That is great and clear explaination. Here is an example of finding which apperture is the main STOP apperture. kzbin.info/www/bejne/d4qmaaurrq5_p5o&ab_channel=Opto-Mechademics

It would be interesting to know whether the lenses are cemented first then coated or vise versa?

event with one lens a stop can exist inside.This how Galilleo improved his telescope

Hey there, Not sure if anyone can help me, but I tried to adapt my Rayxar E50/0.75 to my Fuji X camera. The Rayxar has a flange to focus length of 0.8mm, FX is 17.7mm. Initial solution was to use a rear group of a Helios 44 to increase it to a M42 distance, but I can't really focus that well except for roughly 2 to 4m, of distance. I found some patents of similar lenses to try and work the optics less caveman and more scientific, but IT doesn't seem to be a regular double gauss construction. My question now is, how can I calculate the lens properties and aberration without specialized tools?

Thank you so so much

Answers using wayback machine: (scroll down) 1. An equi-convex fused Silica lens has a 90 radius of curvature. Compute the focal length. Solution: Use the thin lens equation to calculate focal length: Power = φ = 1/F ≈ (n-1) ((1/R1) -(1/R2)) = (n-1)(C1-C2) Fused silica has a visible index of 1.4585 [Source] Caveat: the back radius is negative. If one uses a positive radius of curvature then you are calculating the focal length for a meniscus lens. F ≈ [(n-1) * ((1/R1) -(1/R2))]-1 F ≈ [(1.4585-1) * ((1/90) -(1/{-90}))]-1 = [(1.4585-1) * ((1/90) +(1/90))]-1 = [(1.4585-1) * (2/90)]-1 = 90 / 2 / (1.4585-1) = 45 / 0.4585 F ≈ 98.15 Holy Frioles! that algebra is a mess! 2. A lens has a front convex radius of 200 mm and a rear concave radius of 1000 mm. What is the shape factor? Flip lens and re-compute the shape factor. Solution: in the first incarnation both radii of curvature are positive, since both center of curvatures are to the right. S = (RFront + RBack) / (RBack - RFront) = (200 + 1000) / (1000 - 200) S = 1.5, Positive Meniscus The second incarnation or a flipped lens, both radii of curvature are negative, since both center of curvatures are to the left. S = (RFront + RBack) / (RBack - RFront) = (-1000 + (-200)) / (-200 - (-1000)) S = -1.5, Positive Meniscus Conclusion: Flipping a lens changes the sign of the shape factor. Why do we care about this parameter? It will become more apparent when we begin to study spherical aberration. For a distant star imaged by this lens each lens orientation will have a different amount of spherical aberration and therefore different spot sizes. Aside: "... a distant star..." is referred to as an "infinite conjugate", but I hesitated using overly verbose optical engineering nomenclature:)

Thanks for the presentation. Very interesting.

hello sir, may i know your email, i want to ask some problem in my desing in ansys zemax

Thanks for presentation of this challenging subject.

Thank you sooooo much! It helps me a lot!

Hands off sir, great explanation. Can you please do a video on the design complete from scratch like design of a microscope or telescope?

Thanks for the videos. Wondering if this stuff still relevant or is there some new videos anyone could suggest.

I love how I understand this all

Extremely helpful videos, I appreciate the work!

Grüße, an alle die wegen Farnleitner Referat machen müssen.

Your homework website no longer exists.

amazing job thanks

Amazing

3d artist here. trying to simulate camera lenses in renders. started going down this rabbithole a week ago and now im here. i want this program.

Very interesting to a photographic hobbyist. Never heard of Zemax, so learned several things watching this video. Good pace, too, with right amount of talk. Will look at a few more on the channel. Matt Lee.

Scott. If you shine a HeNe laser into a pool, illuminating the side of the pool, you see a red spot. What if you dunk your head under the water and looked at the same spot? What color would it be? If color was a function of wavelength, then you would see a blue spot instead of red. But if you do this experiment, you still see a red spot, right? So color is not wavelength dependent. It’s frequency dependent. Both wavelength and frequency get smaller by the same amount when light enters a denser medium. You know this, but I had to figure it out. So many people associate color with wavelength, but that’s wrong. Color is a function of frequency. P

thank you

thank you so much

HI, Scott your website is down. Can you share your tutorials...

if I put 3 laser diode in back of a lens of 20mm dia and 300mm focal if the laser diode are at center of the lens do i add the power of the laser diode at focal point

Thanks , interesting optical engineering .

thanks man i always appreacte anyone who take the time to share his lectures/reviews. i needed this review for my exam tmr! thanks again

Nice explanations.. minor error in the slide... visible light is 100s of THz.. Not GHz

Hello

Can anyone suggest a workbook I can work through to learn basic optics ?

🤏

It is a joy to someone with such complete mastery of such a complex topic. I find myself now needing to squeeze the chromatic aberration out of a system and this greatly helped my understanding of how to do it.

Good video ❤️.

Just Perfect!

this helped connect so many dots for me in my head! I've learned these things but havent had it explained in such a linear and clear manner ! Thank you very much.

Thanks for this great resource, for taking you time and share your knowledge.

Feet? Inches? Dude

I'm still unsure how to do #1, can you help? I thought it would be 1/F= (1.5-1)[90-(-90)]. Thanks.

Negative edge thickness is very hard to make.

Dear Scott Website was not opening , Kindly guide

Please design a lens system for aberration minimization and maximum light collection from a point source. Thanks

HI, Scott your website is down. Can you share your tutorials...

As someone doing my own personal research on vision and eyeglasses, this is very useful

These tutorials are amazing! A gentleman and a scholar!

Very clearly explained!

Awesome... but how can we have access to solutions to homeworks?

No one can explain better than this one 👀

First the Hubble, now the James Webb!

I'm just doing an overview again! Now, I'll go back and concentrate on them all!

I'm just doing an overview!