Class 10 Physics (India)
Thin lens formula
Let's derive the thin lens formula which connects the object distance to the image distance. Created by Mahesh Shenoy.
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- how at1:30can we say that two triangles with three equal angles are congruent to each other? as far as i know there is no congruency property for all three angles.(1 vote)
- He says they become similar. Similar triangles have equal corresponding angles and all the corresponding sides are in the same ratio.
Hope this helped!(5 votes)
- @9:09Do the negatives cancel out? I thought you multiply both sides with (-1), so you'd also multiply the big M with (-1), getting a negative magnification. I thought it was kinda neat you could see from the sign in the equation that the image is flipped.(2 votes)
I'm not quite sure if this is the answer to your question.(I have taken your question to be why isn't M multiplied with -1)
The negative of the general formula is not the case here as,
M = v/u = Hi/Ho
F = m*a(force = mass * acceleration)
we won't multiply M with -1 as M is just signifying the value if you know what I mean.
Hope this helps!(2 votes)
- If there would have been a concave lens instead of the convex lens during derivation, the last part (where he uses sign convention on 1/f= 1/v + 1/u) the formula would have come out different (Since the object distance and the image distance both would have been negative in that scenario, the formula would have been 1/f = -1/v - 1/u). So, does the lens formula not work for concave lenses?(2 votes)
- The focal length of a concave mirror is also negative
That would give us -1/f = -1/v -1/u which is the same as 1/f = 1/v + 1/u.
So,the thin lens formula works for both convex and concave lens.(1 vote)
- 8:31in why cant we directly write it as 1/f = 1/v + 1/u as a similar thing of taking "u" as negative can happen in convex mirrors, yet we don't put the negative sign in the formulae of mirrors, so why here in lenses?(1 vote)
- I have a question from the Thin Lenses quiz that is after 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).
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?(1 vote)
- In this video, we take the focal length to be on the right side of the lense, but we could've taken the focal length on the left side as there are two foci. When it comes to derivations and formulas, do we take the focus to be in the direction of the rays?(1 vote)
- 7:42so .. wouldn't the "+" side always be opposite to whichever side the object is placed(1 vote)
- Is there a better way to explain where the minus comes from at7:40ish? The sign convention argument comes across like a last minute rabbit in the hat to make it work.(0 votes)
- No, it should actually be the first-minute rabbit in the hat. When we're dealing with ray optics, our first priority should be always, always, sign conventions. It's a must. So, the minus doesn't come there out of thin air, it's already there according to the sign conventions. Now, why he chose to talk about signs at the last minute, I don't know. It should have been the first thing to talk about.
Hope that helps. :)(1 vote)
say we have a thin convex lenss of focal length F and an object of some height H naught is kept at some distance u from the lens and say we know these values we know the focal length U and that's not the question is can we figure out where the image is going to be exactly and what will be the height of the image so we can draw a ray diagram let's draw a couple of rays and where there is intersecting image is formed and so the goal is to figure out what is the height of that image and how far is that image formed from that lens so these are two things we need to figure out and the trick over here is going to be very similar to what we had done and it comes to mirrors so the idea is we have to find similar triangles which have these as their side lengths and then we use the properties of so similar triangles to make a relationship between them so pause the video for a while and see if you can find some some pairs of similar triangles in this figure alright if you look carefully then you will see these two are similar triangles let's see how if you look at this angle over here that must be the same as this angle they were vertically opposite to each other then they have this angle equal to this angle because they are both right angle triangles and therefore the other angle must also be the same making them similar triangles and also notice the side lengths are exactly what we want and so now we can say that they're rich their sides are having the same ratios and therefore if you take this height and divide by this height so if you take H I and divided by H oh that should be the same as the ratio of any other side and so we can take this now that should be the same as V divided by u so that's equal to V V divided by U so we have found one equation over here and this one equation contains both the unknowns we don't know H I we don't know V so we can't use this just we haven't found our answer yet so we need another equation so we have to find another set of similar triangles and if you couldn't find them before then again I encourage you to pause the video and see if now you can find another set of similar triangles this time try to find one which has F in it as a side length again if you look carefully you'll find these are similar triangles again one angle is 90 degrees they're both equal to each other then you have these vertically opposite angles which are also equal to each other which means the third angle must also be the same and as a result these two triangles are similar so we can write a similar equation for these two triangles as well their height and their side lengths are also in the same ratio again haha the video and try to write a similar equation for these triangles all right let's do that again if you take Hetch I divided by this height which is the same as H oh so if you take H I divided by H oh it has to be the same as this length well what is this length equal to can you see that this total length is V and this is f so this is V minus F V minus F divided by this length which is the same as F and that is equation number two now notice that we have two equations with just two unknowns and so we can eliminate them and you can figure out what is the height of the image and what is the image distance so the geometry is done and now all we have to do is algebra to eliminate the variables and if you look carefully you will see that they have the same mo left-hand sides which means we can directly equate their right-hand sides again one last time try and pause the video and see by creating them the right-hand sides if you can come up with an equation very similar to what we got for mirrors all right let's do this let's make some space we don't need a diagram anymore it's just solving the two equations okay so the right-hand side we can equate them so V divided by U has to be equal to V minus F divided by F and all you have to do is make this equation a little bit pretty so that we can remember this okay so we'll get rid of the denominators we'll multiply the whole equation with you and then with F so we get we get V times F the U cancels out and over here the F cancels out so we get V - minus F times u so just cross multiply it and for the simplification so let's just simplify V times F equals V times u we'll just multiply this minus F times u now to make this equation look pretty just divide the whole equation divide the whole equation with UV and F so let's see what we end up with on the left hand side we V and F cancels out so you end up with 1 over u let's write that down over here so 1 over u equals here v + u cancels out so we are left with 1 over f - over here F and you cancel out which means we are left with 1 over V and if we add 1 or V on both sides to make sure that this looks similar to the mirror equation we can now write 1 over F equals 1 over V plus 1 over u and there we have it that is the equation that connects the focal length the object distance and the image distance okay and similarly once we know V we can now substitute this in any of the two equations we can substitute over here and we can now find what the height of the image is so just like we read it for the mirrors we can define this thing called as the magnification magnification M and that is defined as the height of the image divided by the height of the object that's the definition of the magnification height of the object it tells us how much is the image magnified compared to the object and that turns out to be as you can see just V / u v / u that's the second equation so let's just bring these two equations to the top since we derived this equation for this specific case where we had a convex lenss and red we got a real image these equations may not be general equations they may not work for other cases where we might get say virtual images already we have a concave lens so what to do well we have tackled this situation before the answer is sign conventions we've seen before that if we treat these distances as positions on a graph and if you write these equations with signs then they end up becoming a general formula and that's exactly what we're gonna do over here as well we're gonna use the same Cartesian sign conventions and we're gonna put signs over here and we're gonna generalize them all right just like before how pole of the mirror was chosen as the origin now we're gonna choose the optic center as the origin and then we're gonna choose the incident direction as positive over here in this diagram the right side is the incident direction so that becomes positive direction which means all the positions to the right of the origin are positive positions and all the positions to the left of the origin are negative positions so we have to put these signs signs over here all right so if you look at this equation let's start with the focal length focal length is the position of the principal focus and that's over here and that's positive so this stays positive the image distance is the position of the image which is also on the positive side so that also stays positive and object distance is the position of the object that's on the negative side so that becomes negative and so positive times negative well that becomes negative so our final equation general equation is this now this we can double box it because this now has automatically become the general equation and that's the power of sign conventions now this can be used for any case we want all right let's generalize this now again for sign conventions for the height we're going to choose above the principal axis positive height and below the principal axis is negative fide so let's see the height of the image is negative 5 because that's below the principal axis so there's a negative sign over here the height of the object is above the principal axis that's it's positive V is the image distance so that's the position of the image that says positive U is the position of the object that stays negative now remember the magnification M is defined as Hetch I divided by H oh not negative a child by hetro so this is not our final equation we need to get rid of that negative sign from the height so if you look carefully the two negatives cancel out which means this itself is now the general equation because this now we have used sign conventions so this is general equation so these are the lens equations which are very similar to the mirror equations the only difference between them is the sine of U if you flip this line if you make the u- you end up with mirror equations just check that