Physics - Optics: Lenses (3 of 5) Lens Combinations - Converging & Diverging Lenses

[@shainahkelly9787]

We're having an exam about physics tom and need to understand more of it. Thank you mister michel!

[@MichelvanBiezen]

Glad you found our videos and you find them helpful. 🙂

[@tamassinty]

Now, I do understand the physics of optics, thanks to this crystal clear explanation.

[@jkimt08]

I actually found this example really helpful conceptually. Thank you!!

[@a.c.9478]

Thanks a lot! From a student of University of Padua (Italy) Astronomy course.

[@MichelvanBiezen]

Our pleasure! And welcome to the channel!

[@thecosmos7671]

It's really admirable,I hope you had a good first year.

[@The_moonlit]

I just need to clarify, if the second lens diverged the rays a lot, wouldnt it appear they converge at some point to form a virtual image. So in this example, the second lens basically diverged the rays to be exactly parallel to each other, forming an image at infinity (none at all)

[@ruu8228]

A 10.0-cm tall object is placed 30.0 cm to the left of a converging lens of focal length 20.0 cm. A converging mirror of focal length 20.0 cm is then placed 40.0 cm to the right of the lens. Determine the location and size of the image formed by the converging mirror. So the S2 is also at an infinity?

[@MichelvanBiezen]

The final image will be at 10 cm (in front of the mirror). The final image size will be one third the size of the original object.

[@ruu8228]

@@MichelvanBiezen Thank you! I just got confused about the sign in the denominator of di2.

[@pewpew1645]

Why is there two f1? Is the f1 to the right of the diverging lens meant to be f2?

[@MichelvanBiezen]

There are two lenses with focal lengths f1 and f2. F1 and F1 are the two focal points of lens 1.

[@learnology01]

Why do we take the image formed by the first lens as the object for the second lens?

[@MichelvanBiezen]

That is how you can get the answer. In reality the first image doesn't form since the rays are refracted by the first lens, but it is a good technique to solve the problem.

[@be4913]

How do you draw the ray diagram for the second lens though? I can use the equations to get the right answer but I don't know how draw it

[@MichelvanBiezen]

It is not possible to draw the ray diagrams in the way we find the solution algebraically. Essentially, you take the rays leaving the first lens, and if the second lens is a converging lens, it will cause the image to form more to the left and if it is a diverging lens it will cause the image to form farther to the right or will cause the rays to diverge if the diverging lens is stronger than the first converging lens.

[@AhmedAbdelazim--]

Michael, When u draw the ray diagram for the first image, the image of it (final image) would be formed at 15 cm not infinite .. i think the minus sign u used in s` is wrong .

[@MichelvanBiezen]

The video is correct. Thank you for checking.

[@94chahal]

you sir are great. thank you for helping me out with my finals. if i pass the full credit goes to your great videos.

[@YoyoBear12]

hello, I have a quick question. Since it was determined that the second image distance was infinite, does that mean there won't be a total magnification in this case? just wondering thank you.

[@MichelvanBiezen]

Yoyo Bear If you plug infinity for the image distance into the magnification equation, you will get infinity for the magnification as well. m = - s' /s

[@YoyoBear12]

Thank you so much!

[@mementoaura]

What does this mean in terms of the image we actually see...would it be infinitely magnified?

[@OverlordMD]

Using these lenses in this setup, is it possible to focus an image to be small with a focal point of infinity (or something close to that)? What would happen if the source was directly in front of the lens?

[@webilogIndia]

What happens if lenses are swapped? Still you get the same image system?

[@MichelvanBiezen]

You wll get a different image at a different location. Try it and you will see.

[@webilogIndia]

@@MichelvanBiezen Thank you.

[@LillianZhong]

An object is 30 cm from a concave lens of 15 cm focal length. A convex lens of focal length 10cm is placed 5 cm beyond the concave lens. Find the position and magnification of the final image from the convex lens. I think that there are extra information. Since you have the object distance from the diverging lens, i don't know why they are giving the converging lens. If so, how will I find the distance of the image for the converging lens? Will I use the diverging image as the virtual object for the object length of the converging lens?

[@MichelvanBiezen]

Lillian, A convex lens is a diverging lens, therefore the focal length of the first lens is -15 cm. So the first image can be found at : s1prime = (s1)(f1)/(s1-f1) = (30)*(-15) / (30 - (-15)) = - 10 cm. Which places the 1st image 15 cm in front of the second lens (10 cm + 5 cm) The second image can be found at: s2prime = (s2)(f2)/(s2-f2) = (15) * (10) / (15 - 10) = 30cm That places the final image 30 cm behind the second lens. The magnification = (-(-10)/30) * (-30/15) = (1/3) * (-2) = -(2/3) That makes the final image 2/3 the height of the initial object, it is a real image and it is inverted.

[@aaronbodge431]

Hello Michel. I was wondering how you determine whether s2 is negative or positive. What happens if the observer is on the other side of the lens? Is it that it's negative on on the side with the observer and positive when the observer is on the other side? I'm quite confused on this aspect of the video and would like a more in depth explanation. Thank you so much, you're a great teacher.

[@MichelvanBiezen]

Aaron. It is best to follow this convention: Place the object on the left side of the left lens, and the observer on the right side of the right lens. If the final rays converge on the right side of the right lens, the image is real. If the final rays diverge on the right side of the right lens, the image is virtual.

[@simonburt9990]

Good Afternoon, if I have the size of a light source or original image and the size and distance i want the final image to be, is there a way to work out the focal lengths of 2 lenses I would need? thinking that I would like a 30mm diameter picture if this could be achieved at infinite focus that would be ideal.

[@MichelvanBiezen]

Start with just one lens, and then calculate the focus of that one lens given the object distance and the desired image distance. Then calculate the resulting magnification.

[@robertbhatia7221]

I thought that if an object is placed at the focal point of a diverging lens, the image will be virtual and half way between the focal point ad the lens? I know that if the object is placed at the focal point of a converging lens, the image distance will be infinity, is this correct?

[@MichelvanBiezen]

+Robert Bhatia With a converging lens, place the object just before the focal point and calculate the image position, then place the object just after the focal point and calculate the image position. What can you conclude from that?

[@robertbhatia7221]

+Michel van Biezen THe image position is (+)=Real/inverted if the object is before the focal point and (-)=virtual/upright if it is after the focal point. What i dont understand is what happens when the object is placed at the focal point of a diverging lens?

[@MichelvanBiezen]

+Robert Bhatia With a diverging lens, it doesn't matter. s' = (sf) / (s - f) Since f is negative, the denominator will always be positive and the numerator will always be negative. Therefore the image will always be virtual (in front of the lens), when the object is placed in front of the lens.

[@diakaul7587]

why do we take image of first lens as object for second?? and if we do why don't we draw the ray diagram on other side as we do for the first object of first lens.

[@MichelvanBiezen]

It is difficult to do that since the object for the second lens (which is the image of the first lens) is a VIRTUAL object which makes it difficult to draw rays from that source.

[@kevinjason6697]

Hi, You mentioned that the 'smaller' the focal length of the lens is, the stronger it is. Do you mean the absolute value of the focal length? as f2 is the example above is way smaller than f1 so should be stronger?

[@MichelvanBiezen]

+Kevin Jason Yes, it is the absolute value. Or a better way to look at it: The closer the focal point is to the lens, the stronger the lens is.

[@mementoaura]

So is that true for all lenses? If the object is on the focal point than the image will form at infinity? Can someone help clarify?

[@MichelvanBiezen]

If you look at this playlist you will get a better understanding of that topic: PHYSICS 55.1 LENSES AND MIRRORS UNDERSTOOD

[@mementoaura]

Michel van Biezen will do thank you

[@harimelcentina3422]

Hello sir I would like to ask a question about lens magnification. Is there a way to magnify the image captured by the CCD camera on a microscope using set of Plano convex, Biconvex, and Positive Meniscus lens or any other lens combination? Hope you can help me. Thank you in advance.

[@MichelvanBiezen]

Once the image is captured on a CCD (which is essentially a computer chip), you will need a device that can read the image and project it. (like a computer screen).

[@harimelcentina3422]

yes sir, but my problem is that the image size captured and displayed is far more smaller compared to the image when image when i look through the eyepiece of a microscope.That's why I'm asking if it would be possible to make that image larger by using some combination of lens.Thank you sir..

[@MichelvanBiezen]

There are plenty of electronic gadgets that will do that for you.

[@irshadahmed2879]

Sir. .Is it possible for concave lens to form an image at infinity (like in the Galilean telescope) and it is magnified. As I read in many books that image formed by concave lens is always within its focus. .

[@MichelvanBiezen]

you'll find the answers to those question in this playlist: PHYSICS 55.1 LENSES AND MIRRORS UNDERSTOOD

[@sruveerassathi]

Hello Michel, I was wondering why you made the focal length of the diverging lens negative?

[@MichelvanBiezen]

Sruveera, That is by convention. The focal length of diverging lenses is negative and the focal length of converging lenses is positive.

[@robertktw]

Heres a quick way to remember those lens focal distance... if u see fat middle section of a lens than rest, then it is a converging lens with +f However, skinny middle section of a lens indicates a diverging lens with -f

[@animeaddicted7553]

What if the light comes from the other side ?

[@visalbotrchan1431]

I wonder why the focal length of converging lens is positive ?

[@MichelvanBiezen]

The word "positive" is relative and they could have used any other word. Since parallel rays converge after passing through a converging lens, a converging lens is defined as positive. If we stay consistent with the definitions, all the equations will work out.

[@MsLoversky]

Good point, Mike

[@golammustafa7623]

Thank you sir, it was a great help.

[@prasathr566]

thankyou very much sir for this vidoes

[@thilini22]

Thank you very much

[@FluffyJiaJia]

How can it be infinity if an image is always formed by a diverging lens?

[@FluffyJiaJia]

An image would be formed if the object were to the left of the diverging lens, so why wouldn't it simply because it is on the right? Thank you for your help!

[@MichelvanBiezen]

If after the rays pass through the diverging lens, the rays are parallel to one another, the image will form at "infinity"

[@indrapramana8337]

Thank you sir

[@SAREEYEFILMS]

Help me conceived

[@williamhan2064]

Is it a boy or a girl

[@izzathamjath9945]

Thankyou

[@MichelvanBiezen]

You’re welcome 😊

[@andyquach8437]

Please draw the graph

[@MichelvanBiezen]

That is a good question. We plan on adding that to the "viewer request" list.

[@Beatriz-qw4ix]

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[@sandeepkumar-xh7dh]

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