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
Hello, great video! I have a single question: if the intensity of the light is divided by two when it passes through the polarizer, why isn't the amplitude modified?
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.
google bard solved example 1 correctly, chatgpt - not. i asked definition of Malus law, chatgpt: "I0 is the initial intensity of the unpolarized light incident on the polarizer,". second problem also successfully failed by chatgpt
You wrote at 11:50 that "light reflecting off horizontal surfaces is polarized horizontal" this is in general not correct. It is only correct for the Brewster angle.
Yup, the IB curriculum only prioritized students knowing that it's polarized along the same axis as the surface, not more detail about exactly when that happens. Thank you for pointing that out!
this video made the unit so much easier than it used to be what a life saver
Really easy to follow, makes my physics lab tommorow less intimidating
Extremely useful❤❤❤❤Thank you so much got no words❤❤❤❤
Ur literally a lifesaver. I'm seeing every video you've made to prepare for n23 exams!
Aw so happy they're helpful!
Finally I got some real feelings about this topic! Thank you🎉
Best polarization video I have seen!! Thank you so much for the work
very informative lesson.
Thanks man for helping me for my presentation on Monday thanks for the vid bro!!!!
Ur my saviour and god ! Thanks Andy you absolute gem !
amazing video! i have to do a lab report for a polarization experiment and this helped me so much, thank you
Thank you very much I have my exams tomorrow and this is a huge help
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
Yup they can be thought of as analyzers here!
Excellent work 🥹. Thank you 🙏
Excellent video on Polarisation!! The animations help so much, thank you
This is amazing, this topic has been so difficult for me to understand but you made it so simple.Thank you.
Thank you for this amazing video full of amazing visuals and explanations.
thank you for the video. currently cramming for may 23' ib physics exams in a few days (;-;)
sir mad respect to you! keep up the good work sir!!!
Thank you!!
You're a genius🙌🏻🙌🏻❤❤
It really help me...... Thank u sooo much sir....
Love this man
Thank you very much for this vid!
Unbelievable sir
Thanks Andy
Hey, I was wondering whether the Brewster's Angle is still a part of the syllabus for SL?
Thx! Very clear!
👍👍👏nice 👏👍👍
Hello, great video! I have a single question: if the intensity of the light is divided by two when it passes through the polarizer, why isn't the amplitude modified?
Thank you so much! This is AMAZING
@@AndyMasley Will you be making the HL portions as well?? Your explanations are really thorough. Would love the HL sections as well!
@@AndyMasley Amazing :)
@@AndyMasley Are any aspects of SL left out from your videos yet? something you haven't made?
amazing
well done mate, great work!
Thank you so much for making this
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.
8:55 how do you solve that in scientific calculator?
life saver
the video is awsome but please can you make the audio to stereotype so it is better to listen to in headphones
great
google bard solved example 1 correctly, chatgpt - not. i asked definition of Malus law, chatgpt: "I0 is the initial intensity of the unpolarized light incident on the polarizer,". second problem also successfully failed by chatgpt
I'm enjoying having a RUclips channel in the last few years before AI just teaches everything for us
😭
What happens if unpolarized light goes through an analyzer?
@@AndyMasley thanks!
You wrote at 11:50 that "light reflecting off horizontal surfaces is polarized horizontal" this is in general not correct. It is only correct for the Brewster angle.
Yup, the IB curriculum only prioritized students knowing that it's polarized along the same axis as the surface, not more detail about exactly when that happens. Thank you for pointing that out!
when you square both sides at 6:43, shouldn't it be cos^2 theta instead of cos theta^2? cuz you wrote it there cos theta^2.
Yup correct, I had some reason for putting the exponent at the end that I have since forgot
The explanations are very good but, unfortunately, you speak way too fast...