Unknown Light Source Through Polarizer

In summary, the light from an unknown source with intensity 1.60 × 10−2 W/m2 passed through a polarizer with horizontal and vertical transmission axes, and the intensity of the transmitted light was 8.00 × 10−3 W/m2 when the polarizer was wiggled about. However, the light could also be unpolarized, which is the only answer that makes sense.
  • #1
kkcolwell
3
0
I recently had a homework problem that asked the following. Can someone please explain to me how in the world this can be true?! It makes no sense to me that you can get an intensity that is equal to when light is passed through a vertical transmission axis and a horizontal transmission axis...

When light from an unknown source, with intensity 1.60 × 10−2 W/m2, passes through a polarizer with a horizontal transmission axis, the intensity of the light transmitted is 8.00 × 10−3 W/m2. When the light from the unknown source passes through a polarizer with a vertical transmission axis, the intensity of the light transmitted is again 8.00 × 10−3 W/m2.

These are the choices. To me I would think that maybe it would be unpolarized since the electromagnetic fields are randomly oriented so maybe it could go through both, but that does not seem logical to me...

the light from the unknown source is unpolarized
the light from the unknown source is linearly polarized
the light could be unpolarized or linearly polarized

All I ask is for some clarification because I am finding this section to be very troublesome...
 
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  • #2
If the source were linearly polarised and it just happened to be polarised at 45°, you would get half the power through the polariser in both orientations (but it would need to be 45°. If unpolarised, you would get half power for all orientations. Any of the three choices are reasonable. A2 or A3 could be the answer they're after but the more likely answer would be A1 because it doesn't require a particular pair of orientations. It's a good question because you have to think about it and you would really need to argue a case fro whichever answer you went for.
In real life, you'd wiggle the polariser about a bit and see if the power varied - if not then the answer is Unpolarised.
 
  • #3
From the answers you give, the only possible answer is that the light must be unpolarised. There can be no degree of linear polarisation.

Technically, the light could also be circularly polarised; but since this is not offered as an option, the first answer must be true.

Claude.
 

Related to Unknown Light Source Through Polarizer

1. What is a polarizer and how does it work?

A polarizer is an optical filter that only allows light waves of a specific orientation to pass through. It works by absorbing light waves that are not aligned with its transmission axis, while allowing those that are aligned to pass through.

2. What is an unknown light source?

An unknown light source is any source of light that is not easily identifiable or known to the observer. It could be natural, such as the sun, or artificial, such as a light bulb, but the exact source is not immediately apparent.

3. How can a polarizer help in identifying an unknown light source?

A polarizer can help in identifying an unknown light source by revealing its polarization properties. By rotating the polarizer and observing the changes in the intensity and direction of the light passing through, scientists can determine the polarization characteristics of the light source and potentially narrow down the possible sources.

4. What are some common uses of polarizers in scientific research?

Polarizers have a wide range of applications in scientific research, including in optical microscopy, spectroscopy, and imaging. They are also used in studying the polarization properties of natural light sources, such as the sun, and in materials science to analyze the polarization properties of materials.

5. Can a polarizer be used to block out unwanted light?

Yes, a polarizer can be used to block out unwanted light by rotating it to block the polarized light waves of a specific orientation. This is commonly used in photography to reduce glare and improve the contrast of the image. It can also be used in optical instruments to eliminate unwanted reflections and improve image clarity.

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