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 I_{0}, what is the intensity I_{3}
transmitted by the stack?

Solution:

- Concepts:

Polarization - Reasoning:

For a polarized EM wave incident on a polarizer, I_{transmitted }= I_{0 }cos^{2}θ. Here θ the angle between**E**_{0}and the transmission axis. - Details of the calculation:

After passing through the first polarizer, I_{1 }= I_{o}/2.

After passing through the second polarizer I_{2 }= I_{1}cos^{2}(30^{o}).

Finally after passing through the third polarizer, I_{3 }= I_{2}cos^{2}(60^{o}).

Thus, I_{3 }= I_{o}cos^{2}(30^{o})cos^{2}(60^{o})/2 = 0.094 I_{o}.

(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 90^{o}. - Details of the calculation:

(a)**E**= ε_{0}cos(kz - ωt)**i**describes the EM wave before polarizers are inserted in its path.

Consider two coordinate systems rotated about the z-axis 45^{o}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

E_{0}cos(45^{o})cos(kz - ωt + φ)**i**',

and after the second polarizer we have

**E**= E_{0}cos(45^{o})^{2}cos(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

E_{x}' = 2^{-½}E_{0}cos(kz - ωt),

E_{y}' = -2^{-½}E_{0}cos(kz - ωt).

After one quarter-wave plate we have

E_{x}' = 2^{-½}E_{0}cos(kz - ωt + φ'),

E_{y}' = -2^{-½}E_{0}cos(kz - ωt + φ' + π/2).

After two quarter-wave plates we have

E_{x}' = 2^{-½}E_{0}cos(kz - ωt + φ''),

E_{y}' = -2^{-½}E_{0}cos(kz - ωt + φ'' + π) = 2^{-½}E_{0}cos(kz - ωt + φ'')

**E**= E_{0}cos(kz - ωt + φ'')**j**. The polarization has been rotated by 90^{o}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: E_{x}' = E_{0}cos(kz - ωt), E_{y}' = E_{0}sin(kz - ωt)

LHC polarized light: E_{x}' = E_{0}cos(kz - ωt), E_{y}' = -E_{0}sin(kz- - ωt)

After two quarter-wave plates the field of the RHC polarized wave becomes

E_{x}' = E_{0}cos(kz - ωt+φ), E_{y}' = E_{0}sin(kz - ωt + φ + π) = -E_{0}sin(kz - ωt + φ). The wave becomes a LHC polarized wave.

(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.