Malus Law (derivation and use) + What is Intensity (Polarisation Part 2 - Waves - Physics)

[@carbonator]

I've been sitting here trying to read my textbook for 3 hours trying to understand this. You made this so clear in 12 minutes (+the Part 1) that I'm actually smiling, because, I guess I'm not dumb after all!

[@PhysicsMadeEasy]

haha Carson, no you are not. Being dumb is actually quite a rare feature: too many students think wrongfully that they are. Instead, it can be some teachers that are not that pedagogical, or do not have the resources to spend time understanding and discussing individually with students. All intelligence do not follow the same paths, but can arrive at the same destination.

[@pingu8394]

Thank you much for this video. You have some of the best physics explanations on the internet.

[@PhysicsMadeEasy]

Hi Pingu, I am glad you enjoyed my work and that it is useful to you :-)

[@seanryan6001]

Thank you for posting these physics videos. I’ve been trying to learn some physics on my own. I just think it’s fascinating

[@PhysicsMadeEasy]

You are welcome! and the best thing is that more you advance in your journey learning about physics and more it becomes intriguing and fascinating.

[@gravimagswnforce9123]

same here! physics is very fascinating, addicted and clarifying!

[@ΚωνσταντίνοςΛαζαρίδης-ξ9ι]

thanks

[@roryshannon7629]

Amazing and cool as you promised in the last video! Thanks!

[@PhysicsMadeEasy]

Thank You Rory, I am glad you enjoyed it!

[@toxikarp2063]

Thank you for your efforts sir, I hope you always find happiness in what you do.

[@dadagdadig5344]

Thankyou very much Sir ..... You explains everything so nice , you are very calm and sweet .

[@artranadani8051]

tbh i like how u make things this easy thank u

[@PhysicsMadeEasy]

That's the goal of the channel! Thank you for your feedback Dani!

[@Mark-nu2kd]

really easy to understand the concept with the drawings, Thank you so much

[@PhysicsMadeEasy]

Thank you! I remember spending quite some time on the drawing, so your comment is appreciated!

[@courage936]

underrated channel

[@SamiraMammadova-q3d]

So good explanatiin. Thanks😇

[@surendrakverma555]

Excellent lecture Sir. Thanks 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

[@JOSPH1]

don't understand the last part. malus law is I=Io x cos squared theta but why use 100 as Io not 200 because 200 is the unploarised light

[@PhysicsMadeEasy]

Hello Joseph, When you pass unpolarized light through a polarizer, there is one rule that applies. the intensity is divided by 2. (see video Polarization Part 3 for the proof: kzbin.info/www/bejne/gnXTe4ltqdqDeLMsi=pmwhFS52hMkH0kqE) Malus Law applies on light that is already polarized. In the example at the end of the video, The intensity of the UP light was 200W/m^2, it passes through a polarizer and gets polarized, thus the polarized light coming out I = 100W/m^2. Then this polarized light passes through a second polarizer, and here Malus Law applies (the incident light of I = 100 W/m^2 is already polarized... )

[@JOSPH1]

thank you very much for this clear explanation@@PhysicsMadeEasy

[@notwasi4522]

Thank you so much!

[@zakirhussain-js9ku]

Thank you for Excellent video. I have not understood why amplitude has to be square to get intensity.

[@PhysicsMadeEasy]

Think about the oscillation of a particle in a mechanical wave, like the simple harmonic motion of a spring. The total energy of the spring is proportional to the square of the maximum displacement: when x = xmax, TE = PEmax = (1/2)kx^2. Intensity is related to energy, and amplitude is maximum displacement...

[@elku3775]

excellent.

[@chenu72]

Thankyou sir 🙏 well explained 👏 👍 👌

[@koushik5797]

Finally i understood what is law of malus 😅 By your's extraordinary explanation 😊 By the way i am from india and your video is helpful to me also that is amazing and unbelievable 😂

[@PhysicsMadeEasy]

Thank you for letting me know. I am very glad my work helps you improve your understanding :-)

[@SarahPatel-c7g]

AMAZING

[@mustaphachinoun2580]

Thank you very much, there are light waves in the universe emitted by all objects from space, oceans and the interior of the earth. The question: How can these light waves be captured and then analyzed for % to confirm the fact that this emitted light is C we are looking for.

[@PhysicsMadeEasy]

Hello Mustapha, I am not sure I understand well your question. Whatever the light, it always travels at a speed of c. Are you trying to find a way to determine what fraction of a light coming from multiple objects, comes from what object ? For that, if you know the temperature of the objects and if these are all different, you should be able to figure out at what wavelength each object emits most of its light. So if you make a spectra of the total light, you'll see peaks, and will be to identify the contribution of each object. I recommend you watch my videos about Black Bodies in the suggested order 1 - kzbin.info/www/bejne/gqC1g3holqxmd7ssi=alXB8FrPioNMT-hb 2 - kzbin.info/www/bejne/q5jWp3R6n5aqqtUsi=Zt1-o-wn4FuQ4aMa If this didn't answer your question, fell free to reformulate it.

[@jakubliszka9155]

i love your videos, thanks

[@PhysicsMadeEasy]

Hi Jakub, thank you for your feed back! I am glad you enjoy my work.

[@mrslave41]

5:39 “I~kA^2 [intensity of a given wave 🌊 is proportional to the amplitude squared]” - why?

[@PhysicsMadeEasy]

Hi, In a nutshell: the amplitude of a wave is proportional to the maximum velocity of its oscillators (for example for a given oscillator, v(t) = A*omega (sin (omega * t)). And because the energy carried by the wave (via the movement of its oscillators) is proportional to that velocity squared, the energy is thus proportional to the square of the amplitude. (note that Intensity is proportional to energy). For an EM wave where the amplitude is related to the square of the field strength, while more abstract, the reasoning is similar . The relationship can be derived from Maxwell equations)

[@gravimagswnforce9123]

nice! very well explained!

[@PhysicsMadeEasy]

Thank you gravimag!

[@funnycomplications6523]

You are the best♥️♥️♥️👌👌

[@PhysicsMadeEasy]

:-)

[@durjoysen3866]

Thank you sir

[@PhysicsMadeEasy]

You are welcome Durjoy!

[@OlamideOlarewaju-o4f]

Thanks