# Relationship between object distance and image virtual image is observed. The relationship between the image distances v and u is as shown in Table 1 and Fig. 5. According to Table 1 and Fig. 5, when the. Click here to get an answer to your question ✍ State the relationship between object distance, image distance and radius of curvature of a. Equation 1 defines the relationship between the object distance (o), the focal length (f), and the image distance (i). The image distance is simply.

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.

LENS FORMULA AND MAGNIFICATION

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.

### Object image and focal distance relationship (proof of formula) (video) | Khan Academy

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.

## Converging Lenses - Object-Image Relations

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.

### Converging Lenses - Object-Image Relations

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. A six-foot tall person would have an image that is six feet tall; the absolute value of the magnification is exactly 1. As such, the image of the object could be projected upon a sheet of paper.

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. 