Though I don't have any clear idea of what all this is, Keep them coming! I'm in the process of transferring to be an undergrad in the optics dept. at the UofA and these videos rock!
You may be asking about the difference between focal length and back focal distance. Focal length is a factor to describe how object space maps to image space. Back focal distance is the distance from the last refractive surface to the image. These two parameters can be the same such as a single thin, long focal length lens. More often the are NOT the same. This is the case when the back principle plane is inside the optical system
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
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?
actually i couldn t find any video showing a design like lot s of optics placed on the table and we are trying to see how light propagates along the system. most of video tutorials show how to build an on-axis optical systems.
I think you should mention that all of those rays of light (entering the lens and subsequently focussed by the lens on one point on the sensor) are from ONE point on the subject--in which case they would not be entering the lens as horizontal lines.
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:)
Sorry for the late response...I think you're requesting how to evaluate / compute the system focal length from a lens assembly composed of several lenses of differing focal lengths. I have not addressed how to compute this via equations. Rather I have been lazy and assumed you would put this into ray trace code and use the tools there to find the system focal length. Perhaps I should not be lazy and fully address this in an upcoming video. OR....
@healy1313 - thanks for the kind comments. I have very fond memories of classes at the UofA OpSci... you'll be in great hands! Please email me directly at scott.sparrold@gmail.com - I would love to hear more comments on what else my video series needs. My ultimate goal is get people excited about optics and want to pursue a career in it. Thanks
At what point is the focal length of a lens that is made of several elements. Like a camera lens is say 50mm from what point from the image plane is 50mm. I hope I am not butchering this question to much.
The principle plane is a theoretical surface at which refraction occurs. For a thin lens it is a single plane in the middle. For a thick lens or a series of several lenses there is a front and a back principle plane. I have rambled too long and just need to put another video together:)
@turb0talon Radius is a linear dimension and is intended to be denoted as "R". Sometimes it may be referred to as "Radius of curvature", but I usually try to say simply "radius. Radius is convenient when calculating focal length. The inverse of radius is "curvature" in units of inverse length and denoted as "C". Curvature is convenient to use when calculating sag or power (inverse of focal length) As far as units, I typically work in metric, so "R" is in mm while "C" is 1/mm
So, in your average camera lens, how exactly does focusing work? When I turn my focus ring, what's happening to change the focus? Is a lens moving backward and forward?
So, what kind of optics do I need to create a smaller laser ray(1/4 the original) going out to infinity(no focal point needed) from a Co2 laser tube? Thanks
You may be surprised to learn this isn't possible. Except for Bessel beams, which are only theoretical, all LASER beams diverge. No matter how well you collimate the beam, it will still spread. So a further refinement to your question is required. At what range of distances do you need your beam to be the size you want? If you look up _Rayleigh length_ you will get an idea of what is required.
You are correct! The shape factor as defined by Coddington (Rfront+Rback)/(Rfront-Rback) will give an undefined result or 1/zero when Rfront=Rback. The shape factor is intended to capture the variation of a lens while maintaining a similar focal length. When the front and back radius are equal the lens has little optical power or a very long focal length (it is essentially a window)
Hello, great videos! But I am a bit confused in regards to the coddington shape factor formula. If the radios on both sides is the same, you would end up with a division by zero, and that does not give much sense, unless I am missing something essential here.
Hi, this is Jiarui from OPTIGER, which is a Chinese company focusing on the development and manufacture of optoelectronic products in the field of machine vision. It's headquartered in Tianjin (a northern city in China), and has a branch in Wuhan. Currently, OPTIGER is searching for a professional lens design team which can design zoom lenses for our company. If you are interested in this cooperation, I will send you the detailed technical requirements and parameters forms of zoom lens models, which are for your confirmation of the design cycle and quotation.
Reza Ghaemi TITLE: Introduction to Lens Design: With Practical Zemax Examples AUTHOR: Joseph M. Geary PUBLISHER: Willmann-Bell DATE: August 2002 ISBN: 0943396751 This one is really good.
For homework question 1, I am confused by the information given. First, is the radius of curvature considered "C" or "R". You mentioned one as just "radius" and the other as just "curvature" during the lecture, but use both as the description in the homework. Also, what are the units being used? Thank you.
you probably dont give a shit but does anyone know of a tool to log back into an Instagram account? I was dumb forgot my password. I love any help you can offer me!
Though I don't have any clear idea of what all this is, Keep them coming!
I'm in the process of transferring to be an undergrad in the optics dept. at the UofA and these videos rock!
As someone doing my own personal research on vision and eyeglasses, this is very useful
This guy knows how to explain! And he gets to the bottom of it. Thank you very much.
You may be asking about the difference between focal length and back focal distance. Focal length is a factor to describe how object space maps to image space. Back focal distance is the distance from the last refractive surface to the image. These two parameters can be the same such as a single thin, long focal length lens. More often the are NOT the same. This is the case when the back principle plane is inside the optical system
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
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?
These videos are amazing! Very well explained and to follow, I hope you can keep on posting them! Thank you so much Scott!
actually i couldn t find any video showing a design like lot s of optics placed on the table and we are trying to see how light propagates along the system. most of video tutorials show how to build an on-axis optical systems.
I'm still unsure how to do #1, can you help? I thought it would be 1/F= (1.5-1)[90-(-90)]. Thanks.
Dear Scott Website was not opening , Kindly guide
Thank you as well. That's a good point for our amateur interest. I have subscribed and following Scott. Thank you
Thank you, Scott. Very cool!
I think you should mention that all of those rays of light (entering the lens and subsequently focussed by the lens on one point on the sensor) are from ONE point on the subject--in which case they would not be entering the lens as horizontal lines.
Well explained, very clear...thanks for spreading optics! :)
your explanations are top notch. loved them
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:)
First the Hubble, now the James Webb!
Sorry for the late response...I think you're requesting how to evaluate / compute the system focal length from a lens assembly composed of several lenses of differing focal lengths. I have not addressed how to compute this via equations. Rather I have been lazy and assumed you would put this into ray trace code and use the tools there to find the system focal length. Perhaps I should not be lazy and fully address this in an upcoming video. OR....
@healy1313 - thanks for the kind comments. I have very fond memories of classes at the UofA OpSci... you'll be in great hands! Please email me directly at scott.sparrold@gmail.com - I would love to hear more comments on what else my video series needs. My ultimate goal is get people excited about optics and want to pursue a career in it. Thanks
At what point is the focal length of a lens that is made of several elements. Like a camera lens is say 50mm from what point from the image plane is 50mm. I hope I am not butchering this question to much.
How can I make LED light go straight? What type of lens do I need? Can I use magnifier lenses?
Extremely helpful videos, I appreciate the work!
Quick question, how would light behave in a curved prism? That is, a prism that has an arch.
It would reflect depending not he index of refraction and the angle of the surfaces. since the inside is uniform it doesn't matter if it's bent.
The principle plane is a theoretical surface at which refraction occurs. For a thin lens it is a single plane in the middle. For a thick lens or a series of several lenses there is a front and a back principle plane. I have rambled too long and just need to put another video together:)
@turb0talon Radius is a linear dimension and is intended to be denoted as "R". Sometimes it may be referred to as "Radius of curvature", but I usually try to say simply "radius. Radius is convenient when calculating focal length.
The inverse of radius is "curvature" in units of inverse length and denoted as "C". Curvature is convenient to use when calculating sag or power (inverse of focal length)
As far as units, I typically work in metric, so "R" is in mm while "C" is 1/mm
Good video ❤️.
So, in your average camera lens, how exactly does focusing work? When I turn my focus ring, what's happening to change the focus? Is a lens moving backward and forward?
it moves both ways depending at where you start and where the object of the focus is.
Thanks for your work
So, what kind of optics do I need to create a smaller laser ray(1/4 the original) going out to infinity(no focal point needed) from a Co2 laser tube? Thanks
You may be surprised to learn this isn't possible. Except for Bessel beams, which are only theoretical, all LASER beams diverge. No matter how well you collimate the beam, it will still spread.
So a further refinement to your question is required. At what range of distances do you need your beam to be the size you want? If you look up _Rayleigh length_ you will get an idea of what is required.
You are correct! The shape factor as defined by Coddington (Rfront+Rback)/(Rfront-Rback) will give an undefined result or 1/zero when Rfront=Rback. The shape factor is intended to capture the variation of a lens while maintaining a similar focal length. When the front and back radius are equal the lens has little optical power or a very long focal length (it is essentially a window)
Awesome! Very nice explanation. Thanks man.
The coddington shape factor ,S, doesn't match to the shapes!
What is the Coddington shape factor if there are more than 1 curvatures on one and the same side?
Very good introduction. Thanks
Aws0me...there should be more people like you making things easy for us :)
Great video.
Hello, great videos!
But I am a bit confused in regards to the coddington shape factor formula. If the radios on both sides is the same, you would end up with a division by zero, and that does not give much sense, unless I am missing something essential here.
you're missing signs conventions :)
I learned a lot! Now, on to part 3!
Awesome video!!
Thank you, I work in a lab and one thing I haven't gotten to work on was collimator and I'm getting tested on it.. so this helps
Hi, this is Jiarui from OPTIGER, which is a Chinese company focusing on the development and manufacture of optoelectronic products in the field of machine vision. It's headquartered in Tianjin (a northern city in China), and has a branch in Wuhan. Currently, OPTIGER is searching for a professional lens design team which can design zoom lenses for our company. If you are interested in this cooperation, I will send you the detailed technical requirements and parameters forms of zoom lens models, which are for your confirmation of the design cycle and quotation.
good job ,,, carry on
Thanks a lot. Is there any textbook which could help to design compound lenses?
Reza Ghaemi TITLE: Introduction to Lens Design: With Practical Zemax Examples
AUTHOR: Joseph M. Geary
PUBLISHER: Willmann-Bell
DATE: August 2002
ISBN: 0943396751 This one is really good.
good book. Thanks
@opticsrealm Thank you!
For homework question 1, I am confused by the information given. First, is the radius of curvature considered "C" or "R". You mentioned one as just "radius" and the other as just "curvature" during the lecture, but use both as the description in the homework. Also, what are the units being used? Thank you.
Thanks so much Scott Best Tutorials so FAR
you probably dont give a shit but does anyone know of a tool to log back into an Instagram account?
I was dumb forgot my password. I love any help you can offer me!
@Harry Antonio Instablaster :)
Nerd