Waves and optics
Object image and focal distance relationship (proof of formula) Object Image and Focal Distance Relationship (Proof of Formula)
Object image and focal distance relationship (proof of formula)
- We have been doing a bunch of these videos with these convex lenses,
- where we where we draw parallel rays and rays that go through the focal point,
- to figure out what the image of an object might be.
- What I want to do in this video is to 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 lets see if we can do this.
- 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 here are the focal points.
- They are the focal length away.
- And I did what we always do:
- I drew one parallel ray from the tip of that arrow to the actual convex lens
- and it gets refracted, so it goes through the focal point on the right hand side.
- It goes all the way over there.
- Then I drew a ray that goes through the focal point on the left hand side
- and when it gets refracted it becomes parallel.
- Then it actually intersects with that previous ray, up, right over here,
- and so this gives us a sense of what the image will look like:
- It is inverted, it is real and in this case it is larger than the actual object.
- W\hat I want to do is to come up with a relationship with these values.
- Lets see if we can label them here,
- and then just do a little geometry and a little algebra
- to figure out if there is an algebraic relationship right here.
- So the first number, the distance of the object. That is this distance from here to here.
- Or we can just label it here, since it 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 is 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.
- Then we have the focal distance; the focal length, and that is just this distance right here.
- Or we can 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 am going to draw some triangles here
- so what we can do is; and the whole strategy, I am going to keep looking for similar triangles
- and then trying to find a relationship or ratios that relate these three things to each other.
- So let me find some similar triangles.
- The best thing I can think of to do is to 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.
- If i were to draw the same triangle it would look like this.
- Let me just be clear, this is this triangle right here, I just flipped it over.
- and so if we want to make sure we are keeping track of the same sides
- if this length right here is d sub zero [ d0 ] or d nought or sometimes we can call it d zero, whatever we want to call it
- Then this length up here is also going to be d zero [ d0 ].
- The reason why I wanted to do that is 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 are going to be similar.
- And then we can relate, we can get some ratios of sides
- and then what we can do is show that this triangle over here is similar to this triangle over here
- get a couple more ratios and then we might be able to relate all of these things.
- The first thing we have to prove to ourselves is that the triangles really are similar.
- So the first thing to realise is that this angle right here is definitly the same thing as that angle right over there
- something called opposite angles or vertical angles
- they are on opppsite sides of lines that are intersecting, so they are going to be equal.
- Now the next thing, this comes out of the fact that both of these lines
- this line is parallel to that line over there
- and you can look at, I guess you can call it alternate interior angles
- if you look at the angles again with the parallel lines, with the transversal of parallel lines from geometry.
- We know that this angle, since they are alternate interior angles
- this angle is going to be the same value as this angle, right.
- Cause they are both, you can view this line 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 that this triangle up here has the same
- three angles as this triangle down here.
- So these two triangles are similar, these are both
- and this is really more of a review of geometry than optics
- these are similar triangles.
- They are similar.
- Because they are similar the ratios of corresponding sides are going to be the same
- so d0 corresponds to this,
- they are both opposite that pink angle.
- So the ratio of d0 to d1 is, let me write this over here, so the ratio of d0 to d1 ...
- This is the ratio of corresponding sides, is going to be the same thing.
- Let me make some labels here. That's going to be the same thing
- As the ratio of this side right over here, we call that A
- it is opposite this magenta angle right over here
- that is 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 because side B is opposite the magenta angle on this bottom triangle.
- So that is how we know this side, its corresponding side of the other similar triangle is that one
- they are both opposite the magenta angles.
- So we know that d0 to d1 is as A is to B
- So that is interesting.
- We have been able to relate these two things to these kind of two arbitrary lengths.
- We somehow have to connect these two things to the focal length.
- 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 lets look at this triangle over here.
- Let me put it in a better colour.
- So lets look at this triangle over here that I am highlighting in green.
- This triangle in green.
- and lets look at that in comparison to this triangle that I am also highlighting in green.
- Now the first thing that I am going 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,
- or there is a couple of arguments we can make,
- one you can see that this is a right angle over here, this is a right angle,
- if two angles of two triangles are the same then the third angle also has to be the same.
- So we can also say that this thing, let me do this in another colour, I don't want to be repetitive too much with the colours,
- we can say that this thing is going to be the same thing as this thing.
- Or another way we could have said it is that this line over here,
- which is kind of represented by the lens
- or the lens or the line that is parallel to the lens,
- is parallel to the object over there,
- and you could make the same alternate interior argument there.
- But the other thing is just look, I've 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 are both the hypotenuse of these similar triangles
- So A is to B as, we can say, this base length right here
- it got overwritten a little bit, but this base length right over here is f
- That is our focal legnth.
- As f in this triangle is related to this length in this triangle
- they are both opposite that white angle
- so as f is to this length right over here
- so what is this length
- This whole distance is di, but this length is the whole distance minus the focal length
- this is di minus the focal length
- so A is to B as f is to di minus the focal length
- And again we have a relationship between the distance
- of the object, the distance of the image and the focal length
- And now all we have to do is a little bit of algebra.
- This is equal to this, and this is equal to that
- so this blue thing has to be equal to thist magenta thing
- and now we just have to do some algebra
- So lets do that
- So we got d_0 to d_i is equal to the ratio of the focal length to the difference of the image distance
- minus the focal length.
- And here we just have to do some algebra, lets just simplify this, let's cross multiply it
- lets multiple d_0 times this thing over here
- we get d0<i>di - d0</i>f [we are just distributing it]
- Just cross multiplying, which really is just the same thing as multiplying both sides by both denominators.
- That is going to be equal to di*f.
- and now we can add this term over here to both sides of this equation.
- I am just going to switch to a neutral colour.
- So we get d0<i>di = di</i>f + d0*f.
- So lets see here we could factor out the f, our focal length
- so we get d0*di is equal to f(di+d0)
- and then what can we do, we could divide both sides by f
- So this will become over f
- essentially cancelling it out, and then
- Lets me just re-write what we have here.
- d0*di / f = di + d0
- now lets divide both sides by d0*di
- Cancels out over here,
- So we are left with on the left hand side
- one 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 numerators, is the same thing as
- = di/( d0 x di) + d0/( d0 x di)
- but the di's cancel out, and we just have a one, here the d0s cancel out and you just have a one
- So this is equal to one over the distance of the object
- and this is plus one over the distance of the image.
- So from the get-go this was a completely valid formula,
- we actually had acheved what we wanted.
- But this is a neater formula, you dont have the di's or d0's 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.
- Any way, that is pretty neet how itt came out to be a clean formula.
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