### Polarization

#### Problem:

A unpolarized electromagnetic wave is incident on a series of three linear polarizers, the second with the polarization angle rotated at 30° and the third at 90° with respect to the first polarizer.  If the initial intensity of the unpolarized light is I0, what is the intensity I3 transmitted by the stack?

Solution:

• Concepts:
Polarization
• Reasoning:
For a polarized EM wave incident on a polarizer, Itransmitted = I0 cos2θ.  Here θ the angle between E0 and the transmission axis.
• Details of the calculation:
After passing through the first polarizer, I1 = Io/2.
After passing through the second polarizer I2 = I1 cos2(30o).
Finally after passing through the third polarizer, I3 = I2 cos2(60o).
Thus, I3 = Io cos2(30o)cos2(60o)/2 = 0.094 Io.

#### Problem:

(a)  Assume the electric field of a plane wave propagating in the z-direction oscillates in the x-direction.  What combination of polarizers would transmit the beam so it would emerge with the electric field oscillating in the y-direction.  Assume that perfect linear and circular polarizers are available.  Perfect linear polarizers transmit 100% of the incident wave in one plane of polarization and absorb 100% in the perpendicular plane.  Circular polarizers (quarter-wave plates) retard one plane of polarization ¼ wavelength relative to the perpendicular plane of polarization and hence can change a linearly polarized wave to a right- or left-hand circularly polarized wave and vice versa, depending on the orientation of the linear polarization.
(b)  What combination of polarizers would change a right-hand circularly polarized beam traveling in the z-direction into a left-hand circularly polarized beam.  How?

Solution:

• Concepts:
Polarization, linear and circular polarizers
• Reasoning:
We are supposed to find a combination of polarizers that rotate the polarization axis by 90o.
• Details of the calculation:
(a)  E = ε0cos(kz - ωt)i describes the EM wave before polarizers are inserted in its path.
Consider two coordinate systems rotated about the z-axis 45o with respect to each other as shown below.

Insert linear polarizer with its transmission axis along x' followed by a linear polarizer with its transmission axis along y.  After the first polarizer we have
E
= E0cos(45o)cos(kz - ωt + φ)i',
and after the second polarizer we have
E = E0cos(45o)2cos(kz - ωt + φ)j.
The amplitude of the wave is reduced by a factor of 2 and the power by a factor of 4.
To chance the linear polarization without reducing the amplitude we use two quarter-wave plates.  Assume the quarter-wave plate is oriented so the phase of light polarized along x' is retarded by π/2 with respect to the phase of light polarized along y'.
Then we have before the plates
Ex' = 2E0cos(kz - ωt),
Ey' = -2E0cos(kz - ωt).
After one quarter-wave plate we have
Ex' = 2E0cos(kz - ωt + φ'),
Ey' = -2E0cos(kz - ωt + φ' + π/2).
After two quarter-wave plates we have
Ex' = 2E0cos(kz - ωt + φ''),
Ey' = -2E0cos(kz - ωt + φ'' + π) = 2E0cos(kz - ωt + φ'')
E = E0cos(kz - ωt + φ'')j.  The polarization has been rotated by 90o but the amplitude has remained the same.
(c)  Convention:  When viewed looking towards the source, a right circularly polarized beam at a fixed position as a function of time has a field vector that describes a clockwise circle, while left circularly polarized light has a field vector that describes a counter-clockwise circle
RHC polarized light:  Ex' = E0cos(kz - ωt), Ey' = E0sin(kz - ωt)
LHC polarized light:  Ex' = E0cos(kz - ωt), Ey' = -E0sin(kz- - ωt)
After two quarter-wave plates the field of the RHC polarized wave becomes
Ex' = E0cos(kz - ωt+φ), Ey' = E0sin(kz - ωt + φ + π) = -E0sin(kz - ωt + φ).  The wave becomes a LHC polarized wave.

#### Problem:

(a)  Describe how to prepare a monochromatic plane wave of linearly polarized visible light.
(b)  Describe how to prepare a monochromatic plane wave of circularly polarized visible light

Solution:

• Concepts:
Wavelength selection, polarization, linear and circular polarizers
• Reasoning:
Prisms or gratings are dispersive elements.  Perfect linear polarizers transmit 100% of the incident wave in one plane of polarization and absorb 100% in the perpendicular plane.  Circular polarizers (quarter-wave plates) retard one plane of polarization ¼ wavelength relative to the perpendicular plane of polarization and hence can change a linearly polarized wave to a right- or left-hand circularly polarized wave and vice versa, depending on the orientation of the linear polarization.
• Details of the calculation:
(a)  A plane wave in practical applications is a wave front with a well-defined direction of of propagation, extending over an area larger than the cross-sectional area of an object to be illuminated.
The figure below shows a possible method of preparing a linearly polarized plane wave.

(b)  The figure below shows a possible method of preparing a circularly polarized plane wave.