matrix methods: Optics with matrices

[@royphilip3964]

This is great content....you have my gratitude...

[@dunotpro]

first of all, thank you so much for creating this video, i am very interesting in this methods, even though i did not completely understand what you say ( cause im vietnamese ) but i think this video is very detailed and in a smart way! (sr my poor E)

[@sspickle]

dang xuan du pro Good! Thanks for the feedback.

[@starryfolks]

Wow a different perspective definitely contributes to increase iq. I graduated from high school last year and ever since I've been trying to understand the nomenclature of quantum mechanics. This vid among others makes the math very clear, although I cannot understand some of it lol. Thank you.

[@lukelatto9941]

It's the wink with the nose for me... Thanks for the content!

[@sofimaiqui]

Hey, congratulations for your video, it has helped me in ways you can't imagine. Excuse me, I'm not really sure if I understand what you're doing at 17:25, you say |r4> = R2 |r3>, but to get |r4> you make R2|r3|r3> Why?

[@sspickle]

Hi +sofimaiqui , I'm not sure what's confusing. R2 is the matrix representing the second refraction at the planar surface on the right side of the lens. The radius of curvature of this surface is infinite so the lower left matrix element of R2 goes to zero. The lower right matrix element is 'n' since nl=n and nr=1. The ray vector, r3 is displayed correctly, so the refraction amounts to R2|r3> which comes out to |r4> (just like |r3> but the bottom component is 'n' times bigger than |r3>). Can you be more specific about the trouble?

[@sofimaiqui]

Thankyou for answering. I see my mistake now, I was doing it by my self in my notebook and I didn't supose that the lower left element of R2 was zero, so I got something different from yours. The funny thing is the lower left element goes to zero if you multiply R2 by (r3)^2 also, so I thought you had done that. Thanks again for answering, I get alot of trubble with geometrical optics. P.S. I love your videos!

[@sspickle]

+sofimaiqui Good! I'm really glad you find them useful.

[@fabianleushacke5081]

Thanks for the video Mr. Spicklemire. I still have a question for the special case of a thick lens. Let's say the radius of the front surface curvature is big, for the translation I would have to use t(h) instead of the centre thickness right? Wouldn't it be more precise? Because if the ray is not paraxial the thickness can be varying to the edges quite a lot. Regards

[@sspickle]

If you're far enough from the axis that this matters, then you're probably outside the limits of the thin lens approximation anyway, so you might as well go ahead and do full blown ray tracing.

[@nikola4294]

Thank you, you are a life saver!!

[@mobarakulislam1602]

So much greatful to you

[@petruspetrusuom3342]

This is very good! thank you

[@mathieu246]

Great video. Clear and concise. Would it be possible to upload the python code?

[@jollyjeevan6683]

Height of a parallel ray in a medium of constant refractive index doesn't change right?

[@sspickle]

Right!

[@virustipps]

excellent video , helped me to understand it ty

[@oyemakinwatayelolu3536]

hi! thanks for the video. but i dont seem to get the last translation r5.

[@sspickle]

I'm sorry, what are you getting? A different result?

[@oyemakinwatayelolu3536]

yes and that is beacuse of the 1at the bottom right

[@sspickle]

I'm sorry, that doesn't help. What are you getting?

[@oyemakinwatayelolu3536]

alright....i am getting h(1+f(1-n)/R) in the first column and h((1-n)/R) in the second column after multiplication

[@patrickschanz9805]

very nice viedo! But does anyone know in which literature i can find something to the 'ket' notation in reference to this topic? When i look for the abcd matrices i always find the 'normal' vector notation and not using the 'kets' and operators. Thanks in advance!

[@sspickle]

It was invented by Dirac for use in quantum mechanics, but it is more general than that, so I use it in many "linear algebra" contexts. en.wikipedia.org/wiki/Bra-ket_notation

[@patrickschanz9805]

​@@sspickle Thanks! But do you know some literature that uses abcd matrices in the Bra-ket notation? Because i need to do this for my bachelor's thesis and i want to use this notation. So it would be better to refer to a book that treats this as operators and so on.

[@sspickle]

@@patrickschanz9805 "ABCD" is just a way to refer to any set of 2x2 matrix elements. I'm not aware that it's a formalism of any kind, sorry. What's your thesis about? It may be that you can skip the ABCD and just us Aij or something similar.

[@patrickschanz9805]

@@sspickle Okay, anyway thanks for your help. A part of my work is to develop a lens system for beam shaping for a laser with gaussian mode. And the theoretical calculations i want to do with these matrices.

[@Chathurinda711]

at 16:53 you got "t" this meas thickness of the lens right?

[@sspickle]

+Chathurinda711 Right. The translation step T_2 goes from the left surface of the lens to the right surface through the lens' thickness.

[@Chathurinda711]

Thank you !!!

[@Kuratius]

I think the minus at 12:36 is wrong. You didn't show us where alpha_r actually is, but I assume you meant that theta_l =theta_r+alpha_r. This yields the same formula you presented in the video, except that the sign of alpha_r is different. Is the sign change deliberate? Edit: You also forgot to use subscript at 13:48 for n_r and n_l in some of the equations. Though that's not really an important error, but it continues throughout the entire video.

[@sspickle]

The trouble is that alpha is measured as positive when it's oriented above the horizontal and negative when below. It's essentially the slope of the ray. An incident ray with alpha=0 has alpha

[@Kuratius]

I see. Thank you! Another question: Is it possible that the Refraction matrix at 17:19 is wrong? I tried to derive the matrix for the second refraction, and assuming that the variable names stay the same (so n_r is inside the lens, n_l is outside) I get (n_l-n_r)/(R*n_l) in the lower left corner of the matrix, and not (n_l-n_r)/(R*n_r). I also the get the reciprocal n_r/n_l in the lower right corner (when compared to the first refraction at 16:10 ). I'm assuming the right side of the lens is curved as well. When I multiply the two refraction matrices, I get the identity matrix again, which shouldn't happen. Edit: I found a mistake in my work and now I don't get the identity matrix. I still think your second refraction matrix is wrong, however.

[@sspickle]

n_l, n_r and R refer to the index on the left, right and the curvature of each interface individually. Curvature is measured + if the center is on the right, and - if the center is on the left. So for a double convex lens in air the values of n_l, n_r, and R for the first (leftmost) interface would be 1.0, 1.5 and R>0. The values of n_l, n_r, and R for the second (rightmost) interface would be 1.5, 1.0 and R

[@mahfa7513]

thanks for this video

[@macmos1]

Could you elaborate on the translation part of the second column? I did not understand. How can you not have height, but have an angle?

[@macmos1]

Could you also define what you mean by "translation"? Do you mean a horizontal translation of a ray?

[@macmos1]

+Marco Mosri or do you mean translation along a ray?

[@macmos1]

+Marco Mosri I don't understand how you can have an angle but no height and how that transformation gives you a height.

[@sspickle]

+Marco Mosri This happens any time the ray is propagating horizontally. The angle in question is more or less the slope of the ray (measured in radians) when the slope is small (slope

[@macmos1]

I think I understand. Will brush up on geometric optics and come back to this, thank you

[@Gruemoth]

shouldn't it be plus in the top term of the resulting 2X1 matrix? h(1+ t/R(....))

[@sspickle]

Yeah, I guess so. Unfortunately KZbin no longer allows annotations, and doesn't permit replacing a video, so mistakes live forever. ;-( Anyway, hopefully someone will see your feedback. Luckly the mistake doesn't change the answer.

[@Gruemoth]

at 17:14 you say nl/nr = n for refraction from glass to air and hence the n. Why is it 0 for (nl - nr) / (R nr)?

[@sspickle]

Admiral Hank Dawkins XIII The interface going from glass to air has an infinite radius of curvature so that’s where you get the zero.

[@Frohicky1]

Interestingly also works for Mass Spectrometers :)

[@sspickle]

+Splen borg Of course! ;-)

[@Richjunior1]

You need to explain at the beginning of your video what these abstract vectors correspond to in a ray diagram - e.g. the one you started out with! It's quite confusing to follow you when there is no link offered between the rays in the "2D-ray-diagram-space " and the ket vector you are now talking about!! The abstract objects that you are manipulating here with the operators remind me rather of a car at a specific point (hight) with a velocity (angle). The translation operator is then a time shift, which consequently changes the position (height obove the x-axes) of the car while leaving the velocity of the car unchanged! The velocity of the car is changed by the refraction operator anytime there is a transition between to media. Maybe you should use the word "element" instead of "ray" when adressing the object you are manipulating here. This would eliminate the confusion. Because without the transition operator the objects do not look like "rays" e.g. the lines you are drawing between to states of the objects are actually trajectories of the object and not property of the object itself! Hope I could explain myself accurately enough as english isn't my first language.

[@sunilkchatterjee9463]

Want to see from beginning

[@sspickle]

Is there something missing?

[@am11games7]

can i get the ppt link plz?

[@sspickle]

Wow. This is really old now! I may have it. I'll look around.

[@am11games7]

@@sspickle thank u, will be much appreciated

[@Shock123456789101]

i understand it till 17:13 mark. but why is in the ket vector or |r_3> the C ( if you use the abcd formalism ) ... a "zero" and not a (n-1)/(R)... /E ahhh i see, becasue its planar and you use R = infinite already here so its a zero ... smart men to already use it ^^

[@leechlittle2767]

I thought that this was a D&B track

[@joshdheda8776]

you sound like the voice which does the tutorials in DCS World