Electromagnetism - Polarization

Questions 54-55 concern a plane electromagnetic wave that is a superposition of two independent orthogonal plane waves and can be written as the real part of 

E = x̂ E₁ exp [i(kz −ωt)] +  ŷ E₂ exp [i(kz − ωt + π)]

where k, ω, E₁ and E₂ are real

If the plane wave is split and recombined on a screen after the two portions, which are polarized in the x- and y- directions, have traveled an optical path difference of 2π/k, the observed average intensity will be proportional to

A. E₁² + E₂² 
B. E₁² − E₂² 
C. (E₁+ E₂)² 
D. (E₁− E₂)² 
E. 0

(GR9677 #55)

Solution:

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i(kz − ωt +π )} 

Path difference of 2π/k
E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[k(z + 2π/k) − ωt + π)} 

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[kz − ωt)} · e^i3π 
e^i3π = −1

E = x̂ E₁ e^{i(kz − ωt)}  −  ŷ E₂  e^{i(kz − ωt)} 

Intensity in the x- directions, I_x = |E₁|²  
Intensity in the x- directions, I_y = |−E₂|² 

I_total = I_x + I_y = E₁² + E₂²

Answer: A

Optics - Polarization

A steady of light is normally incident on a piece of Polaroid. As the Polaroid is rotated around the beam axis, the transmitted intensity varies as A + B cos 2θ where θ is the angle of rotation, and A and B are constants with . Which of the following may be correctly concluded about the incident light? 

A. The light is completely unpolarized
B. The light is completely plane polarized
C. The light is partly plane polarized and partly unpolarized
D. The light is partly circularly polarized and partly unpolarized
E. The light is completely circularly polarized

(GR9277 #67)

Solution:

Malus' Law: for plane-polarized light, I = I₀ cos²θ
Trig identity: cos²θ =  ½ (1 + cos 2θ) 

If A = B, the light is completely plane polarized, but A > B > 0

When cos²θ = 0, I = A = constant → unpolarized.

Thus, the light is partly plane polarized and partly unpolarized

Answer: C

Electromagnetism - Dielectric

A dielectric of dielectric constant K is placed in contact with a conductor having surface charge density σ, as shown above. What is the polarization (bound) charge density σ_p on the surface of the dielectric at the interface between the two materials? 

A. σK / (1−K)
B. K / (1 + K)
C. σK
D. σ(1 + K) / K
E. σ(1 − K) / K

(GR8677 #54)

Solution:

Polarization, P = ɛ₀χ_eE = ɛ₀(K − 1)E
Thus, K  1.
If K = 1 → P = 0 and σ_p = 0 → no polarized dielectric

Check answers for K = 1→ σ_p = 0

(A) σ_p = σK/(1 − K) = ∞ → FALSE
(B) σ_p = K/(1 + K) = 1/2 → FALSE
(C) σ_p = σK = σ → FALSE
(D) σ_p = σ(1 + K)/K = 2σ → FALSE
(E) σ_p = σ(1 − K)/K = 0 → TRUE

Answer: E 


Alternative Solution:

In dielectric, E = σ/ɛ = σ/ɛ₀K
Polarization, P = ɛ₀(K − 1)E
σ_p = ɛ₀(K − 1)(σ/ɛ₀K)
= (K − 1)(σ/K)
= σ(1 − K)/K

Optics - Polarization

Unpolarized light is incident on two ideal polarizers in series. The polarizers are oriented so that no light emerges through the second polarizer. A third polarizer is now inserted between the first two and its orientation direction is continuously rotated through 180^o. The maximum fraction of the incident power transmitted through all three polarizers is

A. Zero
B. 1/8
C. 1/2
D. 1/√2
E. 1

(GR8677 #74)

Solution:

Power ~ Intensity

The maximum fraction of intensity is: I = I₀/8 (see GR0177 #51)

→ P = P₀/8

Answer: B

Optics - Polarization

Unpolarized light of intensity I₀ is incident on a series of three polarizing filters. The axis of the second filter is oriented at 45^o to that of the first filter, while the axis of the third filter is oriented at 90^o to that of the first filter. What is the intensity of the light transmitted through the third filter?

A. 0
B. I₀/8
C. I₀/4
D. I₀/2
E. I₀/√2

(GR0177 #51)

Solution:

Polarizer:
The first filter always reduces the intensity of the light to half, I₁=½ I₀
The next filter reduces the intensity by I_n = I_(n-1) cos² θ
θ is the angle with respect to the n^th filter.

Thus,

I₁ = ½ I₀

I₂ = I₁ (cos 45)² = I₁ (½ √2)² = ½ I₁ = ¼ I₀

I₃=I₂ (cos 45)² = ½ I₂ = I₀/8

Answer: B