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### Course: Class 12 Physics (India) > Unit 9

Lesson 6: Refraction in thin lenses- Convex lenses
- Convex lens examples
- Concave lenses
- Paths of light rays through spherical lenses
- Image formation in spherical lenses
- Lens intuition
- Thin lens formula
- Solved example on lens formula
- Using magnification formula for lenses
- Using the lens formula
- Convex and concave lenses
- Thin lenses questions
- Virtual Object
- Thin lens formula - virtual object
- Worked example: Image formed by multiple lenses
- Image formed by multiple lenses
- Lens makers formula
- Solved example: Lens makers formula
- Sign convention: radii of curvature (lens maker's formula)
- Power of lens
- Power of lens
- Thin lenses in contact
- Thin lens sign conventions

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# Thin lenses in contact

Let's figure out the effective focal length when thin lenses are in contact. Created by Mahesh Shenoy.

## Want to join the conversation?

- If the image of lens1 acts as the object for lens2 will the object distance for lens2 be negative since it is considered a virtual object?(3 votes)
- Yes, it will be a virtual object as there is no actual object there. You can watch this video for more clarity- https://youtu.be/_K_jBemKwjs(1 vote)

- I have a question from the Thin Lenses quiz prior to this video (see playlist on left).

Here is the question and solution from the quiz [spoiler]: https://i.imgur.com/odGu9Oj.png

In the solution, the second blue arrow is on the focal point of the eyepiece. Yet, the eyepiece is a converging lens and any image on the focal point will not form an image (see figure 1).

Figure 1-https://i.imgur.com/0t39bZU.png

Should the second blue arrow be towards the right of the eyepiece focal point so that a magnified image is provided? (see figure 2).

Figure 2 - https://i.imgur.com/LDU0cAB.png

I have attempted to correct this problem in figure 3.

Figure 3 - https://i.imgur.com/O5iOGe4.png

Is this correct or am I missing something?(3 votes) - I've one question. In my textbook, there's a formula given for the total magnification
**m**obtained by the combination of thin lenses in contact as the product of magnification (m1 x m2 x m3 x ...) of individual lenses.`m = m1 x m2 x m3 x ....`

Can you please tell me the logic or the derivation behind it. Please.(3 votes)- I saw your question and I came to this derivation myself-

Let ho be the object height and the image height be hi'(height of image but not the actual image as there is another lens). Then the second lens would treat that image as an object so now the object height for lens 2 is hi' and the image height(actual image) is hi. Now,

M1=hi'/ho

M2=hi/hi'

and M(total)=hi/ho

All by M = (height of image)/(height of object).

Now M(total)= M1 x M2

That I'll leave up to you!!

HAPPY LEARNING :))(1 vote)

## Video transcript

suppose we have a thin lens of focal length F 1 and we put in contact with it another thin lens of focal length say F 2 then these two thinner lenses in contact can be thought of as one giant effective lens and so the question is what will be the focal length of this effective lens how will it be connected to f1 and f2 so let's draw our principal axis and since we're dealing with very thin lenses over here we can totally neglect their thickness and so we are assuming that their optic centers are at the same point so to figure out the focal length of this effective lens we'll do what we always do we just shoot parallel rays of light and see where they get focused so if we should parallel rays of light on this and find out where the Rays get focused that point itself will represent the effective principal focus and then that length will represent the effective focal length so let's go ahead and shoot two parallel rays of light if you look at this area along the principal axis then it is passing through the optic center and so this way will go undeviated and this ray of light is going to bend twice once here and once over here and in such cases to draw ray diagrams what we gonna do is first we're going to neglect this second lens we're going to assume only the first lens exists so in the absence of the second lens where would this ray go well since this ray is parallel to our principal axis it has to pass through its principal focus and the rays of light would have been focused at this point and so the image would have been formed at this point but that is neglecting the second lens so if you bring the second lens into the picture this ray of light will bend further and we'll draw that in a while but the important thing is that this point which was the image due to the first lens now becomes the object for the second lens and we've seen this before this is the general way in which we tackle problems when we have multiple reflections or refractions we always take the image in the first stage as the object so this point let me just write that down so this point over here that point is now going to be the object for our second lens so let's write that that's going to be the object for lens two for lens two and now where will this ray go we'll do that now let's only concentrate on the second lens and let's only concentrate on the incident rays on the second lens now if you look at the incident rays carefully you can see that the incident rays themselves are converging rays of light and our lens is going to converge it even further now if the incoming rays were parallel those rays of light would have been converged at this point since the incoming rays themselves are converging they will get converged even closer to over lens can you see that even close are somewhere even closer to a lens and so the final image is formed somewhere over here somewhere at this point and so if you now look at the complete picture what we are seeing is that the pal rays of light are eventually getting focused at this point which means if you think from that effective lens point of view then we can say that this itself is where the pal rays of light are being focused and so this must be the effective principal focus so call this as the effective principal focus and this would be the effective focal length and we need to figure out how much that is so again if you bring back the two lenses how do we figure out this length well if you look at the second lens again one more time just focus on that second lens we've already seen that this is the object for the second lens because that's where the incident rays of light on this lens appear to meet then after refraction the rays of light meet over here that means this is the image for our second lens so this would be image image for lens to lens to this part over here would be the image and we also know what's the focal length of the second lens f2 that means we know all three so we can connect them directly by using the lens formula so highly encourage you to pause the video over here and see if you can use the lens formula and connect them alright let's do this lens formula tells us one or the focal length is equal to one over the image distance V minus minus one or you so let's apply this for lens two so for lens to the focal length is f2 so 1 over f2 that's going to be equal to 1 over the image distance the image is this and so the image distance itself is the effective focal length so 1 over f effective minus 1 or the object distance and the object distance is this point which is the focal length of the first lens so that's F 1 1 over F 1 and we have connected them and so to figure out what this is all we have to do is add 1 or f1 on both sides so 1 or effective F effective is going to be since you're adding you on both sides you'll get 1 over F 1 1 or F 1 plus 1 over F 2 and there we have it that is the connection between the effective focal length and the individual focal lengths and here's the interesting thing instead of just two lenses what if you had a third lens over here with some say focal length F 3 then you can continue this now you can treat this as the object for the third lens and then the new image would be now the new effective focal length again you can use a lens formula and then you will find then you will find that the effective overall effective focal length could be 1 over F 1 times 1 over F 2 plus plus 1 over F 3 and I highly encourage you to try that exercise yourself and you would see that we can continue this if use for lenses we can just keep on doing 1 over F 4 for n lenses we can just keep on doing that but the important thing to remember is while using this formula the focal length of converging are positive but for a diverging lens the focal lens will become negative that's the only thing to take care of and of course in practice we cannot use this formula for n lenses because eventually when we put too many lenses together the effective lens becomes thick enough and so our approximations fail so of course theoretically you can use it for n lenses but can be only used as long as we are using the thin lens approximations