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

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.

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?

@@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

@@mip0 and what's so good of lecture being live, is it really worth it to organize time of large group of people over outdated concept of live lecture?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

Congratulations! That took a lot of hard work.

We Thank you .Kind Sir.

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

clear as a linse

Glad it was helpful! 🙂

Happy to help!

Good luck with your project.

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

-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.

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

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

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).

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

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?

@@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.

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.

Glad you liked it

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

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

Most welcome! And welcome to the channel!

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.

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.

Great! Glad you found our videos! 🙂

Glad it helped! 🙂

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

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

@@MichelvanBiezen That would be pretty helpful!

*mirrors

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.

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

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.

Michel van Biezen thank you :D

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.

Glad it was helpful!

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!

@@MichelvanBiezen Thanks

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

We have videos on that in the playlists on lenses.

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.

+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.

+Michel van Biezen Thanx... :)

Take a look at this video: Physics - Optics 2 (1 of 15) The Thin Lens Equation: Intro kzfaq.info/get/bejne/pdqbZ7Gp1bC2oo0.html&index=1

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?

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.

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

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

Ok sir thank you..

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.

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.

It is the second equation with all + signs.

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

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.

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.

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.

@@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

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

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

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.

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).

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

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

Welcome to the channel!