the normal to the surface. n₁ sin θ_P = n₂ sin θ₂ n₁ sin θ_P = n₂ sin ( 90° - θ_P ) n₁ sin θ_P = n₂ cos θ_P tan θ_P = n₂ / n₁ (Brewster's Law)

A property of light is that it can be polarized. Unpolarized light points in all sorts of direction, while polarized light has its E vectors going in the same direction.

Malus' Law

Where I₀ is the intensity of unpolarized light, and I is the intensity after polarization:

For polarized light going at angle θ to the polarizer: I = I₀ cos² θ

For unpolarized light, since Since the average value of cos² θ is ½ then: I = ½ I₀

Brewster's Angle (θ_P)

When a ray travelling through a medium with index of refraction n₁ strikes a second medium with n₂, then the angle θ_P is when the reflected ray and refracted ray have an angle of 90° between them. The angle θ_P is formed between the incident ray and the normal to the surface.

n₁ sin θ_P = n₂ sin θ₂

n₁ sin θ_P = n₂ sin ( 90° - θ_P )

n₁ sin θ_P = n₂ cos θ_P

tan θ_P = n₂ / n₁ (Brewster's Law)

Practice problems

The critical angle for total internal reflection at a boundary between two materials is 40.0°. What is Brewster's angle at this boundary? Give two answers, one for each material.

Critical angle: sin θ_c = n₂ / n₁Brewster's angle for first material: tan θ_P = n₂ / n₁Brewster's angle for second material: tan θ_P = n₁ / n₂

sin 40.0° = n₂ / n₁n₂ / n₁ = tan θ_P for first material
0.64278761 = tan θ_Pθ_P = 32.7° for first material

sin 40.0° = n₂ / n₁n₁ / n₂ = tan θ_P for second material
0.64278761 = 1 / tan θ_P1.55572383 = tan θ_Pθ_P = 57.3° for second material

Two polarizers are oriented at 50° to one another. Light polarized at an 30° angle to each polarizer passes through both. What is the transmitted intensity?

I = I₀ cos² θ

I₀ = Initial intensity
I₁ = Intensity after first polarizer
I₂ = Intensity after second polarizer
θ₁ = 30° is incident light angle to first polarizer
θ₂ = 50° since the second polarizer is rotated 50° relative to the first polarizer.

I₁ = I₀ cos² θ₁ = I₀ cos² 30°
I₂ = I₁ cos² θ₂ = I₀ cos² 30° cos² 50°
I₂ = I₀ 0.309881933
I₂ / I₀ = 0.309881933 = 31% transmitted intensity

Two polarizers are oriented at 70° to each other and plane-polarized light is incident on them. If only 10% of the light gets through both of them, what was the initial polarization direction of the incident light?

I = I₀ cos² θ

I₀ = Initial intensity
I₁ = Intensity after first polarizer
I₂ = Intensity after second polarizer (10% of I₀)
θ₁ = Incident light angle to first polarizer
θ₂ = 70° since the second polarizer is rotated 70° relative to the first polarizer.

I₁ = I₀ cos² θ₁I₂ = I₀ cos² θ₁ cos² 70°
I₂ / I₀ = 10% = .10
.10 = cos² θ₁ cos² 70°
0.854863217 = cos² θ₁0.924588134 = cos θ₁θ₁ = 22°