Converging Lenses - Object-Image Relations
In this article, I explain the correlation between the magnification factor and the focal lenght / focal distance of Object outside of the focal distance – real image. Treating the camera lens as a simple thin lens (this is an approximation), then the image distance, v and the object distance, u, are related by. Perhaps you noticed that there is a definite relationship between the image the magnification is the ratio of the height of the object to the height of the image. As the object distance approaches one focal length, the image distance and.
And so this gives us a sense of what the image will look like. 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 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.
Object image and focal distance relationship (proof of formula) (video) | Khan Academy
These are similar triangles. Similar-- I don't have to write triangles. 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 d 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.
Object image and focal distance relationship (proof of formula)
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. The object is located between 2F and F When the object is located in front of the 2F point, the image will be located beyond the 2F point on the other side of the lens.
Regardless of exactly where the object is located between 2F and F, the image will be located in the specified region. The image dimensions are larger than the object dimensions. A six-foot tall person would have an image that is larger than six feet tall. The absolute value of the magnification is greater than 1. The object is located at F When the object is located at the focal point, no image is formed. As discussed earlier in Lesson 5the refracted rays neither converge nor diverge.
After refracting, the light rays are traveling parallel to each other and cannot produce an image. The object is located in front of F When the object is located at a location in front of the focal point, the image will always be located somewhere on the same side of the lens as the object. Regardless of exactly where in front of F the object is located, the image will always be located on the object's side of the lens and somewhere further from the lens.
The image is located behind the object. In this case, the image will be an upright image. That is to say, if the object is right side up, then the image will also be right side up.
In this case, the image is enlarged; in other words, the image dimensions are greater than the object dimensions. The magnification is greater than 1.
Finally, the image is a virtual image. Light rays diverge upon refraction; for this reason, the image location can only be found by extending the refracted rays backwards on the object's side the lens. The point of their intersection is the virtual image location.
It would appear to any observer as though light from the object were diverging from this location. Any attempt to project such an image upon a sheet of paper would fail since light does not actually pass through the image location. It might be noted from the above descriptions that there is a relationship between the object distance and object size and the image distance and image size.
Starting from a large value, as the object distance decreases i.
At the 2F point, the object distance equals the image distance and the object height equals the image height. As the object distance approaches one focal length, the image distance and image height approaches infinity.
Finally, when the object distance is equal to exactly one focal length, there is no image. Then altering the object distance to values less than one focal length produces images that are upright, virtual and located on the same side of the lens as the object. Finally, if the object distance approaches 0, the image distance approaches 0 and the image height ultimately becomes equal to the object height.