Angelo The Io in Malus' law refers to the intensity of light that is already polarized. This polarized light then passes through an analyzer. In other words, it is the intensity passing through the analyzer. To polarize the light in the first place, a polarizer must be used and this would result in a 50% drop in intensity.
I watched all your videos on this topic, and they really helped me to construct a better understanding. You also clarify a number of concepts which are kind of vague in my physics class. Thank you sooooooo much!
I am confused aout the graphic at the 1 minute 40 second mark: Specificially, I am looking at the hand-drawn electric field shown in green. I am having problems understanding how this is perpendicular to the direction of motion. The angle looks more llikee 45 degree. What am I missing here?
The electric field oscillates within the green plane. It points towards and then away from the common axis of all the planes (black line running left to right across the screen) The wave motion is along the common axis . The oscillations are away from and towards the common axis, and are hence, perpendicular to the direction of motion.
Thank you for another amazing video which explains polarization so well. You mentioned polarization by crystals but didn't say how it works. I think this is the same as double refraction. Can you say briefly how this works. Thanks again.
Ranny Ran Check how your teacher wrote their answer. There is a factor of 1/4 due to the second polarizer, and there is a factor of 1/2 for the light being polarized by the first. The total intensity drop is 1/8.
Sir, I got 7 thanks to you and now 2 years later I still refer to these in uni. Thank you for these amazing videos
Very good to hear!
Bengi Ağçal
What tactics did u use to study effectively? Can you guide
All the best,
You are the reason I am doing well in IB physics! Thank you so much! You are a life saver.
Great to hear!
Angelo
The Io in Malus' law refers to the intensity of light that is already polarized. This polarized light then passes through an analyzer. In other words, it is the intensity passing through the analyzer. To polarize the light in the first place, a polarizer must be used and this would result in a 50% drop in intensity.
@D D It is diagonally polarized after travelling through the diagonal polarizer.
i've just been using all your videos sir and it is really really helping me in understanding everything that i missed out in class!! tysm!!
Great to hear. Thank you!
Thank you so much! You explain it better than every text book haha!
Excellent ! The exercise supports your work perfectly!
I watched all your videos on this topic, and they really helped me to construct a better understanding. You also clarify a number of concepts which are kind of vague in my physics class. Thank you sooooooo much!
Happy its helping!
thank you very much!
Great explanation. Thank you so much.
Your videos are extremely informative, thank you so much!
Glad you like them!
Thank you very much!
excellent video
Thanks for the comment!
Do you have a video where you go into more depth about the intensity equations at 10:55
No. Do you have a question about the physics?
Super EXCELLENT
I don't get the part on 2:50 why can't there be diagonal wave propagations?
malus's law is no longer in the new syllabus right?
It is currently included in the syllabus. In fact, most exams have one question in which it is used.
I am confused aout the graphic at the 1 minute 40 second mark:
Specificially, I am looking at the hand-drawn electric field shown in green.
I am having problems understanding how this is perpendicular to the direction of motion.
The angle looks more llikee 45 degree.
What am I missing here?
The electric field oscillates within the green plane. It points towards and then away from the common axis of all the planes (black line running left to right across the screen) The wave motion is along the common axis . The oscillations are away from and towards the common axis, and are hence, perpendicular to the direction of motion.
It is much appreciated that you took the time to reply.
Thanks for alll your great IB physics videos! They are a wonderful resource for so many of us.
Thank you for another amazing video which explains polarization so well. You mentioned polarization by crystals but didn't say how it works. I think this is the same as double refraction. Can you say briefly how this works. Thanks again.
Yes, essentially the two orthogonal planes of polarization are refracted by different amounts, so you get two polarized beams.
@@donerphysics Awesome. Thanks so much for taking the time respond.
Magical
much obliged, as always
nice!
Thank you! Cheers!
Amazing work! You're saving my ass out here.
Happy its helping!
Thank you sir dancealot
You are welcome
hey,,my teacher wrote on his lecture that the answer for the last question in your video is Io/4 instead of Io/8... im confused...
Ranny Ran
Check how your teacher wrote their answer. There is a factor of 1/4 due to the second polarizer, and there is a factor of 1/2 for the light being polarized by the first. The total intensity drop is 1/8.
Ranny Ran i also got confused first but then realised that it is cos^2 so it is (1/2)^2 = 1/4
1/4 * intensity/2
ah okay! thank you!
Chris Doner hmm i think my teacher made a mistake.. thank you for reminding !!
Ranny Ran In the hectic life of a teacher, lots of mistakes are made.