In the case the last image to appear in front of the lens, the light has to go through the lens once more, therefore solution is not complete for this setup, you need to form a third image through the lens!
It would be true if the numbers were slightly different, such as if the distance between the lens and mirror were greater or if the image was reflected off axis like with a Newtonian telescope.
I tried a third calculation of this and the 3rd image ends up virtual as well, in front of the lens and even larger than I2... do you guys think it will ever end?
if 3rd image form through lens, then 3rd image forms in between lens and mirrow from the distance of 3,69cm behind lens..then 3rd image will be object of mirrow and then, this will be repitiation again and again. Therefore we can not think about of 3rd image.
It actually does. That is why the second image appears where it does. If it didn't go through the lens a second time, the second image would appear farther to the left.
I finished this series too. But, l didn't get this example. My intuitions ask why we didn't calculate for the 3rd image. Rays should go through the lens again. I mean we found the first image because rays interact with the lens, then we found the second image oriented by the concave mirror because the mirror intercepts the rays and rays were reflected back. The second was somehow a virtual image according to the mirror. However, now the image must occur left side of the lens which means rays should have interact with the lens again. I desire to replace the lens slightly left, so the problem can be solved once and for all. :D Anyway thank you professor. I get the main point what you want us to understand.
a convex lens of focal length 36 centimetre is placed in front of a convex mirror 12 centimetre when a pen is placed at a distance of 36 centimetre from the convex lens it is found that it coincides with it with its own inverted image formed by the lens and the mirror find the focal length of the mirror
Hi there, I hope you see this as I have a question. How come you had given f2 as one of our foci's but never ended up using it on the diagram? Are the foci distances relative to the eye? Thank you!
That method works well to solve the problem algebraically, but it is not what is actually going on. It is difficult to draw the actual ray diagram. After the rays pass through the first lens, they are then reflected off the mirror's surface without forming an image behind the mirror.
@@MichelvanBiezen Also, is the reason the rays do not pass through the lens a second time, because we're sending two incident rays at different angles and being observed by the human eye? Or is it just how it is?
The lens and mirror cause the rays to bend in such a way that the rays appear to be emanating from the virtual image. Note that a virtual image isn't real and the rays just appear to be coming from the (non-existing) image.
Hello sir, I have a doubt. Why won't the rays reflected by mirror pass through lens for the second time? I'm pretty sure that it does and we need to apply the lens formula for the second time.
Nisarg, Rest assured that the presentation on the video is correct. The image of the first lens becomes the object of the mirror. Note that image 1 = object 2 is behind the mirror. The mirror will then affect the light rays in such a way to move the final image to its indicated location.
+Michel van Biezen I think he means that the light travels in this order 1. Through the lens 2. Bounces off mirror 3. Back off the mirror into the lens again
Great lecture as always, but I have been taught by the instructor at my institution that in cases similar to this (meaning that Image 2 is formed in front of the lens and the eye is seeing from the front), the image 2 should be used as 2nd source to be used to find the 3rd ("final"- a conflicted term here) image. Would you kindly make a clarification?
Where the first image forms depends on the position of the object, and the focal length lens 1. Since there is a mirror, the observer will ultimately need to sit on the other side in order to see anything.
I'm with destroyer and our teacher, so, shouldn't the video show the effects of the lens on the image that's reflected through it? Isn't this example the principle of a Casergrain or some such astronomy telescope? Maybe the casergrain lens has a center hole through which the rays pass and the lens doesn't refract the rays? Thanks for this series of videos. I'm wanting to use the information on convergent and divergent lens combinations to build some lenses from surplus parts. The formulas and explanations are very well done and easy to follow. When using multiple negative or positive lens combinations is the same process just repeated for each lens until they're all considered? Thanks again
Many combination lenses contain multiple lenses and yes, the same technique can be used for multiple lens combinations, one lens at a time until all lenses have been considered. Keep in mind that this is a simple technique ignoring what happens within each lens. There is a more advanced technique where a matrix is developed where each transition from air to lens, lens to air, lens to lens, and within each lens is considered. For most cases the simple technique explained here in the video is adequate.
Thank you for the reply. The simple method will suffice, expecially for using surplus lenses. If I find the need someday I'll research the matrix method. Your videos are helpful and appreciated.
What sign convention are you using? All things in the passage of incident rays is negative whereas those to the other side are positive,you have some other convention
Yes, this can be done with many lens systems. However that said, there is another method to determine the propagation of light through multi-lens systems. I will be preparing some videos on that this summer.
Hello, so wont the final image have to go through the lens once more after it is reflected from the mirror? Why are we only considering the lens when the light rays are moving towards the right but with the reflected rays(coming FROM the right) the lens is ignored?
The image that is seen by the observer on the left is an apparent image that appears to originate from behind the mirror. Thus the image you appear to see are the rays that are reflected by the mirror (as if the lens isn't there). It seem very strange, but that is how that works.
So you ended up with a positive 13.1 cm. Am I mistaken in thinking that it should be 13.1 along the negative x-axis(to the right). Just like the previous video? Did something change between that video and this one besides the mirror. I thought that a positive image distance means to the right not the left?
***** That is correct for lenses, but not for mirrors. A positive image distance for a mirror means the image is in FRONT of the mirror. In this case that means to the left of the mirror.
Michel van Biezen Thank you Michel I actually watched more of your videos and I made that connection. I have watched a lot of your videos for E&M they were very well done, amazing work!
can somebody explain to me how to get image 2 through ray diagram. ive tried mirror then f2front and f2back then mirror and for some reason im not getting the image on the right spot
Tracing rays through a double lens system or a lens-mirror system is very difficult. (we use the thick lens technique and matrices to work out the details - see the thick lens videos). It is easier to calculate the location and size of the first and second image and then surmise the rays from that.
Can you do an example where you use an object 100 to the left of a convex lens that is 100cm to the left of a convex mirror? f[lens]=80,f[mirror]=-50. I am following every step but my final image solves on the wrong side according to my text book.
The first image (from the lens) will be 400 cm to the right of the lens, which places it 300 cm to the left of the mirror. Which means that the object distance to the mirror = - 300 cm. Using that you'll get an image distance for the mirror of -60 cm which places it to the right (behind) the mirror. (virtual image)
According to my calculations, the final image would be 160 cm to the right of the lens. (Either the book's answer is wrong, or we are making an error, I'll have to think about it some more).
You can work the problem backwards with the final image 160 cm to the left of the lens (260 cm to the left of the mirror) which makes it a final real image. That places the original object at 183 cm to the left of the lens.
I personally think it just Won't stop there. The 2nd image will become s3 and thus form another image and so on. In this way, we will get infinitely many images. Please sir answer if I'm wrong
could you do a video on lens combinations in which you draw out the rays from the mirror/second lens to show that the image from the ray diagram corresponds to the image we expect from our calculations?
+Morgan Lyttle Those are difficult to do with the current technique. I am planning on making some more advanced optics videos where we consider thick lenses and that is the proper technique to draw all the rays.
I meant the image of the mirror, s'2 after it passes through the mirror, is on the right of the lens, what then? does it stays the same? or we reffer to it as a new object of the lens?
Ah, yes, you did ask about the image of the mirror. It depends how big the lens is and how you look at it. If you look through the lens to observe the image (instead of looking directly at the image from a slight angle), then yes, you'll have to work it a third time.
Michel van Biezen So just to be sure, if I look through lens I always need to do it for the third time or only if s'2 is between the lens and the mirror? How can I tell? Thank you.
Hi Sir, I was just wondering for this problem why the object distance given, 50cm (S1), is positive. Wouldn't it be positive if the eyeball was on the right, if the person has to look through the lens first? I could be wrong, but it seems like the eyeball should be on right for this problem instead of on the left.
We use the same equation for the lens and for the mirror. There are some differences in the rules to draw the rays and how the focal length is defined in the equation. (Also the sign for s and s' is different between the mirror and lens)
Since the rays converge and form an image in front of the mirror (on the left side), the image can be seen by an observer in front of the mirror and is therefore real
Violinsheetmusicsblog, At some point I will add a series of videos with a more advanced technique for multiple lenses. After you watch several of the videos with 2 lenses you should be able to use the same technique for 2 diverging lenses. Just use the first image as the object for the second lens.
Okay, got it. I'm sure I can do it since it's basically the same procedure each time. It would be nice to see an alternative method to doing these problems. Thanks!
Michel van Biezen thanks for your reply. I actually meant if you changed the lens system to get that particular scenario. But anyway, maybe a better question: in general, when do you define object distance as -ve and when do you define it as +ve?
1. Refraction from lens 2. Reflection from mirror 3. Refraction from lens is missing Because distance between mirror and lens is 10cm and image formed after reflection at approx 13cm....
Sir, is Image 2 real or virtual relative to the original object (O1)? I`m asking because I see that image 2 appears behind the lens, i.e between the eye and the lens.
Trymore, Yes the final image is real. The best way to determine that is so see how the final image is formed. In this case the rays are reflected off the mirror and CONVERGE down to a single point where the image is formed. If the rays DIVERGE, then the image will be virtual.
hey can anyone help on this question. A person with near sighted eye has near and far points of 16cm and 25cm respectively. (a) assuming a lens is placed 2cm from the eye, what power must the lens have to correct this condition? b) Suppose that contact lenses placed directly on the cornea are used to correct the persons eye. What is the power of the lens required in this case, and what is the new near point? thank you anyone who has time to help!
Hannah, For part a, you want the image from an object placed at infinity to appear 25 cm away from the eye (the far point) which will be 23 cm in front of the lens. That makes the image virtual and therefore the image distance will be negative. (1/f) = (1/s) + (1/s') (1/f) = (1/infinity) + (1/-23cm) f = -23 cm and its power is P = 1/-0.23m = -4.35 diopters. For part b, the lens is directly on the eye, therefore the image distance will be 25 cm from the lens. (1/f) = (1/infinity) + (1/-25cm) f = -25 cm and its power is P = 1/-0.25m = -4.00 diopters The new near point is found by using the equation (1/s) +(1/-16cm) = (1/-25cm) s = 44.4 cm which is now the new near point.
+Peter C You know the image is real, when you place the object on the left side of the lens (or combination of lenses) and the image appear on the right side (or other side). If the image appears on the same side, as the original object, it is a virtual image. If the magnification is positive, then the image is upright.
Peter C if an image is formed at the same side of object in case of lens then that image is virtual and vice versa ! If an image is formed in the same side side of object in the case of curved mirror , then the image is real and vice versa !! I hope you got that
this is being taught the same way I was taught in physics 45 years ago. unfortunately it is no less 'unclear'. what is missing is teaching how to DESIGN a lens system for a particular PURPOSE. otherwise all you can do is try this and try that like a blind squirrel looking for a nut. these type lectures should be preceded by examples without all the algebra - showing different systems of lens and how they behave. then proceed to the same examples with the algebraic methods. disappointing.
I feel the same way. It looks to me that it would be better to buy a magnifying glass at 100x on Amazon for $10.00 rather spend hundreds for a possible .33 magnification. Does not make sense at all. It CV oils be that it's missing something or lacking something. Anyway, thank you for sharing, but it left me empty in the end.
In the case the last image to appear in front of the lens, the light has to go through the lens once more, therefore solution is not complete for this setup, you need to form a third image through the lens!
It would be true if the numbers were slightly different, such as if the distance between the lens and mirror were greater or if the image was reflected off axis like with a Newtonian telescope.
I tried a third calculation of this and the 3rd image ends up virtual as well, in front of the lens and even larger than I2... do you guys think it will ever end?
If the second image formed is a virtual image, will the system form 3rd image?
U are right
if 3rd image form through lens, then 3rd image forms in between lens and mirrow from the distance of 3,69cm behind lens..then 3rd image will be object of mirrow and then, this will be repitiation again and again. Therefore we can not think about of 3rd image.
Thanks professor! You're helping me to pass the exam at the Univeristy of Padua (Italy) for the Astronomy course!
Happy to help!
It actually does. That is why the second image appears where it does. If it didn't go through the lens a second time, the second image would appear farther to the left.
SO thankful people like you exist, your videos rock!
The marker wants to rest in peace, ahhaha. I loved this series.
Thank you. Glad you enjoyed it.
I finished this series too. But, l didn't get this example. My intuitions ask why we didn't calculate for the 3rd image. Rays should go through the lens again. I mean we found the first image because rays interact with the lens, then we found the second image oriented by the concave mirror because the mirror intercepts the rays and rays were reflected back. The second was somehow a virtual image according to the mirror. However, now the image must occur left side of the lens which means rays should have interact with the lens again. I desire to replace the lens slightly left, so the problem can be solved once and for all. :D Anyway thank you professor. I get the main point what you want us to understand.
DId you solve your question? Do you calculate a third image?
shouldn't the image distance be -ve according to the sign convention.
a convex lens of focal length 36 centimetre is placed in front of a convex mirror 12 centimetre when a pen is placed at a distance of 36 centimetre from the convex lens it is found that it coincides with it with its own inverted image formed by the lens and the mirror find the focal length of the mirror
Thank you for posting this series! I love the extra practice with many different examples!
Hi there, I hope you see this as I have a question. How come you had given f2 as one of our foci's but never ended up using it on the diagram? Are the foci distances relative to the eye? Thank you!
That method works well to solve the problem algebraically, but it is not what is actually going on. It is difficult to draw the actual ray diagram. After the rays pass through the first lens, they are then reflected off the mirror's surface without forming an image behind the mirror.
@@MichelvanBiezen Also, is the reason the rays do not pass through the lens a second time, because we're sending two incident rays at different angles and being observed by the human eye? Or is it just how it is?
The rays simply pass through the first lens and are reflected by the mirror in the second position.
Since the eye is on the same side than the 1st object, shouldnt s1 equal -50 instead of 50?
No, because when an image appears on the other side compared to the object, s1 is positive.
Why the focal length of concave mirror is considered positive . By sign convention it should be negative
how focal length of concave mirror can be positive?
Concave lenses have a positive focal length.
dont we have to recalculate the new image distance after the mirror since it goes again through the 1st lens>?
The lens and mirror cause the rays to bend in such a way that the rays appear to be emanating from the virtual image. Note that a virtual image isn't real and the rays just appear to be coming from the (non-existing) image.
can you draw the rays bounces off from the mirror
Hello sir, I have a doubt. Why won't the rays reflected by mirror pass through lens for the second time? I'm pretty sure that it does and we need to apply the lens formula for the second time.
Nisarg,
Rest assured that the presentation on the video is correct. The image of the first lens becomes the object of the mirror. Note that image 1 = object 2 is behind the mirror. The mirror will then affect the light rays in such a way to move the final image to its indicated location.
+Michel van Biezen I think he means that the light travels in this order 1. Through the lens 2. Bounces off mirror 3. Back off the mirror into the lens again
Nisarg Tike do an experiment and try to measure it :p
Great lecture as always, but I have been taught by the instructor at my institution that in cases similar to this (meaning that Image 2 is formed in front of the lens and the eye is seeing from the front), the image 2 should be used as 2nd source to be used to find the 3rd ("final"- a conflicted term here) image. Would you kindly make a clarification?
Where the first image forms depends on the position of the object, and the focal length lens 1. Since there is a mirror, the observer will ultimately need to sit on the other side in order to see anything.
I'm with destroyer and our teacher, so, shouldn't the video show the effects of the lens on the image that's reflected through it?
Isn't this example the principle of a Casergrain or some such astronomy telescope? Maybe the casergrain lens has a center hole through which the rays pass and the lens doesn't refract the rays?
Thanks for this series of videos. I'm wanting to use the information on convergent and divergent lens combinations to build some lenses from surplus parts. The formulas and explanations are very well done and easy to follow.
When using multiple negative or positive lens combinations is the same process just repeated for each lens until they're all considered?
Thanks again
Many combination lenses contain multiple lenses and yes, the same technique can be used for multiple lens combinations, one lens at a time until all lenses have been considered. Keep in mind that this is a simple technique ignoring what happens within each lens. There is a more advanced technique where a matrix is developed where each transition from air to lens, lens to air, lens to lens, and within each lens is considered. For most cases the simple technique explained here in the video is adequate.
Thank you for the reply. The simple method will suffice, expecially for using surplus lenses.
If I find the need someday I'll research the matrix method.
Your videos are helpful and appreciated.
Sir how did you derived the formula s'1=s1×f1/s1-f1 in 1:56
You start with the basic equation: 1/f = 1/s + 1/s' and you solve it for s'
Michel van Biezen Ok sir
Michel van Biezen thank you sir
Thank you so much sir. Your video are helping me so much for my final exam preparation
What sign convention are you using? All things in the passage of incident rays is negative whereas those to the other side are positive,you have some other convention
This is the standard convention of rays. With double lenses, you need to work through the process as shown in the video.
@@MichelvanBiezen ok thank you
hello, i wonder if the final image should be again an object with respect to the lens ,so that a third image will be formed ?
Yes, this can be done with many lens systems. However that said, there is another method to determine the propagation of light through multi-lens systems. I will be preparing some videos on that this summer.
Thank you very much for your videos! Helped me understand so much!
Wow thanks a lot 👏 and I had understood the topic
Glad to hear that
Hello, so wont the final image have to go through the lens once more after it is reflected from the mirror? Why are we only considering the lens when the light rays are moving towards the right but with the reflected rays(coming FROM the right) the lens is ignored?
The image that is seen by the observer on the left is an apparent image that appears to originate from behind the mirror. Thus the image you appear to see are the rays that are reflected by the mirror (as if the lens isn't there). It seem very strange, but that is how that works.
Note that you are not seeing a "REAL" image but you are seeing a "VIRTUAL" image that is not real.
Ah ok that is interesting. I appreciate your response thanks
Sir how did you know that distance between mirrors and lens is 10 cm in 3:31
That was a "given" quantity in the problem.
Michel van Biezen ok sir
So you ended up with a positive 13.1 cm. Am I mistaken in thinking that it should be 13.1 along the negative x-axis(to the right). Just like the previous video? Did something change between that video and this one besides the mirror. I thought that a positive image distance means to the right not the left?
*****
That is correct for lenses, but not for mirrors.
A positive image distance for a mirror means the image is in FRONT of the mirror.
In this case that means to the left of the mirror.
Michel van Biezen Thank you Michel I actually watched more of your videos and I made that connection. I have watched a lot of your videos for E&M they were very well done, amazing work!
can somebody explain to me how to get image 2 through ray diagram. ive tried mirror then f2front and f2back then mirror and for some reason im not getting the image on the right spot
Tracing rays through a double lens system or a lens-mirror system is very difficult. (we use the thick lens technique and matrices to work out the details - see the thick lens videos). It is easier to calculate the location and size of the first and second image and then surmise the rays from that.
Thanks Michel! Your videos really helped me throughout my physics series!
I was wondering does this work the same with a convex mirror instead?
The method is the "same" but you will get a different result.
Can you do an example where you use an object 100 to the left of a convex lens that is 100cm to the left of a convex mirror?
f[lens]=80,f[mirror]=-50. I am following every step but my final image solves on the wrong side according to my text book.
The first image (from the lens) will be 400 cm to the right of the lens, which places it 300 cm to the left of the mirror. Which means that the object distance to the mirror = - 300 cm. Using that you'll get an image distance for the mirror of -60 cm which places it to the right (behind) the mirror. (virtual image)
Michel van Biezen Yes, this is my result. But book says the final Image will be 160cm to the left of the lens.
According to my calculations, the final image would be 160 cm to the right of the lens. (Either the book's answer is wrong, or we are making an error, I'll have to think about it some more).
You can work the problem backwards with the final image 160 cm to the left of the lens (260 cm to the left of the mirror) which makes it a final real image. That places the original object at 183 cm to the left of the lens.
I personally think it just Won't stop there.
The 2nd image will become s3 and thus form another image and so on.
In this way, we will get infinitely many images.
Please sir answer if I'm wrong
could you do a video on lens combinations in which you draw out the rays from the mirror/second lens to show that the image from the ray diagram corresponds to the image we expect from our calculations?
+Morgan Lyttle
Those are difficult to do with the current technique. I am planning on making some more advanced optics videos where we consider thick lenses and that is the proper technique to draw all the rays.
what if the image of the mirror is between the lens andthe mirror? will we need to pass through the lens again to know where we'll see it?
No, that then becomes the object for the mirror in front of the mirror.
I meant the image of the mirror, s'2 after it passes through the mirror, is on the right of the lens, what then? does it stays the same? or we reffer to it as a new object of the lens?
Ah, yes, you did ask about the image of the mirror. It depends how big the lens is and how you look at it. If you look through the lens to observe the image (instead of looking directly at the image from a slight angle), then yes, you'll have to work it a third time.
Michel van Biezen thank you! You were very helpful!
Michel van Biezen
So just to be sure, if I look through lens I always need to do it for the third time or only if s'2 is between the lens and the mirror? How can I tell?
Thank you.
thank you , it means a lot . keep up the good work !
Thank you Anna. I appreciate the feedback.
Hi Sir, I was just wondering for this problem why the object distance given, 50cm (S1), is positive. Wouldn't it be positive if the eyeball was on the right, if the person has to look through the lens first? I could be wrong, but it seems like the eyeball should be on right for this problem instead of on the left.
+Deonna Flanagan
The eye ball cannot be on the right, because that would place it behind the mirror and you wouldn't see anything.
@@MichelvanBiezen yet since the eye is on the same side as the object shouldn't it be -50?
how you can use mirror formula for lens.
We use the same equation for the lens and for the mirror. There are some differences in the rules to draw the rays and how the focal length is defined in the equation. (Also the sign for s and s' is different between the mirror and lens)
sir please solve dis question
Can someone please explain how the s2 or q2 (13.11) is a real image? I thought real image would be if p>f?
Why is it not a virtual image?
Since the rays converge and form an image in front of the mirror (on the left side), the image can be seen by an observer in front of the mirror and is therefore real
I see! Thanks
Can you have 2 diverging lenses in a setup like these video series?
Violinsheetmusicsblog,
At some point I will add a series of videos with a more advanced technique for multiple lenses.
After you watch several of the videos with 2 lenses you should be able to use the same technique for 2 diverging lenses. Just use the first image as the object for the second lens.
Okay, got it. I'm sure I can do it since it's basically the same procedure each time. It would be nice to see an alternative method to doing these problems. Thanks!
If the image created behind the lens was virtual, would you still say that the object distance for the mirror was -ve?
With this lens, the image behind the lens can only be real with an object on the left side.
Michel van Biezen thanks for your reply. I actually meant if you changed the lens system to get that particular scenario.
But anyway, maybe a better question: in general, when do you define object distance as -ve and when do you define it as +ve?
1. Refraction from lens
2. Reflection from mirror
3. Refraction from lens is missing
Because distance between mirror and lens is 10cm and image formed after reflection at approx 13cm....
The video is correct as is.
Very easy to follow. Thank you!
Sir, is Image 2 real or virtual relative to the original object (O1)? I`m asking because I see that image 2 appears behind the lens, i.e between the eye and the lens.
Trymore,
Yes the final image is real.
The best way to determine that is so see how the final image is formed.
In this case the rays are reflected off the mirror and CONVERGE down to a single point where the image is formed.
If the rays DIVERGE, then the image will be virtual.
Oh, yes, I understand. Thank you Sir.
thanks a ton sir, was having a lot of problems in problems of this type
Nice video but brightness is the problem
focal length of concave mirror is ,-ve...
Aman Tak no its positive.
It depends on your sign convention
hey can anyone help on this question.
A person with near sighted eye has near and far points of 16cm and 25cm respectively. (a) assuming a lens is placed 2cm from the eye, what power must the lens have to correct this condition? b) Suppose that contact lenses placed directly on the cornea are used to correct the persons eye. What is the power of the lens required in this case, and what is the new near point?
thank you anyone who has time to help!
Hannah,
For part a, you want the image from an object placed at infinity to appear 25 cm away from the eye (the far point) which will be 23 cm in front of the lens. That makes the image virtual and therefore the image distance will be negative. (1/f) = (1/s) + (1/s')
(1/f) = (1/infinity) + (1/-23cm)
f = -23 cm and its power is P = 1/-0.23m = -4.35 diopters.
For part b, the lens is directly on the eye, therefore the image distance will be 25 cm from the lens.
(1/f) = (1/infinity) + (1/-25cm)
f = -25 cm and its power is P = 1/-0.25m = -4.00 diopters
The new near point is found by using the equation
(1/s) +(1/-16cm) = (1/-25cm)
s = 44.4 cm which is now the new near point.
how u know it was real and inverted?
+Peter C You know the image is real, when you place the object on the left side of the lens (or combination of lenses) and the image appear on the right side (or other side). If the image appears on the same side, as the original object, it is a virtual image. If the magnification is positive, then the image is upright.
Peter C if an image is formed at the same side of object in case of lens then that image is virtual and vice versa ! If an image is formed in the same side side of object in the case of curved mirror , then the image is real and vice versa !! I hope you got that
Thanks for the post
Thank you :)
You are welcome.
sir you r awesome
this is being taught the same way I was taught in physics 45 years ago.
unfortunately it is no less 'unclear'.
what is missing is teaching how to DESIGN a lens system for a particular PURPOSE.
otherwise all you can do is try this and try that like a blind squirrel looking for a nut.
these type lectures should be preceded by examples without all the algebra - showing
different systems of lens and how they behave. then proceed to the same examples with
the algebraic methods.
disappointing.
I feel the same way. It looks to me that it would be better to buy a magnifying glass at 100x on Amazon for $10.00 rather spend hundreds for a possible
.33 magnification. Does not make sense at all. It CV oils be that it's missing something or lacking something. Anyway, thank you for sharing, but it left me empty in the end.
Thnx sir
Great¡
virtual object