Unpolarised light: Orientation of the E-field

[palkia]

Homework Statement

How does this picture represent a unpolarised light?

Homework Equations

The Attempt at a Solution

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I thought light waves were perpendicular to the direction of propagation so if it has diagonal components then isn't that not perpendicular to the wave direction

If it is a unploarised light then can we resolve an unpolarised light into two polarised light wavesIs unpolarised light mix of different polarised light in different directions?

 

[Orodruin]

You are misinterpreting the image. All of the vectors in the yellow disk are perpendicular to the direction of propagation. The yellow disk itself is perpendicular to the direction of propagation. (In three dimensions, there are two directions that are perpendicular to any given direction.)

 

[palkia]

are these wavesof a single light wave or of multiple light waves

 

[NascentOxygen]

The incident light is a mix of lots of waves, they favour no particular polarisation.

 

[palkia]

The incident light is a mix of lots of waves, they favour no particular polarisation.

So I can think of unpolarised light as conissting of many polarised lights perpendicular to the direction of wave direction...?
Can we do components of these polarised waves in any two perpendiular direction?

 

[haruspex]

So I can think of unpolarised light as conissting of many polarised lights perpendicular to the direction of wave direction...?

Yes, but bear in mind that an individual photon does not need to be polarised exactly in line with the filter to get through. The rule is that if the angle between the polarisations is θ then the chance of getting through is cos²(θ), but having got through it will now behave as though exactly aligned with the filter.

Can we do components of these polarised waves in any two perpendiular direction?

I think you can in terms of the quantum wave function (I may be corrected by a quantum theorist on the thread) but not in terms of the chances of getting through the filter, so not in terms of the intensity that gets through. The reason is that cos² behaviour.

 

[rude man]

So I can think of unpolarized light as consisting of many polarized lights perpendicular to the direction of wave direction...?
Can we do components of these polarized waves in any two perpendicular direction?

A polarized wave can be resolved into any two assigned perpendicular directions (axes).

 

[haruspex]

A polarized wave can be resolved into any two assigned perpendicular directions (axes).

So how does this work...
Suppose a light beam is polarised at angle θ anticlockwise from the x-axis and the filter is set at angle φ to that axis.
Without resolving into components, the transmitted intensity factor is cos²(θ-φ).
If I first say that the incoming beam is equivalent to a fraction f(θ) parallel to the x-axis and a fraction f(θ') normal to it, where θ+θ'=π/2, then the transmitted intensity should be
f(θ)cos²(φ)+f(θ')sin²(φ).

Is there a function f that makes the two expressions the same? Looks to me that cos² is the only possibility, but that gives cos²(θ)cos²(φ)+sin²(θ)sin²(φ) whereas cos²(θ-φ) expands to cos²(θ)cos²(φ)+sin²(θ)sin²(φ)+2cos(θ)cos(φ)sin(θ)sin(φ).
Note in particular what each gives for θ=φ.

 

[rude man]

So how does this work...
Suppose a light beam is polarised at angle θ anticlockwise from the x-axis and the filter is set at angle φ to that axis.
Without resolving into components, the transmitted intensity factor is cos²(θ-φ).
If I first say that the incoming beam is equivalent to a fraction f(θ) parallel to the x-axis and a fraction f(θ') normal to it, where θ+θ'=π/2, then the transmitted intensity should be
f(θ)cos²(φ)+f(θ')sin²(φ).

Is there a function f that makes the two expressions the same? Looks to me that cos² is the only possibility, but that gives cos²(θ)cos²(φ)+sin²(θ)sin²(φ) whereas cos²(θ-φ) expands to cos²(θ)cos²(φ)+sin²(θ)sin²(φ)+2cos(θ)cos(φ)sin(θ)sin(φ).
Note in particular what each gives for θ=φ.

I was referring to the E field vector.The E field vector is the fundamental parameter. Vectors can be resolved, intensities (which are not vectors) only indirectly so (by squaring the resultant E field).

When it comes to this kind of problem I prefer to think in terms of E vectors rather than intensities. Then the resultant E vector can be squared to get the corresponding intensity. And E vector magnitude goes as cos and sin. Probably just habit.

 

[haruspex]

I was referring to the E field vector.The E field vector is the fundamental parameter. Vectors can be resolved, intensities (which are not vectors) only indirectly so (by squaring the resultant E field).

When it comes to this kind of problem I prefer to think in terms of E vectors rather than intensities. Then the resultant E vector can be squared to get the corresponding intensity. And E vector magnitude goes as cos and sin. Probably just habit.

Ok, I think I see how it works for intensities.
If the incoming light has intensity I we can resolve that into vectors of magnitude I^½cos(θ) and I^½sin(θ). The magnitudes that get through the filter then follow a cos rule, not a cos² rule, so are I^½cos(θ)cos(φ) and I^½sin(θ)sin(φ). These are magnitudes of vectors aligned with the filter, so we can add them directly to get magnitude I^½cos(θ-φ).
Finally squaring to get the transmitted intensity: Icos²(θ-φ), as required.