Electromagnetism - Superposition

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 E₂ = E₁, the tip of the electric field vector will describe a trajectory that, as viewed along the z-axis from positive z and looking toward the origin, is a

A. Line at 45^o to the + x-axis
B. Line at 135^o to the + x-axis
C. Clockwise circle
D. Counterclockwise circle
E. Random path

(GR9677 #54)

Solution:

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

with 
E₂ = E₁ = E
e^iπ = −1

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

tan θ  = a / (−a) = −1

tan 45  = 1
tan 135 = tan 315 = −1

Answer: B


Note: 
e^iϕ = cos ϕ + isin ϕ
e^iπ = cos π + isin π =  −1 + 0 = −1

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

Electromagnetism - Conductor

A plane-polarized electromagnetic wave is incident normally on a flat, perfectly conducting surface. Upon reflection at the surface, which of the following is true? 

A. Both the electric vector and magnetic vector are reversed.
B. Neither the electric vector nor the magnetic vector is reversed.
C. The electric vector is reversed; the magnetic vector is not.
D. The magnetic vector is reversed; the electric vector is not.
E.  The directions of the electric and magnetic vectors are interchanged.

(GR9277 #36)

Solution:

Perfect conductor: 

Its charge and current are distributed on the surface so that external fields can not penetrate it.

Boundary Condition (BC) on the surface of perfect conductor:

• E_tangential = 0
• E_normal ≠ 0 
• B_tangential ≠ 0
• B_normal = 0 

EM wave: E ⊥ B (perpendicular to each other).

If E normal to the surface of conductor, B is tangential.
BC: E_normal ≠ 0 and B_tangential ≠ 0
Only E is reversed.
B is not reserved, it's parallel to the surface.

Answer: C

Electromagnetism - Conductor

An electromagnetic plane wave, propagating in vacuum, has an electric field given by E = E₀ cos (kx−ωt) and is normally incident on a perfect conductor at x = 0, as shown in figure. Immediately to the left of the conductor, the total electric field E and the total magnetic field B are given by which of the following?

	E	B
A.	0	0
B.	2E0 cos ωt	0
C.	0	2(E0/c) cos ωt
D.	2E0 cos ωt	2(E0/c) cos ωt
E.	2E0 cos ωt	2(E0/c) sin ωt

(GR0177 #61)

Solution:

Perfect conductor: 

Its charge and current are distributed on the surface so that external fields can not penetrate it.

Boundary Condition (BC) on the surface of perfect conductor:

• E_tangential = 0
• E_normal ≠ 0 
• B_tangential ≠ 0
• B_normal = 0 

EM wave: E ⊥ B (perpendicular to each other).

If E normal to the surface of conductor, B is tangential.

BC: E_normal ≠ 0

E_initial is reflected by the conductor with the same magnitude but opposite direction.

Total E = E_initial − E_reflected = 0

(B), (D), (E) FALSE

BC: B_tangential ≠ 0

B is not reflected, it's parallel to the surface.

Total B ≠ 0

(A) FALSE

Answer: C