Love this lecture series - concise, informative and highly conceptual. Thanks!

I am so happy each time I google a precise subject and find one of your videos talking exactly about what I am looking for !

Fabuloso , no dejes de hacer estos videos, es un gran aporte para la cultura y los Físicos del mundo.

Excellent video for my level, thanks a lot

a little nervous when watching this lecture, for there are so many concepts. However, i suggest we shold watch this video closely ,and better more times , it is really informative and enlightening

at 25:07 , why do you suppose that the upper bundle(green) creates a shallower divergence and the marginal bundle (yellow) a bigger divergence resulting in smaller and bigger blur respectively? shouldn't the divergence of each of these bundles be the same?

@SanderKonijnenberg

8 ай бұрын

Could you explain why the divergence of the bundles would have to be the same? It seems to me that once you define your exit pupil plane and your image plane, the divergences of the bundles through each image point follow automatically. If the distance between the two planes is finite, I'd think the bundles would necessarily have different divergences.

@sammyapsel1443

8 ай бұрын

@@SanderKonijnenberg just looking it from a geometrical perspective, if i have an aperture stop and i draw a cone that goes through this stop, i can simply choose the angle of this cone exiting the aperture stop meaning that it would result in the same "divergence". Otherwise, I'd be interested to learn why you suppose that the divergence becomes smaller as you direct the cone more towards the axis.

@SanderKonijnenberg

7 ай бұрын

@@sammyapsel1443 Given the two constraints 1) The rays in a cone must fill the exit pupil, 2) The rays in a cone must intersect at a single image point, I don't see a way to draw the rays any differently than I have. And from the traced rays, the divergences follow.

at 24:12, you defined m.r. for on axis points, however i saw that in this lecture (MIT course in Optics) kzread.info/dash/bejne/kZ5_qsejm5ebg8o.html , at 26:15,the professor defines it for an off-axis point, is there a significant difference?

@SanderKonijnenberg

7 ай бұрын

At 27:08 the professor says 'I can define them for any point I like'. I think it makes sense: according to his definition, any ray that hits the margin of the aperture stop is a marginal ray. However, many other sources (including Wikipedia en.wikipedia.org/wiki/Ray_(optics) ) require that the marginal ray goes through an on-axis object point.