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- Object image and focal distance relationship (proof of formula)
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Object image and focal distance relationship (proof of formula)
Object Image and Focal Distance Relationship (Proof of Formula). Created by Sal Khan.
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- I have learnt the formula of focal length (1/f = 1/di-1/do).Is it wrong?(8 votes)
- no the formula 1/f=1/di- 1/do is correct, because it is based on the sign covention for lenses where object distance (do) is always taken as negative for all real objects, but sal's formula is also not wrong because he did not stick to the sign convention ( he took the object distance as positive)
for eg -
for a covex lens, if f=10cm do=15 di=X
according to ur formula (1/f=1/di-1/do) which is based on sign convention
1/f =1/di -1/do
f is positive(10)
do is negative (-15)
so , 1/10 =1/di - (-1/15)
1/10= 1/di + 1/15
1/di= 1/10-1/15 =1/30
take the reciprocal of i/di
di=30 cm (it is positive)
now we take salman's formula
1/f= 1/di +1/do (remember we are not taking sign conventions we are simply putting the values)
1/10= 1/di +1/15 (not applying sign convention)
1/di=1/10 -1/15 =1/30
we take the reciprocal of 1/di
and di = 30 cm
thus both the formulas are correct ! :)(24 votes)
- Why does he keep saying D-NOT while writing Do when the subtitles say D0?(2 votes)
- He's saying d-naught. Naught is another term for nothing or zero "0".(16 votes)
- Is this formula applicable to convex mirrors? If it is then can you do a video on its proof(4 votes)
- No, the lens was a convex lens but for this to be same we need to have a concave mirror(11 votes)
- How can I solve this formula? Equation for focal point: formula: 1/u +1/v= 1/f(2 votes)
- If you mean solving for a particular variable, here you go:
1/u + 1/v = 1/f
f/u + f/v = 1
f + fu/v = u
fv + fu = uv
Now that there aren't any confusing denominators, we can solve for any variable:
fv + fu = uv
f(v + u) = uv
f = uv / (v + u)
fu + fv = uv
fv = uv - fu
fv = u (v - f)
u = fv / (v - f)
Hope this helps!(7 votes)
- What's a "Focal Plane"?(4 votes)
- A plane passing through the principal focus and at right angles to the principal axis of a spherical mirror is called the focal plane.(4 votes)
- Does this equation work with convex and concave mirrors as well?(2 votes)
- There is a difference between solving a convex lens and a concave lens(2 votes)
- if the object is between the lens and second focal length...how would the image be??(4 votes)
- In this case, an erect virtual image of the object would be formed to the right of the object. The image would also be bigger than the object.
However, this isn't the conventional way of imagining the way you would place the lens and the object. After placing the object between the lens and the second focal length, you could view it from the other side, as if you turned the table around. Now the object seems as if it's placed between the first focus and the lens. Then you can proceed with doing it like it's always done!
The first and second focus are actually equal. You can obtain their value using the Lens Maker's Formula (I hope I'm not wrong. If I am, someone please correct me).(2 votes)
- FOCAL LENGTH...NEVER CONSTANT!
According to our science textbook focal length(f) of a mirror/ lens is equal to half its radius of curvature(R).Which means that there is a specific focal length for a given lens/mirror(f=R/2).
But in science lab when we perform the experiment to find the focal length of a lens/mirror we get different focal lengths for the same mirror/lens for different objects(that we observe)...this means that the focal length of a mirror/lens is never constant...so the formula f=R/2 is contradicted!
In fact...what i think is that the ray of light parallel to principle axis never pass through the focus[what our science textbook states (f=R/2)] after reflection/refraction!
is it correct?(3 votes)
- 1) real lenses do not exactly follow the formula that we devise for an ideal lens. Real lenses never produce perfectly focused images
2) possibly, you are seeing some diffraction.
3) if you are not using monochromatic light then the focal length is not exactly the same for all wavelengths(3 votes)
- hey,is concave lens can be a treatment in myopia?(2 votes)
- Of course! As the concave lens diverge light rays, the diverged rays fall on the eye retina having faulty lens. Those lens then converge the rays to
- in lens distance of object is taken negative or positive(2 votes)
- Good ? Either form can be used with positive or negative lenses and predicts the formation of both real and virtual images....(2 votes)
We've been doing a bunch of these videos with these convex lenses where we drew parallel rays and rays that go through the focal point to figure out what the image of an object might be. But what I want to do in this video is actually come up with an algebraic relationship between the distance of the object from the convex lens, the distance of the image from the convex lens usually on the other side and the focal length. So let's see if we can do this. And just to save you the pain of having to watch me draw straight lines, I drew this ahead of time. And so we can imagine this green thing right here is the object. This is the object, and these two little pink points right here are the focal points. They're a focal length away. And I did what we always drew. I drew one parallel ray from the tip of that arrow to the actual convex lens, and then it gets refracted, so it goes through the focal point on the right-hand side and goes all the way over there. And then I drew a ray that goes through the focal point on the left-hand side. And then when it gets refracted, it becomes parallel. And then it actually intersects with that previous ray right over here. And so this gives us a sense of what the image will look like. It's inverted. It's real. In this case, it is larger than the actual object. What I want to do is come up with a relationship with these values. So let's see if we can label them here. And then, just do a little bit of geometry and a little bit of algebra to figure out if there is an algebraic relationship right here. So the first number, the distance of the object-- that's this distance from here to here, or we could just label it here. Since this is already drawn for us, this is the distance of the object. This is the way we drew it. This was the parallel light ray. But before it got refracted, it traveled the distance from the object to the actual lens. Now, the distance from the image to the lens, that's this right over here. This is how far this parallel light ray had to travel. So this is the distance from the image to the lens. And then we have the focal distance, the focal length. And that's just this distance right here. This right here is our focal length. Or, we could view it on this side as well. This right here is also our focal length. So I want to come up with some relationship. And to do that, I'm going to draw some triangles here. So what we can do is-- and the whole strategy-- I'm going to keep looking for similar triangles, and then try to see if I can find relationship, or ratios, that relate these three things to each other. So let me find some similar triangles. So the best thing I could think of to do is let me redraw this triangle over here. Let me just flip it over. Let me just draw the same triangle on the right-hand side of this diagram. So this. So if I were to draw the same triangle, it would look like this. And let me just be clear, this is this triangle right over here. I just flipped it over. And so if we want to make sure we're keeping track of the same sides, if this length right here is d sub 0, or d naught sometimes we could call it, or d0, whatever you want to call it, then this length up here is also going to be d0. And the reason why I want to do that is because now we can do something interesting. We can relate this triangle up here to this triangle down here. And actually, we can see that they're going to be similar. And then we can get some ratios of sides. And then what we're going to do is try to show that this triangle over here is similar to this triangle over here, get a couple of more ratios. And then we might be able to relate all of these things. So the first thing we have to prove to ourselves is that those triangles really are similar. So the first thing to realize, this angle right here is definitely the same thing as that angle right over there. They're sometimes called opposite angles or vertical angles. They're on the opposite side of lines that are intersecting. So they're going to be equal. Now, the next thing-- and this comes out of the fact that both of these lines-- this line is parallel to that line right over there. And I guess you could call it alternate interior angles, if you look at the angles game, or the parallel lines or the transversal of parallel lines from geometry. We know that this angle, since they're alternate interior angles, this angle is going to be the same value as this angle. You could view this line right here as a transversal of two parallel lines. These are alternate interior angles, so they will be the same. Now, we can make that exact same argument for this angle and this angle. And so what we see is this triangle up here has the same three angles as this triangle down here. So these two triangles are similar. These are both-- Is really more of a review of geometry than optics. These are similar triangles. Similar-- I don't have to write triangles. They're similar. And because they're similar, the ratios of corresponding sides are going to be the same. So d0 corresponds to this. They're both opposite this pink angle. They're both opposite that pink angle. So the ratio of d0 to d1-- let me write this over here. So the ratio of d0. Let me write this a little bit neater. The ratio of d0 to d1. So this is the ratio of corresponding sides-- is going to be the same thing. And let me make some labels here. That's going to be the same thing as the ratio of this side right over here. This side right over here, I'll call that A. It's opposite this magenta angle right over here. That's going to be the same thing as the ratio of that side to this side over here, to side B. And once again, we can keep track of it because side B is opposite the magenta angle on this bottom triangle. So that's how we know that this side, it's corresponding side in the other similar triangle is that one. They're both opposite the magenta angles. So we know d0 is to d1 as A is to B. As A is to B. So that's interesting. We've been able to relate these two things to these kind of two arbitrarily lengths. But we need to somehow connect those to the focal length. And to connect them to a focal length, what we might want to do is relate A and B. A sits on the same triangle as the focal length right over here. So let's look at this triangle right over here. Let me put in a better color. So let's look at this triangle right over here that I'm highlighting in green. This triangle in green. And let's look at that in comparison to this triangle that I'm also highlighting. This triangle that I'm also highlighting in green. Now, the first thing I want to show you is that these are also similar triangles. This angle right over here and this angle are going to be the same. They are opposite angles of intersecting lines. And then, we can make a similar argument-- alternate interior angles. Well, there's a couple arguments we could make. One, you can see that this is a right angle right over here. This is a right angle. If two angles of two triangles are the same, the third angle also has to be the same. So we could also say that this thing-- let me do this in another color because I don't want to be repetitive too much with the colors. We can say that this thing is going to be the same thing as this thing. Or another way you could have said it, is you could have said, well, this line over here, which is kind of represented by the lens, or the lens-- the line that is parallel to the lens or right along the lens is parallel to kind of the object right over there. And then you could make the same alternate interior argument there. But the other thing is just, look. I have two triangles. Two of the angles in those two triangles are the same, so the third angle has to be the same. Now, since all three angles are the same, these are also both similar triangles. So we can do a similar thing. We can say A is to B. Remember, both A and B are opposite the 90-degree side. They're both the hypotenuse of the similar triangle. So A is to B as-- we could say this base length right here. And it got overwritten a little bit. But this base length right over here is f. That's our focal length. As f in this triangle is related to this length on this triangle. They are both opposite that white angle. So as f is to this length right over here. Now, what is this length? So this whole distance is di, all the way over here. But this length is that whole distance minus the focal length. So this is di minus the focal length. So A is to B as f is to di minus the focal length. And there you have it, we have a relationship between the distance of the object, the distance of the image, and the focal length. And now we just have to do a little bit of algebra. If this is equal to this and this is equal to that, then this blue thing has to be equal to this magenta thing. And now we just have to do some algebra. So let's do that. So we got d naught, or d0 I guess we'd call it, to di is equal to the ratio of the focal length to the difference of the image distance to the-- the focal length to the image distance minus the focal length. And now here, we just have to do some algebra. So let's-- just to simplify this, let's cross multiply it. So if we multiply d0 times this thing over here, we get d0 di. I'm really just distributing it, minus d0 f. I'm just distributing this d0, just cross multiplying, which is really just the same thing as multiplying both sides by both denominators, or multiplying the denominators twice. Either way, that's all cross multiplying is. That is going to be equal to di times f. And now, we can add this term right over here to both sides of this equation. I'm just going to switch to a neutral color. So we get d0 di is equal to-- I'm just adding this to both sides. When you add it to the left side, it obviously cancels out. Is equal to di f, this thing over here, plus d0 f. And then, let's see. We could factor out an f, a focal length. So we get d0 di is equal to f times di plus d0. And then, what can we do? We could divide both sides by f. So this will become over f. This becomes over f. Essentially, canceling it out. And then, I just don't want to skip too many steps, so let me just rewrite what we have here. So these cancel out. So we have d0 di over f is equal to di plus d0. And now, let's divide both sides by d0 di. So 1 over d0 di. Divide this side by d0 di. It cancels out over here. And so we are left with, on the left-hand side, 1 over the focal length is equal to this thing over here. And we can separate this thing out. This thing over here, this is the same thing-- we just separate out the numerator. Is the same thing as di over d0 di plus d0 over d0 di. But di over d0 di, the di's cancel out. We just have a 1. Here, the d0's cancel out. You just have a 1. So this is equal to 1 over the distance of the object. And this is plus 1 over the distance the image. So right from the get-go, this was a completely valid formula. We actually had achieved what we wanted. But this is a neater formula. You don't have the di's repeat and the f repeated. Right here, we have an algebraic relationship for a convex mirror that relates the focal length to the distance of the object and the distance of the image. Anyway, I think that's pretty neat how it came out to be at least a pretty clean formula.