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27.8 Polarization  (Page 2/19)

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The figure shows a slender arrow pointing out of the page and to the right; it is labeled direction of ray (of propagation). At a point on this ray, eight bold arrows point in different directions, perpendicularly away from the ray. These arrows are labeled E.
The slender arrow represents a ray of unpolarized light. The bold arrows represent the direction of polarization of the individual waves composing the ray. Since the light is unpolarized, the arrows point in all directions.
The figure shows a slender arrow pointing out of the page and to the right that is labeled direction of ray. At the left end of the ray are eight blue arrows emanating from a point on the ray. These arrows are all in a plane perpendicular to the ray and are symmetrically oriented in the perpendicular plane. They are labeled E. Farther to the right on the same ray is a thin rectangle labeled polarizing filter that is in the plane perpendicular to the ray. This filter has seven vertical lines that are equally spaced on its surface. It also has a vertical double headed arrow on its surface that is labeled axis. Still farther along the ray is a single blue double headed arrow oriented vertically that is labeled E and direction of polarization.
A polarizing filter has a polarization axis that acts as a slit passing through electric fields parallel to its direction. The direction of polarization of an EM wave is defined to be the direction of its electric field.

[link] shows the effect of two polarizing filters on originally unpolarized light. The first filter polarizes the light along its axis. When the axes of the first and second filters are aligned (parallel), then all of the polarized light passed by the first filter is also passed by the second. If the second polarizing filter is rotated, only the component of the light parallel to the second filter’s axis is passed. When the axes are perpendicular, no light is passed by the second.

Only the component of the EM wave parallel to the axis of a filter is passed. Let us call the angle between the direction of polarization and the axis of a filter θ size 12{θ} {} . If the electric field has an amplitude E size 12{E} {} , then the transmitted part of the wave has an amplitude E cos θ size 12{E"cos"θ} {} (see [link] ). Since the intensity of a wave is proportional to its amplitude squared, the intensity I size 12{I} {} of the transmitted wave is related to the incident wave by

I = I 0 cos 2 θ , size 12{I=I rSub { size 8{0} } "cos" rSup { size 8{2} } θ,} {}

where I 0 size 12{I rSub { size 8{0} } } {} is the intensity of the polarized wave before passing through the filter. (The above equation is known as Malus’s law.)

This figure has four subfigures. The first three are schematics and the last is a photograph. The first schematic looks much as in the previous figure, except that there is a second polarizing filter on the axis after the first one. The second polarizing filter has its lines aligned parallel to those of the first polarizing filter (i e, vertical). The vertical double headed arrow labeled E that emerges from the first polarizing filter also passes through the second polarizing filter. The next schematic is similar to the first, except that the second polarizing filter is rotated at forty five degrees with respect to the first polarizing filter. The double headed arrow that emerges from this second filter is also oriented at this same angle. It is also noticeably shorter than the other double headed arrows. The third schematic shows the same situation again, except that the second polarizing filter is now rotated ninety degrees with respect to the first polarizing filter. This time, there is no double headed arrow at all after the second polarizing filter. Finally, the last subfigure shows a photo of three circular optical filters placed over a bright colorful pattern. Two of these filters are place next to each other and the third is placed on top of the other two so that the center of the third is at the point where the edges of the two filters underneath touch. Some light passes through where the upper filter overlaps the left-hand underneath filter. Where the upper filter overlaps the right-hand lower filter, no light passes through.
The effect of rotating two polarizing filters, where the first polarizes the light. (a) All of the polarized light is passed by the second polarizing filter, because its axis is parallel to the first. (b) As the second is rotated, only part of the light is passed. (c) When the second is perpendicular to the first, no light is passed. (d) In this photograph, a polarizing filter is placed above two others. Its axis is perpendicular to the filter on the right (dark area) and parallel to the filter on the left (lighter area). (credit: P.P. Urone)
This schematic is another variation of the schematic first introduced two figures prior. To the left of the vertically oriented polarizing filter is a double headed blue arrow oriented in the plane perpendicular to the propagation direction and at an angle theta with the vertical. After the polarizing filter a smaller vertical double headed arrow appears, which is labeled E cosine theta.
A polarizing filter transmits only the component of the wave parallel to its axis, E cos θ size 12{E"cos"θ} {} , reducing the intensity of any light not polarized parallel to its axis.

Calculating intensity reduction by a polarizing filter

What angle is needed between the direction of polarized light and the axis of a polarizing filter to reduce its intensity by 90 . 0% size 12{"90" "." 0%} {} ?

Strategy

When the intensity is reduced by 90 . 0% size 12{"90" "." 0%} {} , it is 10 . 0% or 0.100 times its original value. That is, I = 0 . 100 I 0 . Using this information, the equation I = I 0 cos 2 θ size 12{I=I rSub { size 8{0} } "cos" rSup { size 8{2} } θ} {} can be used to solve for the needed angle.

Solution

Solving the equation I = I 0 cos 2 θ size 12{I=I rSub { size 8{0} } "cos" rSup { size 8{2} } θ} {} for cos θ size 12{"cos"θ} {} and substituting with the relationship between I size 12{I} {} and I 0 size 12{I rSub { size 8{0} } } {} gives

cos θ = I I 0 = 0 . 100 I 0 I 0 = 0 . 3162. size 12{"cos"θ= sqrt { { {I} over {I rSub { size 8{0} } } } } = sqrt { { {0 "." "100"I rSub { size 8{0} } } over {I rSub { size 8{0} } } } } =0 "." "316"} {}

Solving for θ size 12{θ} {} yields

θ = cos − 1 0 . 3162 = 71 . 6º. size 12{θ="cos" rSup { size 8{ - 1} } 0 "." "316"="71" "." 6°} {}

Discussion

A fairly large angle between the direction of polarization and the filter axis is needed to reduce the intensity to 10 . 0% size 12{"10" "." 0%} {} of its original value. This seems reasonable based on experimenting with polarizing films. It is interesting that, at an angle of 45º size 12{"45"°} {} , the intensity is reduced to 50% size 12{"50"%} {} of its original value (as you will show in this section’s Problems&Exercises). Note that 71 . 6º size 12{"71" "." 6°} {} is 18 . 4º size 12{"18" "." 4°} {} from reducing the intensity to zero, and that at an angle of 18 . 4º size 12{"18" "." 4°} {} the intensity is reduced to 90 . 0% size 12{"90" "." 0%} {} of its original value (as you will also show in Problems&Exercises), giving evidence of symmetry.

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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