Why does a polariser cut the intensity of unpolarised light in half? (Polarisation Part 3 - Physics)

[@PhysicsMadeEasy]

Errata: as some of you guys noticed, there is a little calculation mistake at 9:25. The primitive of cos(2x), is of course (1/2)sin(2x)... It doesn't have any effect on the reasoning or the final result. That is why I didn't spot it in time ;-)! Sorry about that!

[@natharistaats]

Finally I came to the right place!! thanks to you sir it's as clear as crystal to me now !! The google and other channels were utterly inadequate to teach this to me but you were able to do this wonderfully!! Brilliant sir..❤

[@PhysicsMadeEasy]

Thank you for your kind words Nathari, I am glad you enjoyed my work :-)

[@natharistaats]

@@PhysicsMadeEasy ❤😊hope for more fascinating videos sir!

[@TN-id4yx]

Outstanding explanation. I recently discovered your channel and it really helped in clarifying my basic concepts.Sir, please make videos of topics from Indian physics textbook for grade 12. Your teaching style is brilliant and makes me understand the concepts to the core.

[@PhysicsMadeEasy]

Thank you so much TN for these encouraging words. You know, Physics is the same whatever the textbook, I prefer to focus on the basics that can be useful to students from all programs :-). But. if you have a specific question, you can always ask me ;-)

[@TN-id4yx]

@@PhysicsMadeEasy Thank you very much.

[@EmpyreanLightASMR]

I'm sure OpenStax textbook tells us a polarizer cuts unpolarized light in half and I probably missed it somewhere, but I had to dig to find an explanation and this video is AMAZING. I still wish I were smart enough to understand when to just integrate something. You're given the intensity of light equation, but I wish I knew I could actually integrate it over an interval (and even average it) to find the final result. Like, even after three calc classes, I'm still not on that level of comprehension. Beautiful though, thank you.

[@PhysicsMadeEasy]

Hi, Integration is used when you want to sum something (a quantity A) across another quantity (B) on which A depends. Simple example to illustrate: A spring. You know that if you want to extend a spring by x meters, you need to pull with a force F = k*x where k is the spring constant (stiffness of the spring). How much energy do you need to provide to the spring to do so? You could just multiply F by x, to find the work you did, but that would be incorrect, because F depends on x... So you integrate W = Int(Fdx) with F = kx, that gives you W = Int(kxdx) By applying integration rules, you find: W = k*Int(xdx) = kx^2/2, which is the formula of the potential energy of the spring when extended by x.. I hope this helps you understand when to use integration

[@АлексГудев]

Great video!! Loved the physics explanations on the energy in the beginning :) By the way, you can also do it without the integral - just note that the average of cos2θ in [0,π] is zero due to symmetry of cos2θ. Then average(cos²θ)=average(½ + ½cos2θ) = ½ :) Found your video googling precisely the question in the video title, glad that I found it! Looked at the other videos in the series, found great explanations there, too. Thank you!

[@PhysicsMadeEasy]

Great point, thanks for sharing! And thank you for your kind words, I am glad my work helps you!

[@goodVibes-un6ml]

Thank you so much 😊 great explanation!

[@PhysicsMadeEasy]

You are warmly welcome. Thank you for the good vibes ;-)

[@aashishlearningbeastkarki9552]

Thanks a lot Sir. Love from Nepal.

[@AmitKumar-oq6qu]

Finally I found this derivation.... Loved this video, you are awesome sir......

[@PhysicsMadeEasy]

Thank you Amit!

[@cindyyang]

Great explanation! Happy I found this video~ greetings from Taiwan 😊

[@PhysicsMadeEasy]

Thank you, greetings from France :-)

[@jlpsinde]

Very good!

[@Nobody-y1m]

9:59 the intergral of cos(2x) = sin(2x)/2, not 2sin(2x)

[@PhysicsMadeEasy]

Absolutely, Well spotted. You are not the first one to notice that... I didn't notice it at first because it does not interfere with the final result... I pinned an errata comment on that matter, so thanks for your feedback :-)

[@yuzon3754]

integral of cos2x is 1/2sin2x, not 2sin2x! but otherwise, great video thanks!

[@PhysicsMadeEasy]

Ooops... you are correct, well spotted! Thank you Zain. Luckily, this error does not invalidate the proof (the terms with the sin end up being equal to zero anyway).

[@physicslab5787]

Very nice

[@ElVerdaderoAbejorro]

Why do you have to take the average while taking the integral? Isn't the Intensity that passes through simply the sum of all intensities between 0 to pi? I understand tha regardless if you take the average or not, the result will still be the same, but I would like to understand the physical reasoaning on why you are averaging the integral.

[@PhysicsMadeEasy]

To try and make some physical sense out of this, consider each polarisation plane intersecting a screen. For each polarised ray, you get a line. By integrating this line over the angle, you get the surface of a disk. If you wanted to know the total energy or power received by that disk, you would just sum the energy or power contribution of all polarised beams. But here, we are considering an intensity (a surface density of energy), this is why you need to take an average, As an analogy: consider 1 liter-samples of sea water from 3 different oceans and measure they concentration in salt: Atlantic: 35 g/Kg, Pacific: 40 g/Kg, Baltic 30 g/Kg. Add them all in a bucket. To get the total amount of salt, you just add 35+40+30 =105g. But what would be the salt concentration in the bucket ? You would add all concentrations, and then take the average (35+40+30)/3 = 35g/Kg , wouldn’t you? That’s what we do with intensity: because the energy is not deposited on the same surfaces and that we are dealing here with "energy concentrations", we need to average the sum to get the intensity on the full disc. I hope this helps!

[@EmpyreanLightASMR]

@@PhysicsMadeEasy awesome explanation

[@MijanurRahman-uk3dv]

Awesome😊

[@adhihcmw]

Can U do Some tutorials in English based on Indian grade 12 physics lesson by lesson so it's very help full

[@PhysicsMadeEasy]

Hello Adith, The idea already crossed my mind because I realised how Indian students are highly motivated. All curricula are different, and that reflects in the exams. So if I decided to produce a course for Indian Grade 12 in Physics, I would need to first study that curricula in detail to make sure that what I teach is concordant with what you can expect at an exam. And that takes time... And you know how time is a very rare commodity! So instead, I focus on the roots of Physics which are common to all curricula. With this channel, I try to help students get the basics. And from there, students can easily adapt to their respective physics classes because they understand the root concepts.

[@adhihcmw]

@@PhysicsMadeEasyI agree...I Indian schools like mine teaches just the theory whithout showing am example or anything they just say with Thier mouth ... So our basics in physics is bad...our aim is to solve physics promblem buy instead of teaching in base of promblem like teaching theory with an question example they are not doing that...I like tutorials which first give a question based in the topic and then to solve that question what we need to study that should be Teached....I really hate physics because I was not taught well..but when I saw your videos I find it entertaining.....and sad truth is there no videos in KZbin with good tutorials

[@RehemaAkoth-i4s]

What is polarization

[@PhysicsMadeEasy]

Hi, there is a video in my channel that answers this question. "What is polarised light? ": kzbin.info/www/bejne/i2mpkqGQqpmLZ6ssi=nnIyZ9QGXIJ_kzy5

[@zedliu5945]

Hi, why for the integration part, it's pi not 2*pi, since the angles are all around 360 degrees?

[@sagartripathi4826]

With respect to axis of polarizer, the rays can't make any angle greater than 180 degrees.

[@PhysicsMadeEasy]

Hi Ze, Sagar responded correctly. If you rotate the polarizer by 180 degrees, you are in the same configuration...

[@_BhagavadGita]

cheers

[@BYMYSYD]

I find this explanation still unsatisfactory. Why? Because simply saying unpolarized light is composed of a random distribution of linearly polarized plane waves makes a contradiction. Since electromagnetic fields follow the rules of vector addition, and the distribution is kept rather symmetric (since puting unpolarized light through a polarizer always gives 1/2 Intensity independent of rotation), you would expect the average of the E field to be zero. But it's not. I'm looking for an unpolarized classical solution to Maxwell's equations, whos intensity is cut in half when passing through a linear polarizer.

[@PhysicsMadeEasy]

Hello Mike, Thank you for your comment. It is difficult explaining this without illustrations But I’ll try, at least maybe this will give you a path for reflection. First, I am sure you know, but, just to make sure: an EM wave is the oscillation of an electric field vector: If you fix a position, and observe the evolution with time of the E Field vector at this position, you will observe the oscillation of its magnitude in a plane perpendicular to the direction of its propagation. Why do you think the integration was from 0 to Pi and not 2Pi, because when you rotate the polarizer by 180 degrees, it is the same wave that passes through: the same vector. You do not have any vector opposite that would cancel it in any given plane. I hope this helps Edouard

[@BYMYSYD]

@@PhysicsMadeEasy I have trouble with the second paragraph. You say you imagine a plane fixed to a particular location perendicular to the direction of motion (the wave vector k). The trouble arises when you are trying to model UNpolarized light by saying that it is a bunch of plane waves with random polarization vectors. This means that if you were to calculate the electric field in that fixed plane, you'll end up with zero. It's like asking yourself what is the average velocity of all the gas molecules in a stationary box...zero.

[@BYMYSYD]

@@PhysicsMadeEasy Here. This might have the answer I'm looking for. What do you think? www.osapublishing.org/viewmedia.cfm?r=1&rwjcode=josa&uri=josa-35-8-525&html=true

[@PhysicsMadeEasy]

Hello Mike, You are correct. Your analogy with the gaz moledules made me realise that the model I use to describe unpolarised light is conceptually incorrect. Such model imposes that when the relative phases are perfectly random, that the light is perfectly monochromatic and that all amplitudes of all rays are rigorously identical then, saying that unpolarized light is the sum of polarized rays, means that unpolarized light would become… no light at all… The integration leads to the right answer, but the model is incorrect. The reason it provides the right result is that I integrate the intensity (the square of the electric field vector), the same way we get a non zero average speed by taking the rms of the velocity. I did a bit of digging, and found out a model that remains classical, and prevent the nullification of the resultant electric field strength vector. It does so by involving a time component. Instead of seeing unpolarised light as a distribution of polarised rays in space, we can do so in time: As a classical model, unpolarised light can be seen as a polarized ray that fluctuates in polarisation on extremely small time scales (check the answer of Wolpertinger to this thread for more details: physics.stackexchange.com/questions/177572/what-does-electric-field-of-unpolarized-light-look-like-when-measured). Thank you si much for your input on this one, you made me evolve! :-)

[@BYMYSYD]

@@PhysicsMadeEasy Yeah. I spent my entire undergraduate education thinking that the unpolarized light model was just a bundle of pane waves with random polarization angles (just as my instructors and professors told me..and it got my through the exams thinking that). And later learning the quantum model further reinforced that belief, since the quantum model was a more solid theory, so there was no point to find anything wrong with the classical model despite knowing that adding random vectors results in zero. I didn't realize there was something wrong with the way classical unpolarized light was taught the school system until about 10 years later. It wasn't until I remembered that when Maxwell, and Einstein used EM theory (last quarter of the 19th century), they didn't have a quantum model, but certainly they understood that Maxwell's solutions required vector fields, and there was such thing as unpolarized light. So there had to be a working classical model.

[@mcneelynorman1]

Savage!

[@PhysicsMadeEasy]

what's savage about it?

[@mcneelynorman1]

@@PhysicsMadeEasy Where I am from savage can be thought of as, incredible. And I thought this video was incredible.

[@PhysicsMadeEasy]

@@mcneelynorman1 Thank you 🙂