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Polarization

Module by: Paul Padley. E-mail the author

Summary: We discuss the polarization of light waves.

Polarization

Recall that we can add waves so lets take a plane wave traveling in the z direction and break it into components. E x = E 0 x cos ( k z ω t ) ı ̂ E x = E 0 x cos ( k z ω t ) ı ̂ E y = E 0 y cos ( k z ω t + δ ) ̂ E y = E 0 y cos ( k z ω t + δ ) ̂ Where δ δ is some arbitrary phase between the two components. The electric field for the wave is E = E x + E y E = E x + E y Now there are a number of different cases that arise.

If δ = 0 , 2 π , 4 π , δ = 0 , 2 π , 4 π , Then we can write the field as E = ( E 0 x ı ̂ + E 0 y ̂ ) cos ( k z ω t ) E = ( E 0 x ı ̂ + E 0 y ̂ ) cos ( k z ω t )

Figure 1
Figure 1 (Drawing1.png)
Polarization of light for the case E = ( E 0 x ı ̂ + E 0 y ̂ ) cos ( k z ω t ) E = ( E 0 x ı ̂ + E 0 y ̂ ) cos ( k z ω t )

The field is linearly polarized, that is the E field lies along a straight line.

Likewise If δ = π , 3 π , 5 π , δ = π , 3 π , 5 π , Then again it is linearly polarized, but is now "flipped" E = ( E 0 x i ̂ E 0 y j ̂ ) cos ( k z ω t ) E = ( E 0 x i ̂ E 0 y j ̂ ) cos ( k z ω t )

Figure 2
Figure 2 (Drawing2.png)
Polarization of light for the case E = ( E 0 x ı ̂ E 0 y ̂ ) cos ( k z ω t ) E = ( E 0 x ı ̂ E 0 y ̂ ) cos ( k z ω t )

Circularly polarized light is a particularly interest example. Let δ = π / 2 δ = π / 2 and E 0 x = E 0 y = E 0 E 0 x = E 0 y = E 0

Then E x = E 0 cos ( k z ω t ) ı ̂ E x = E 0 cos ( k z ω t ) ı ̂ E y = E 0 cos ( k z ω t π / 2 ) ̂ E y = E 0 cos ( k z ω t π / 2 ) ̂ E y = E 0 sin ( k z ω t ) ̂ E y = E 0 sin ( k z ω t ) ̂ or E = E 0 [ cos ( k z ω t ) ı ̂ + sin ( k z ω t ) ̂ ] E = E 0 [ cos ( k z ω t ) ı ̂ + sin ( k z ω t ) ̂ ]

The direction of E E is changing with time. For example consider the case at z = 0 z = 0 . Then

E = E 0 [ cos ( ω t ) ı ̂ + sin ( ω t ) ̂ ] E = E 0 [ cos ( ω t ) ı ̂ + sin ( ω t ) ̂ ]

E = E 0 [ cos ( ω t ) ı ̂ sin ( ω t ) ̂ ] E = E 0 [ cos ( ω t ) ı ̂ sin ( ω t ) ̂ ]

In this case the electric field undergoes uniform circular rotation. For light coming out of the page, it will have the motion shown in the figure

Figure 3
Figure 3 (Drawing3.png)
Polarization of light for the case E = E 0 [ cos ( ω t ) ı ̂ sin ( ω t ) ̂ ] . E = E 0 [ cos ( ω t ) ı ̂ sin ( ω t ) ̂ ] . I, and most physicists would call the Left Hand Circular polarization (LHC). The thumb points in the direction of the light ray and the fingers curve in the direction of rotation. (This is known as the Angular momentum convention, optical scientists will use the "Optical" convention which is opposite - we will stick to the angular momentum convention.)

Suppose δ = π / 2 δ = π / 2 then you get E = E 0 [ cos ( k z ω t ) ı ̂ sin ( k z ω t ) ̂ ] E = E 0 [ cos ( k z ω t ) ı ̂ sin ( k z ω t ) ̂ ] This now has the opposite rotation, it is Right Handed.

The most general case of polarization has δ δ arbitrary and E 0 x E 0 y E 0 x E 0 y and is elliptically polarized.

Figure 4
Figure 4 (Drawing4.png)
Elliptical polarization

Natural light

Natural light is emitted with the E E field in a mixture of random directions. This in unpolarized light. At any particular instant E E has a particular direction, but that direction changes rapidly and randomly.

Malus' Law

A polarizer is a device that takes incident natural light and transmits polarized light. For example a linear polarizer will take incident light and select only that component of the light that has its E E field lined up along the transmission axis. Suppose there is a linear polarizer that transmits light along a particular axis. This is followed by a second linear polarizer that has its transmission axis at a different angle with θ θ being the angle between the transmission axes. Since I E 2 I E 2 then at the second polarizer I ( θ ) = I ( 0 ) cos 2 θ I ( θ ) = I ( 0 ) cos 2 θ where I ( 0 ) I ( 0 ) is the Irradiance of the light hitting the second polarizer.

Lets consider unpolarized light hitting a linear polarizer. Then half the Irradiance will get through. If this is followed by a second polarizer at 90 0 90 0 then no light will pass through the second polarizer. Now what happens if a third polarizer is placed between them?

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