Physics 53 Polarization (3 of 5) Three Polarizers
Vložit
- čas přidán 28. 04. 2013
- Visit ilectureonline.com for more math and science lectures!
In this video I will show you how to calculate the final intensity of the ray after it passes through 3 polarizers. - Komedie
For anyone wondering, Malus' law works only for polarized light.
This was indeed pretty interesting! My intuition definitely led me to believe there would be 0 light.
Thank you sir, keep the videos coming, it helps students like me. May God bless you!!!
I know all this must hold up for a the photon (quantum) explanation?
Thank you! Good explanation.
Counter-intuitive resoult to me. I especially like this stuff in science. Thanks for the video!
Yes, The first time I saw this, I had a hard time believing this myself. When I finally had a chance to study more advanced optics it became clear how it worked. That is the exciting part of science, discovering for yourself how everything in nature works.
Michel van Biezen I can kind of guess why the three polarizers in this video work like this - it feels to me like since they are set up so that there is a 45° angle difference, they will never be able to completely block out any of the x-and y-components, but only parts of them.
In other words, if you have two polarizers, and they are perpendicular to each other, then the first one will completely block one component and the second one will completely block the second component, but you cannot completely block any of them if the polarizers differ by 45°.
Kinda hard to explain, but I think it works a bit like this on some level.
@@Peter_1986 it is because polorasiers actualy rotate the light.
Amazing Thanks Sir!!!!
Thank you for the video. It helped me a lot :)
The math checks and the experiment confirms it... but it's so counterintuitive it still bothers me a bit :P
How come there is a factor of 1/2 for intensity I_1 but not for I_2?
Oh and thanks for these videos! These helped me with engineering dynamics last semester and now this is helping with E&M :)
+CheeseTomatoe When unpolarized light passes through the first polarizer, the intensity drops by 1/2, because one of the directions is completely blocked. When passing through the second polarizer, the amount of Intensity reduction depends on the angle between the two polarizers. (note that the light going through the second polarizer is already polarized).
Hi, I have a problem with my LCD display. The display was monochrome. After removing the polarizing film and rotating it by 90 degrees, the colors have changed. The black has changed to white and white to black. After removing the next filter (transparent, which was under the polarizing filter) and re-applying the polarizing filter, the colors change - yellow, purple, green ..... and I can not use black. I am asking for a hint of what's stuck under the polarizing filter, what should I look for to "fix" my display. On my channel there are videos that show the problem :( Thank you in advance for a thank you.
What if the third polarizer is moved in place of second. The intensity of light is 0, but in application if you put the third polarizer at an angle of 45 degrees it becomes bright again. 0*cos^2(45). (i2 = 0.)
Nice presentation. Please reply, can we say 0-degree polarised image is a non-polarised image?
It is not so much the angle, as a beam of unpolarized light will have components oscillating in all directions in an equal amount. Once such light goes through a polarizer, the all the components of the various portions of the beam perpendicular to the polarizer will be blocked.
Hello, I am trying to replace a poliriser on a device, but the colours are inverted, any thoughts?
Hello and thanks for the explanation, I got a question, If the light is already polarized (vertically for example)and it goes through a first polarizer with and angle, does the intensity would be half? or in that case we have use the angle formed between the vertical polarized light and the polarizer?.
In that case you DON'T have to divide by half. Just use the formula to find the intensity (as if it is the second polarizer).
I got it now! Thx a lot sir!!
How many molecules are there in 1m^3 gas in the standard state? (Pressure is 1.013•10^5 Pa, Temperature is 273 K)
At STP conditions 1 mol occupies 22.4 liters. (There are a thousand liters in a cubic meter)
what if the first polarizer is tilted through an angle theta then what will be the intensity of light is it half of the given intensity or not?
Assuming that the light is not polarized when it reached the first polarizer, the intensity will drop to 1/2 of the original intensity when passing through the first one, regardless of the angle of orientation.
Michel van Biezen really you mean angle does not matter for first polarizer
No, only the angle between the first and the second polarizer.
Unpolarized light falls on two polarizing filters so oriented that no light is transmitted. If a third polarizing filter is placed between them, light still cannot be transmitted.is it right or wrong?
Wrong, if the two polarizing filters are perpendicular to one another, not light will make it through the second polarizing filter. But if another filter is placed between the 2 at an angle between 0 and 90 degrees (not including 0 or 90) then light will pass through the third filter.
So one could graph the resultant intensity based off of the 2nd polarizer as cos^2(pi/4)*cos^2(pi/2-x)*cos^2(x)?
The best way to check is to calculate it and see if you get the same answer as in the video.
@@MichelvanBiezen I did and I got the same result, just wanted to comment tbh
So if we would rearange those filters: first, third, second, then no light would pass?
If you set the second polarizer at an angle of 90 degrees relative to the first one, no light will pass through. (theoretically)
Hello, thanks so much for the explanation, I'm incredibly grateful.
I have a question; this thing has really stumped me...
I have two polarizers at an angle of 90°. I'm supposed to put a polarizer in between at such angle so that the final light intensity is 1/4 of the initial.
I have been calculating this for a few hours now and I came to a conclusion, that there is no solution to this... (the highest intensity I could reach was at 45° angle and was equal to 1/8). Thank you so much for your answer and videos - like I said I'm really grateful.
You are correct. Read the problem carefully. Are they referencing the initial intensity (prior to entering the first polarizer) or relative to the the intensity leaving the first polarizer.
@@MichelvanBiezen Hello, thanks for your answer and time.
They are referencing the intensity prior to entering the first polarizer.
Now if the beam approaching the first polarizer was already polarized, this should work out for the intensity to be 1/4 of the initial.
It is not explicitly stated whether the initial beam is polarized or not, however, they do state that the electric intensities of the initial light relative to the axes are equal (Ex=Ey). This should further imply, that the light is unpolarized, since if it were polarized, the intensity relative to one of the axes should be 0.
Are my assumptions correct?
Thanks for your answer
Yes, that is a good way of looking at it.
@@MichelvanBiezen All right. Thank you sooo much! :)
Excellent explanation.
Glad it was helpful!
The third polariser is at 90' to the first polariser though shouldn't the end result/intensity be 0?
+Evultz Strangely enough, no.
Its some quantum mechanics bs :D
thanks a ton mate
You are welcome.
why?
I think the filters breakdown the field, try to use polarized filters that reflects 50% polarized light, like polarized mirror. I believe those filters breakdown and mergers the field The light must be there, the filters just changes to a spectrum we are not able to see it, using mirroring polarized erases the light we are not able to see it. The mirror and polarization must be in the same layer.
It is not really a believe. The science behind polarization is well understood.
@@MichelvanBiezen this must be proven by experiment only, I think 50% of the light are hidden by changing the spectrum to invisible light waves, the electromagnetic waves appears again when we add another filter. By having the mirror, it will completely removes 50% visible light away from the setup. We can not use the mirror and polarized filter in two layers they must be combined in one layer. I think light has more like waves behave when passes through the filter, just like double slit experiment. Thank you for your response!
I would suggest that you watch some of the videos on electromagnetic radiation which would help in the understanding of the properties of E&M, during interference, diffraction, polarization, etc.
Why/how does this happen?
Herp. The intensity of light is a function of the electric field oscillations squared. The oscillations of unpolarized light are in all directions. When light is polarized, the oscillation magnitudes are decreased due to the cancellation of the components perpendicular to the polarizer. If a second polarizer is placed in the path of the light at an angle less than 90 degrees, then some of the electromagnetic oscillations will survive (the components parallel to the polarizer) so there will still be some intensity getting through the second one.
I found an article
www.theimagineershome.com/blog/?p=39
Unrelated but:
Nice tie, bro
Thanks
Why is I1=0.5 I0? Cos 0=1 so I1=I0
You don't apply that formula for light passing through the first polarizer. If unpolarized light passes through the first polarizer, all the component of each wave perpendicular to the polarizer will be blocked and all the components of each wave parallel to the polarizer will pass through so that 1/2 of all light will go through.
I don't think this is correct. There should be a mention of the change in theta for each equation and he simply just plugs in the angle instead of delta theta...
It is correct.
Michel van Biezen so why don't you plug in for theta two (45-45)? All I'm saying is I had 0 for theta one 30 for theta two and 60 for theta three and I had to plug in an account for the change of theta. So I plugged in 0 30 and 30. Whereas following you I would have plugged in 0 30 and 60.
the angles that he wrote ARE the change in theta. relative to the first polarizer, the angles are 0, 45, and 90 degrees. he already took the difference - 45 degrees between each.
it's not working in real life
It is. Try it!