Polarization of Light and Malus's Law - IB Physics

This lecture covers:
-The difference between polarized and unpolarized light
-The two ways light can be polarized (polarizers and polarization by reflection)
-The intensity of polarized light
-Malus's Law and analyzers
-Proving Malus's Law
-Sunglasses as a real world example of polarization
-Graphs of intensity vs. angle for polarizers and analyzers

0:00 Summary of the Lecture
0:13 Unpolarized vs. Polarized Light
1:57 Notation for Unpolarized vs Polarized Light
2:15 How Light is Polarized
3:14 How Polarization Affects Intensity
3:48 Analyzers
5:07 The Axis of Transmission
5:27 Proof of Malus's Law
7:27 Example Problem 1
8:57 Example Problem 2
9:55 Example Problem 3
11:16 Sunglasses as an Example of Polarization
12:23 Intensity vs. Angle Graphs
12:40 Example Graph 1
14:15 Example Graph 2
14:58 Example Graph 3
16:07 Example Graph 4
16:37 Example Graph 5
17:30 Summary of the Lecture

[emeraldbeast]

this video made the unit so much easier than it used to be what a life saver

[mtasim]

Dont know why i dont get recommended such good videos :(
Best video explanation on yt! thank you sir!

[nataliadelgado]

Ur literally a lifesaver. I'm seeing every video you've made to prepare for n23 exams!

[merrymusic]

Extremely useful❤❤❤❤Thank you so much got no words❤❤❤❤

[jennyzhang]

Best polarization video I have seen!! Thank you so much for the work

[lukascupic]

Really easy to follow, makes my physics lab tommorow less intimidating

[KenCorpuz-fmry]

Thanks man for helping me for my presentation on Monday thanks for the vid bro!!!!

[mashanatanzon]

Hello, Great video ! At minute 11:43, you say that the reflected polarized light off horizontal surfaces will then reach the glasses in which Malus's Law is used, respective to the angle. Therefore, glasses are analysers right since polarizers off the surfaces already polarised the light. Thank you

[juniorcyans]

Finally I got some real feelings about this topic! Thank you🎉

[shreya]

Sir, at 8:53, you seem to get the answer as 18.8W/m^2. When I calculate it, it comes out us -25. Cos(30)^2 = -1 and this times 25 = -25. Can you explain how you did your calculations?

Same for 9:43, the intensity before light goes through the analyser is 120 W/m^2 but 30/Cos(60)^2 is coming to be 30 for me.

[kateanf]

amazing video! i have to do a lab report for a polarization experiment and this helped me so much, thank you

[pratishparswani]

Ur my saviour and god ! Thanks Andy you absolute gem !

[mrkhunt.]

Excellent video on Polarisation!! The animations help so much, thank you

[bagoffood]

Thank you very much I have my exams tomorrow and this is a huge help

[stephenhidalgo]

8:55 how do you solve that in scientific calculator?

[ore]

This is amazing, this topic has been so difficult for me to understand but you made it so simple.Thank you.

[ninadlunge]

Thank you for this amazing video full of amazing visuals and explanations.

[aaronchoo]

sir mad respect to you! keep up the good work sir!!!

[shamelsinha]

Hey, I was wondering whether the Brewster's Angle is still a part of the syllabus for SL?

[musicme-e]

thank you for the video. currently cramming for may 23' ib physics exams in a few days (;-;)