How the intensity of light is halved when unpolarised is passed through polariser???
Answers
Why is the intensity of unpolarized light halved when it passes through a polarizer?
When I received an invitation to answer this question, I struggled with the decision to contribute. After all, I am neither a mathematician nor physicist. But a more pressing concern was not related to my credentials. I can often explain physical phenomena clearly and succinctly. The problem in this case, is that my mental mnemonic for thinking about polarized filters would not satisfy a high-school physics teacher. It is just the intuitive way in which I find it easy to predict results.
Incidentally, the Law of Malus calculates the fraction of light that passes through a polarized filters with a given orientation relative to each other. But I don’t think of rules and laws as an explanation of “WHY”. The question “why” is not typically in the realm of mathematics.
Here, then, is my lay explanation. It may not satisfy you…
Think of a series of many sine waves emanating from a dipole antenna—or a person pushing waves down a jump rope. The sine waves are broadcast at one wavelength, but with random polar orientation. Some waves propagate sideways or at some other angle relative to the sky.
Now, suppose that you create a slit aperture to align the waves (or jump rope energy), forcing it to be at one particular angle—for example, at 45 degrees from vertical. If you place the aperture at a null crossing, you will not affect the waves that pass your device, but if you place it at a peak, you will effectively polarize the waves.
But an aperture is a passive and resistive device. It passes only (Sine 45°)^2 = 50% of the overall energy component along one axis—or along any particular angular orientation. Why? Because the integrated area of single-sine waves distributed over all angles, is 1/2. That is, if the sine waves represent potential energy waiting to be attenuated, half the energy is available to an attenuator at orientation of ‘x’ and half is available to an atenuator at orientation 180-‘x’.
But if you insert a polarizer at 45° between the two perpendicular polarizers, you will pass 25% of the light. This is because you have created a two step filter. Each filter in the series attenuates ½ of the light. Even though the interim stage is columnated, it is 45° off-axis to the third filter.
[I find the insertion of a ¼-wave plate to be far more fascinating].
I am not a mathematician. This is only my intuitive basis for interpreting polarization and attenuation. I really doubt that this mental mnemonic would satisfy a homework or quiz answer for high-school physics class. For that, you would need either better math or a chain of postulates and theorems.
Ellery Davies is a frequent contributor to Quora.
He is also co-chair of and chief editor at .
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Originally Answered: Why is the intensity of unpolarized light half when unpolarized light passes through a polarizer?
This is a very good question that I was always confused about until I actually figured it out. The source of your confusion is probably that you think only a SINGLE VERY SPECIFIC orientation is able to pass through the lens. This is a very common misconception when learning about them. The light wave doesn’t have to be in that exact orientation to come through the lens. That one orientation is where it passes through the lens UNPHASED. As the wave rotates farther away from that specific orientation, light still passes through, but diminishes in intensity, more so at greater angles until you reach the normal, the only orientation where no light actually travels through the lens. However, once through, their orientations become turned to the favored orientation of the lens. This happens because the vector components of the light in the favored orientation can still get through the lens.
Here’s a little thought experiment to put that into application. If you were to have a polarized lens on top of another polarized lens turned perpendicular to the other one, no light will pass through. This is because the one orientation of light waves that could make it through the first lens are exactly elimintated by the second lens. However, if you were place a third lens between those two lens at an angle to the other two lens, light would pass through all three lens. That’s because a certain degree of light is able to pass through the first one depening on the orientation of the wave when it enters the system. Then, the vector components of the new wave change so that they are oriented in the way that was favored by the first lens. That orientation has vector components in the direction of the new favored orientation of the second lens. Those components allow the light to pass through the second lens with a fraction of its original intensity. The same thing happens between the second and third lens, and so, voila! Light could pass through all three lens, even though they couldn’t pass through the first two by themselves.