It's pretty bad when folks on You Tube teach something better than in the classroom. Thanks for your free lectures you have saved lot of headaches for most of us students.

@ismailb4334

6 жыл бұрын

Calling him one of the "Folks on KZread" is quite disrespectful. He is a teacher too.

@mip0

6 жыл бұрын

No it isn't. First of all, before you're confident enough to upload your explanation on youtube like this you're probably quite good. So you're simply comparing average teachers to an exceptionally good teacher. It's not "pretty bad" to be worse than someone exceptionally good. Secondly, when you're doing it in a classroom you're doing it live. Of course it's not gonna be as good as if you can redo it as many times as you want. But it's true that sometimes it might be more efficient for you to study on your own than to go to a lecture. It varies a lot between people.

@kshinji

4 жыл бұрын

why? why assume we need classic classroom? just because people employed there can't reinvent the concept or find a different purpose for their time? I don't see a point why we have so many people putting so much effort in face to face education when they all could put that effort in creating materials, from textbooks and courses, through exercise bases and lots more, redact them better and better over the year and reduce face to face time to 10-20% where pupils and students actually look for it?

@kshinji

4 жыл бұрын

@@mip0 what do you mean "before", no amount of time will make someone a good teacher if that's not meant to happen as is the case for most, often because they don't really care

@kshinji

4 жыл бұрын

@@mip0 and what's so good of lecture being live, is it really worth it to organize time of large group of people over outdated concept of live lecture? to keep your life convenient? why are you more important than then the students who may have a bad day for learning, may prefer learning in the evening, learning from books? let people do stuff their own way and test them, why so scared of people doing things their own way and their methods being objectively evaluated? why be afraid of comparing your way of teaching on a level field with whatever your students prefer? one thing i hate the most in classic education - enforcing the process, micro-managing the education of every individual. with large focus on not letting be proven wrong about this idea, because that would cause outcry and a reform and many people who are wasting student's time have a lot to lose on real and good change in education

When parallel rays pass through a converging lens, the rays come together (they converge) When parallel rays pass through a diverging lens, the rays move away from each other (diverge) When parallel rays reflect from a converging mirror, the rays come together (they converge) When parallel rays reflect from a diverging mirror, the rays move away from each other, (diverge)

As written several weeks ago, thanks for your help; I passed the exam of Optics (Experimental Physics 2) of the course in Astronomy, University of Padua, Italy (the university where Galileo taught!) Thanks again, Antonio

@MichelvanBiezen

3 жыл бұрын

Congratulations! That took a lot of hard work.

Adorable work on Physics Lectures of the whole series.

Thank you for the feedback.

This was so helpful. Thank you so much for teaching me equations that weren't even mentioned in my class.

You helped me understand a semesters worth of physics in less than 12 minutes! You're physics god haha

Years later this is still useful. Thank you!

Hello Michel. I have some questions and hope you can help me. What will happen to the final image if we put the second lens behind (which means on the right-hand side) of the first image? And how will the light rays go?

Thank you for the kind words. I am glad this is helpful.

@yaminireddy5157

4 жыл бұрын

We Thank you .Kind Sir.

@darkmatter2640

3 жыл бұрын

why did u take 65 as -65 even it is right of the 2nd lens

Bro you're the best! Keep up the good work , your vids are awesomeeeeeeeee

my greatest online lecturer ever

I have an exam tomorrow morning, and I didn't really have an idea on how to solve this until I saw your videos. Kudos from the Philippines :D

Thank you for explaining it so vividly.

Thanks for being alive

Hi Do you know what does focal length means on a camera lens specifically?

Useful.Been a great help already.

I have enjoyed your lectures on optics. I have a camera lens I want to re-purpose as a projector lens. After following the equations and making measurements, I noticed that the focal length on the lens does not match the actual. My lens says f=42mm but it consistently measures 28mm. The lens is from an old 35mm camera. I was wondering if you have an explanation for this discrepancy? I did a brief search on the web and it seems these focal lengths may not have the same definition. If they are defined differently it may make for an interesting lecture.

Thank you very much for this lecture, this helps also to understand certain things you couldn't put together back in school

@MichelvanBiezen

2 жыл бұрын

Glad it was helpful! 🙂

Another banger Michel, thank you!!

How is there a minus in finding magnification in the first image?? At 4:00

I have an experiment I want to conduct, using lenses focusing through fiber optic image bundles to wrap around a solid pillar, effectively making it optically transparent, as theoretically you should be able to look through one lens and see the image created by the other lens projected from one focal point to the other through the fiber optic. This lecture helped me understand the mechanics so I can better plan for my project. Thank you

@MichelvanBiezen

Жыл бұрын

Good luck with your project.

this really helped me completing my assignment thanks a lot

Your demonstration is formidable!

Whoah! This actually makes me want to study physics, not pretend to study, like I did it at school..and university :) Thanks a lot!

@MichelvanBiezen

3 жыл бұрын

Happy to help!

i was searching Code V. some how i got here. Sir thank you for teaching.

Clear as crystal

@emmazimmermann9963

3 жыл бұрын

clear as a linse

This is really well done. Thank you very much for your video :)

Really amazing lecture. Thanks so much!

Pretty good man . Hats off 🤗

Very nice explanation. Thank you sir!!

Thank u for this video.. Its really helps me..

Very good teacher! I am jealous of your students. Great!

THANK YOU SOOOOOOOOO MUCH, SIR. FOR ALL THE FREE LECTURES. REALLY REALLY HELPFUL.

@MichelvanBiezen

9 жыл бұрын

Thank you Hannah, It is good to know these videos are helping.

Excellent explanation!

I have an acromath 400 mm the + 165 mm the - 2.5 diopters ( It matches my left eye prescription"is it - 400 mm or not?' " ) I have some issues going to the ophthalmologist ! There is no doubt that I can use a formula to calculate the closest negative focal length to my real UNSOLVED diopter power of the same left eye and make sure I do not have either a macular degeneration or retina detachment ( ? ) I do have many lenses and I want to utilize them before I go into the dark ! Some years ago you helped me ratifying the magnification of a lens 10 inches focal length as 1 X thanks for that (back then) And thanks for this video and thanks for you answer if you feel like to respond !

@MichelvanBiezen

Жыл бұрын

-2.5 diopters does indeed equate to a focal length of -400 mm. If you are experiencing some issue with your eyes, I would recommend that you see an opthamologist. If you don't receive good advise, do see another one for a second opinion. All the best.

@javiermancheno8531

Жыл бұрын

@@MichelvanBiezen Thanks Sir I think that I imagined your response !

can't we just use formula to find effective focal lengths and then find the Img distance using the lens formula?

I have a question about creating the second (final) image. I was able to get the same calculations as you but my image appeared 15.29 cm on the left side of the second convex lens instead of the right. You said at @6:40 that the final image is right to the lens because di2 (distance of image #2) is a positive 15.29cm. I have from my notes that if di = + it is a real image and that in a lenses the image will appear on the opposite side of the lens. You compare that because the magnification M= + it is upright in comparison to the first image. So why are you comparing the magnification of image 2 to image 1 but comparing the final image (image 2) if it is virtual or real in terms of the original object? As well, when I drew out my diagram using the rules of going from the top of an object to parallel to when it hits a lens going through its focal point I got that the cross over for the final image goes through the left side of the second lens.

@MichelvanBiezen

6 жыл бұрын

The position of the second image is difficult to determine because it is difficult to determine the change in direction after the rays pass through the first lens and are now diffracted by the second lens. That is why I recommend that one draws the first image and calculates the position of the second image as shown in the video

@yaminireddy5157

4 жыл бұрын

Also...is this the new Cartesian sign convention?

Nice explanation. How object distance is negative in second case. It should measure from optic center and compare with incident ray

Thank you so much homie i Really appreciate the help :)

For the second lens, the initial distance is given a negative sign (-65 cm), but why then is the resulting image distance positive if it falls on the same side?

@MichelvanBiezen

7 жыл бұрын

For lenses the left side is the positive side for the object and the right side is the positive side for the image.

@themesaregreat

7 жыл бұрын

Thanks for the response-- so it just another convention? If the object begins on the right side, it will be negative? (and vice versa for image).

@MichelvanBiezen

7 жыл бұрын

It is indeed a convention. With lenses the image is expected to be on the other side of the lens relative to the object.

@joelip4396

6 жыл бұрын

Does the sign of the image location tell you if the image is real or virtual as long as you stick with the convention throughout the whole problem?

@sreerammadhavanv3604

4 жыл бұрын

@@themesaregreatIt is actually like if the distance that we are measuring is in the direction of light rays it is positive. Otherwise, it is negative. For e.g., object distance is measured from pole. The direction is opposite to that of light rays. So it becomes negative.

Sooo helpful!!! You made physics actually seem fun :-)

Thank you. You are a life saver sir! literally!

amazing ..thanks a lot sir

very good diagrams and explanations…..

Very helpful video. Thank you

i still dont get it, how can you get I2 = I1(cos teta)square. and also how can you assume that I1 is half than I0

well done

Which lenses are use

What happens if the object i positioned between the first focal-point and the first lens? From calculations I'll get that the first image is virtuell and upright while the second image will be reell and upright, or have i totally missed the physics??

@MichelvanBiezen

7 жыл бұрын

The first image appears correct (according to your description), the position and orientation of the second image depends on a number of factors such as focal length of the second lens, position of the first image, etc.

this is sorta what im taking in high school but i dno what he started talking about when he got to the equations!

Thank you so much sir! From India :)

@MichelvanBiezen

2 жыл бұрын

Most welcome! And welcome to the channel!

Beautiful explanation. Thank you.

@MichelvanBiezen

2 жыл бұрын

Glad you liked it

thank u this made so much sense

@MichelvanBiezen

Жыл бұрын

Great! Glad you found our videos! 🙂

sir, so will there be two images or just the second one and the first one is just a help for the second's calculation? thanks

@MichelvanBiezen

5 ай бұрын

There is only the final image. We use the "first image" in order to determine where the final image will appear.

Thank you very much. This is very usefull

very clear, Tks !

So the final image is smaller than the original object? I'm trying to figure out how to achieve the highest magnification for a homemade microscope by combining laser pointer lenses. I've already tried this with 1 (works great!) and even two (lenses touching) and it certainly made the image larger but less clear and I don't believe it was by any means doubling the previous magnification Any idea how I would be able to figure out the right distance to place the two lenses apart? Let's assume the focal length 9 Also how do the lenses inside a laser point work? Are they concave or convex or compound? I really don't know much about this But I'm trying to learn

@MichelvanBiezen

8 жыл бұрын

+Faraz Bolourian Look in this playlist: PHYSICS 59 OPTICAL INSTRUMENTS There is a video about a microscope and how to place the lenses.

géométrical optics please can you do Under Title in French

You save me ! Thank you ! 😊

@MichelvanBiezen

Жыл бұрын

Glad it helped! 🙂

sir why dont we take 50 as negative in the first case

How do you derive the equivalent focal length of the 2 lens combination?

@MichelvanBiezen

4 ай бұрын

We have videos on that in the playlists on lenses.

Mr. Michel there's a question. The formula for magnification says (-image dist./object dist.) if the image distance is negative for a convex lens then why lens equation says 1/f = 1/p + 1/q why it doesn't say 1/f=1/p - 1/q ? where p and q are the distances of object and image from the lens respectively.

@MichelvanBiezen

6 жыл бұрын

The equation remains the same regardless of what lens (or mirror) you have. But when you substitute the actual numerical value, you will substitute in a negative value if q is negative.

thank you so so much!!

Great videos, you're a very clear, concise teacher. Everything you worked out makes sense, but I'm struggling with the intuitive side of things: I'm trying to imagine this scenario as a viewer, and I can't help but think that the position the image depends on the position of the viewer. For example, if I look through a lens at an object, the appearance of that object will change depending as I move closer or further away from the lens. How is that all factored into this? or what am I missing?

@MichelvanBiezen

9 жыл бұрын

Matt, Traditionally the object is on the left side of the lens and the observer is on the right side of the lens. There are 2 kinds of images, real and virtual. The real images can be seen from anywhere as long as a surface is placed at the location where the image is formed. A virtual image can only be seen by looking into the lens and the virtual image will appear (although there isn't a real image there), the virtual image is seen because the rays appear to be coming from there.

Professor, do you mind explain that why the ray didn't go through the center of the lens 1, but go to the F1 on the left of lens 1 instead?

@MichelvanBiezen

6 жыл бұрын

Many rays go through the various parts of the lens. But to figure out where the image is, you only need to draw the 2 rays.

@meganyu9233

6 жыл бұрын

Thank you for your kind reply, and have a nice day!

how can i design a zoom telescope?

really helpful for my hight schl physics

but what if the distances between the two lenses is more than from what you were explaining? like, maybe second the lens is somewhere behind the first image produced by the first lens

@MichelvanBiezen

7 жыл бұрын

Then the object to the second lens (which is the image of the first lens) becomes a real object as seen by the observer on the other side of the second lens. This would work just fine and you would use the same technique to find the answers.

@flabbergabber

7 жыл бұрын

Michel van Biezen thank you :D

There are some ways to find the path of the light rays like adding parallel lines with the light rays, but honestly my professor didn't explain them too well and I don't know how to use them well.

@MichelvanBiezen

2 жыл бұрын

Yes, it can be done, but it is challenging. Perhaps we should do a few videos on that.

@Ana_leaf

2 жыл бұрын

@@MichelvanBiezen That would be pretty helpful!

THANK YOU A LOT!!!

Pardon me sir? Do you have a video about combining two concave lens?

@MichelvanBiezen

3 жыл бұрын

I believe we have on convex and one concave combination example (the method is the same regardless what the lenses are)

@jimmytorreon9615

3 жыл бұрын

Ok sir thank you..

how do i calculate the focal length of a lens with one side covered with a transparent medium of diffenrent refraction index( lets say water) ?

@MichelvanBiezen

7 жыл бұрын

Take a look at this video: Physics - Optics 2 (1 of 15) The Thin Lens Equation: Intro kzread.info/dash/bejne/oaiflrCxoK24orw.html&index=1

@jeremymah5704

7 жыл бұрын

thanks for the tipp but in my case does that mean that the water surface on that ones side forms another lens but with both sides bulged to the same direction or can i directly use the formula you provided in the video?

@MichelvanBiezen

7 жыл бұрын

That equation does take into account the radius of curvature of both sides of the lens, but not if the indices of refraction are different on both sides of the lens. At that point we have to use the tick lens equations.

good day sir what is the reason behind making S2 negative?

@MichelvanBiezen

Жыл бұрын

The convention is that if the object is in the front of the lens (to the left of the lens), the object distance (s) is positive. If the object is on the back side of the lens, (to the right of the lens), the object distance (s) is negative.

What if S1' was 30cm? The second image would be on infinity? Is there a way to show that with equations?

@MichelvanBiezen

10 жыл бұрын

Cuchilloc, That is a very good question. And here is how you will answer your own question. Find the image of the first lens when S1 = 29 cm Then repeat that with S1 = 29.9 cm You'll find that a virtual image will form to the left of the lens farther and farther away. In the limit when S1 approaches 30 cm below the image will go to infinity at the left side of the lens. Then you do it again for S1 = 31 cm. Then for S1 = 30.1 cm. You will find as you get closer and closer to 30 cm a real image will form far to the right and in the limit as S1 approached 30 cm from above, the image will go to infinity on the right side of the lens. Therefore there is a discontinuity at 30 cm and no image can form.

I dont understand why if s2'(15,3cm) < f2(20cm) you draw the final imatge after f2, i just dont see it.

in the case of 2 lenses, how do we know that the first image acts as the object for the second lens?

@MichelvanBiezen

8 жыл бұрын

+Nabarun Kashyap That is just the method that we use to solve this type of problem. In actuality, this first image doesn't really exist. That is because after the rays pass through the first lens and were caused to bend to a converging path, they reach the second lens and bend some more. There are techniques (that are more advanced) that trace the rays through the lenses and solve the problem that way, but it is easier to assume the the "image of the first lens, becomes the object of the second lens.

@nabarun83

8 жыл бұрын

+Michel van Biezen Thanx... :)

Great video :) i have a question. Is there difference between converging or diverging LENSES to converging or diverging MIRROR? I am just confused. Thanks :)

@markuy6799

9 жыл бұрын

*mirrors

Very helpful, thanks

@MichelvanBiezen

3 жыл бұрын

Glad it was helpful!

Why does the image of the 1st lens act as the object for the 2nd one?

@MichelvanBiezen

6 жыл бұрын

In actuality, after the light rays refract as they pass through the first lens, they reach the second lens and refract some more as they pass through the second lens before producing an image. (assuming they produce a real image behind the second lens). To make it easier to determine the location, size, and orientation of the final image we use the "trick" to assume the image of the first lens to be the object of the second lens. As it turns out, it will give us the correct results. If the rays are traced both ways, they will come together at the same point.

Sir can u tell about the equivalent focal length of two convex lenses.. I'm getting negative in my case? Is it possible?

@MichelvanBiezen

3 жыл бұрын

It is possible. It depends on the location of the object, the focal length of the 2 lenses, and the distance between the lenses

@bhavanapal6034

3 жыл бұрын

Ohk if it's negative equivalent focal length. Then can you say something about the image? Like how it will be?

Hi, Thank you for the great Explanation. But In Iraq, We use this law: 1÷f=1÷u + 1÷v To find the distance between The Image and the len. u: represents the distance between the object and the len. v: represents the distance between the image and the len. Instead of the law used in this video.

@MichelvanBiezen

2 жыл бұрын

That is the exact same equation. If you solve that equation for v algebraically you'll get the equation used in the video. Welcome to the channel!

@EveryThing-zj1us

2 жыл бұрын

@@MichelvanBiezen Thanks

Why does the beam change direction when going through a lense?

@MichelvanBiezen

10 жыл бұрын

Light will refract (bend) when it traverses a boundary from one medium to another and the index of refraction changes. (See videos on refraction and Snell's law). The shape of the lens takes advantage of that and thus causes light to bend in order to achieve the images desired.

Dear sir, When i use efl = f1*f2/(f1+f2-d) = (30*20)/(30+20-10) = 15. then use S = 50, i got I = 50*15/(50-15) = 21.42. whey the EFL is not the result of this video? And I notice if d =0, then the S'_from_ELF and S'_From_direct_calculate_by_your_method will the same

@MichelvanBiezen

3 жыл бұрын

Your approach is correct, except for the placement of that "equivalent" lens. If you place the equivalent lens at the location of f1, the image will NOT be at the correct place as you have found out. if you place the equivalent lens a the location of f2, again you will get the wrong answer. If you place it exactly between the two your answer will be close (s will be = 55 cm), but not exact. That is because f1 and f2 are different and the effect of each lens independently is different. If f1 and f2 were the same then the equivalent lens would be placed exactly in the middle.

@grantkobe9

3 жыл бұрын

@@MichelvanBiezen Dear Sir, Thanks for your response! I do not except you will response so soon due the vidoe have 6 years old. I am very appreciate it, and search other lens lesions you teach in the net. Help me a lot. 1.1 By your suggestion, i calculate the actually x in this case. when s1 =50cm, f1 = 30cm, f2 = 20cm. d = 10cm ,then x = 4.6212 .( Lens1- x(the EFL position)- Lens2). 1.2 When I change S1 to 70 cm. I found the x is move to 4.84cm. 1.3 If set object to infinite, s1' = -30cm, x = 5cm. it happen d/2! 2.1.If f1 == f2, x still not be d/2, for example, make s1 = 50, f1 = f2 =30. x = 3.3642 2.2 when i set s1 to infinite, x = 4. 2.3 i use octative to check the detail i calculate is stated bellow, if you have interest. ( 1.1 imgur.com/sye48w5 1.2 imgur.com/XE81xHi 1.3 imgur.com/hBfE7yZ 2.1 imgur.com/KOHWoJ1 2.2 imgur.com/edXDO3W 2.3 imgur.com/f5Wscc8 ) The reason I search for ELF is when I checked our vendor's lens spec, I found use ELF to calculate the Depth of Field. I explain the DOF formulate to my collage. He said it should not be right. 3p or 4p lens should have more complicated formulate. After study your teach and suggestion from you. I think he is right in some way. Since the setup position of ELF lens will change in between f1 ... fn for a n lens, if the object distance is far away, for example S1 >10f), the dof may be right, if the object distance is near the lens. For example, 1.x -5.x f distance, we can not only use EFL to calculate, because the actual ELF position must take consideration. Are this conclusion right? Or i have missing something else. Terry

You are a GREAT great teacher. Thank you so much!

I have a question, is object two real or virtual? I had a lab in class we had a whole debate about it and still don't know.

@MichelvanBiezen

Жыл бұрын

The confusion stems from the assumption that the TECHNIQUE of solving this problem is what ACTUALLY is happening with the light rays, which is incorrect. What is actually happening is the following: Only place the first lens there and see how the first image is formed due to the refraction of the light through the lens bending the light to a single point where the first image is formed. Then place the second lens there which will cause the light to refract more and cause the light to bend to the final (and only image) closer to the 2 lenses.

@Michel van Biezen, you made a wrong assumption in this video. The forth ray of light, completing the second image, isn't supposed to go in a random (or approximative) direction. This is because when the ray has left the first lens it's going to be parallell to the normal, which means that passing through the second converging lens, it has to be bent towards the fokal point. Now you'll see that the calculated answers (from the equations) will represent the second image much better. No offense intended, just a heads up! ;)

@MichelvanBiezen

8 жыл бұрын

+Anton Carlsson The assumptions and the calculations in this video are correct.

can we calculate combined focus of both the lenses

@MichelvanBiezen

6 жыл бұрын

Yes 1/f = (1/f1) + (1/f2)

Sir , is the formula of lens 1/f= 1/v -1/u or 1/f=1/v+1/u. I thought it's the first one but in the video u have written the second one. Could u pls clear my doubt.

@MichelvanBiezen

2 жыл бұрын

It is the second equation with all + signs.

When you say behind, you mean relative to the object, not the viewer correct? Sorry I am a bit confused as to why you went from +65cm to =65cm, thanks!

@MichelvanBiezen

10 жыл бұрын

Lili, That is a good question. It is easy to get confused. By definition, the object is placed in FRONT of the lens which here is drawn to the left of the lens. By definition, the observer is placed BEHIND the lens which here is drawn to the right of the lens. So when the image of the first lens appears BEHIND the first lens, it will be REAL image to the right of the lens. That image then becomes the object of the second lens, so therefore, the object of the second lens will be BEHIND (to the right) of the second lens and thus it will be a VIRTUAL object and the object distance for the second lens will be negative.

Thank your, sir!!!!

U r best

This saved me from stress from lecturers that find it out to explain well

For the second lens the object(I 1) and the image(I 2) are on the same side of the lens. How is it possible?? 🤔🤔🤔

@MichelvanBiezen

5 жыл бұрын

Even with a single diverging lens, the object and image are on the same side. With 2 lenses the "first image" is just is used as a method to find the final image. (There actually is only one image, which is the final image).

@gamingleaf3169

5 жыл бұрын

Thank you so much for answering my question. It really helped me

doesnt the minus means the image is virtual not inverted?

@MichelvanBiezen

7 жыл бұрын

That depends on which minus sign you are talking about. If it is the one for the magnification, then it indicates that the image is inverted relative to the object.

I'm still having trouble understanding why I2 is a real image and POSITIVE 15.3 if the image is formed behind the lens also? Didn't we make the object distance aka 65 cm negative in the step before? Any help would be much appreciated thank you!

@MichelvanBiezen

10 жыл бұрын

Sruveera, Maybe this can help. If you ignore the second lens, the first lens will converge the rays to a single point causing a real image. The second lens will simply cause the rays to converge sooner (or more to the left) but still causing a real image. If the rays converge to the right of the lens or lenses, the image is real. If the rays diverge after passing through both lenses, the image will be virtual.

@sruveerassathi

10 жыл бұрын

I think I understand the big picture but I just want to understand the sign conventions again. Why is 65 (aka the distance from the 2nd lens to the first image) is made negative? But the second image that's formed is made positive even though it is on the same side? Is it because we're looking at the second image distance with respect to the first object on the other side? Then why not do the same with the first image to the second lens? I hope that made sense...I'm just trying to figure out how to be consistent and not mix up signs. Thank you!

@MichelvanBiezen

10 жыл бұрын

Sruveera Sathi Sruveera, That is by convention. A normal situation is where the object is on the left side of the lens and the image is a real image on the right side of the lens, making it a real image. Since the "object" of the second lens (which is really the image of the first lens) is on the RIGHT side of the second lens, that makes that a "virtual" image. That means that the rays DO NOT go from the VIRTUAL OBJECT to the right side of the 2nd lens to the 2nd lens. Those rays DO NOT exist. But if we take the first image to be the 2nd object of the second lens, the mathematics will work out and give us the correct answer. Therefore we make the object distance of the second lens NEGATIVE. IN ACTUALITY, THE RAYS DO NOT COME FROM THE SECOND OBJECT, BUT ARE SIMPLY REFRACTED MORE STRONGLY BY THE SECOND LENS, PLACING THE FINAL IMAGE CLOSER TO THE FIRST LENS THAN IT WOULD IF THE SECOND LENS WASN'T THERE.

@sruveerassathi

10 жыл бұрын

Michel van Biezen Ah I get it now. I needed to realize that the rays coming from the first image aka the virtual object do not actually exist! Thanks so much sir for your time! These videos are my life saver as I'm preparing for the MCAT--cannot thank you enough.

@hughnguyen425

9 жыл бұрын

Michel van Biezen Thank you , Prof Biezen, this explanation really makes me understand this clearer.

i think i should stop listening to my physics teacher she don't know how to teach students. thank you so much

Thanks from INDIA🙋🙋👍👍👍

@MichelvanBiezen

4 жыл бұрын

Welcome to the channel!

What happens with combination of converging lens and Concave mirror

@MichelvanBiezen

8 жыл бұрын

+Amit Sharma You work it out the same way. The image of the lens becomes the object of the mirror.

@Michael Van Biezen would you mind helping me understand something about lenses? I'm trying to build a certain type of lens, but I need a better grip on how I should go about it. Would really appreciate some help understanding. I dunno if you are able to message me on here or not though...

@MichelvanBiezen

5 жыл бұрын

What kind of information are you looking for?

@Tommo020788

5 жыл бұрын

@@MichelvanBiezen thanks for replying mate. Im actually trying to make a lens for my phone (huawei mate 20 pro), so straight off the bat its hard already because the phone already has a lens on it which isnt specified. So theres a lot of guess work and experimentation needed to find the right fit. But basically what im not sure about is how a macro lens is generally built. Do macro combination lenses usually consist of a biconvex lens at the front, then a plano concave one in the middle to counter aberration, followed by a plano convex lens at the back? Also, how would the focal length of each lens then come into play in making that work? Say if we were trying to have a focal length of 5cm in front of the lens. I know I am probably missing a lot of necessary information, but im trying. Ive been watching your videos and learning too, but havent seen one on macro lenses particularly. Any ideas on how i might determine what lenses would suit my phone lens through experimentation, without the manufacturer giving up its exact specs? Cos the manufacturers of these lenses seem to want to keep them as closely guarded secrets.

@MichelvanBiezen

5 жыл бұрын

Good lenses are usually manufactured with compound structures with up to 7 lenses placed together. This is done with computers and precision machinery and impossible to duplicate by hand. Typically lenses on phones are not that complicated but you'll have a hard time to get the specifications from the manufacturer on how they are built.