Now, forget the diagram. Pretty simple. We’re going to understand what this mirror formula is. 3) Axis Magnification: When object lies along the principle axis then Take some of the triangles, take the ratio of the sides and what happens? And if you have already been accustomed to what’s called the Cartesian coordinates, then we know something. focal length of the mirror and this formula is valid in all situations for all So with that in mind the other thing we want to say is all distances are measured from the origin or the pole. Just calculating out of pure math. Mirror Wouldn’t it be nice if we were able to do that? Those two quantities are going to be equal. Right? So now what we are going to do is just define some distances. Let’s take one particular specific case and let you generalize to the other cases. So we all love formulas don’t we? So all these distances are what we are going to try and connect because that’s what we mean by finding out where the image is. So now with this knowledge in our hand, we can jump into and try to find out where image is formed even without any pencil, paper, drawing, nothing. 4) Areal Magnification: If 2D – object is placed with its plane We’re looking for something that relates to those quantities. A mirror formula may be defined as the formula which gives the relationship between the distance of image v, distance of object u, and the focal length of a mirror. To know more, visit Byju's.com The object is beyond 2f. See most of us sit together and decide okay this is how we’ll do it. The sign convention for spherical mirrors follows a set of rules known as the “New Cartesian Sign Convention”, as mentioned below: a. \right)}^{2}}={{\left( \frac{f}{f-u} \right)}^{2}}={{\left( \frac{f-v}{f} So now let us see what we can do? We’ve a very simple looking expression right now and divide the whole thing by a product. spherical mirror gives the relative extent to which the image of an object is So now that we have the sign convention in place what do we do to use it? So let’s see what happens now. In fact most of the countries in the world do it in the right side. It would be pretty pissing off if we all wrote A’s in our own ways and then we had to translate stuff. So what do we need though? And why did we need to do that? Anything you want to know, why do I need it? Magnification may also be determined by taking the ratio of the distance from the image to the mirror and the distance from the object to the mirror: magnification, m = image distance = di object distance d0 Either of the above formula may be used to determine magnification. It could either be this way or it could be that way. spherical mirrors for all positions of the object. You know that this will connect, image distance u, object distance v, and focal length f. Which means that if you know any two of these, you can get the third one. It’s going to be pretty irritating if you think about it. From the origin, which is the pole of our mirror, everything else becomes very simple and it’s just this. Just like that. But what if you’re even more lazy? Now, at some length and the image might be at some other distance. So where is the image forming? The principal axis is taken as the x-axis of our coordinate system. So what we do? Right, becomes a very simple thing. We’ll drive on the left side of the road. There is a formula for the magnification but, unfortunately, it is not always applied correctly. The distance from the object to the mirror is called u. We’ll call it u, the object distance. So we found one particular set of similar triangles. Yeah. Divide the whole thing, the entire equation by uvf. Anything right is positive, anything left is negative. Given where the object forms we should be able to tell where the image will form. The pole (p) of the mirror is taken as the origin. One term cancels. And the other ray that’s going through the center of curvature is going to come back that way. Yeah, those are the two. 3) For most users a 3X or a 5X mirror works very well. Or let us see what happens and where the image forms by our known method. its axial magnification is: $$m=\frac{I}{O}=\frac{-\left( {{v}_{2}}-{{v}_{1}} And you’ll learn this. It’s the place from which we start all out measurements. Right? Two angles same for two triangles, what does that mean about those two triangles? It may be written as, where, v = Distance of image from pole of mirror u = Distance of object from pole of mirror and, f = Focal length of the mirror (1/v) + (1/u) = (1/f). Two triangles that seem to be similar. So let’s take, now you can watch there and you can see, I want you to notice that there are two triangles there that have a very special property. c. The object is always placed on the left side of the mirror which implies that light falling from the object on the mirror is on the left-hand side. If you know where the image is it will tell you where the object is. But the first thing we need before we talk any math is to define something called a convention. That’s the first question to ask right? All of us get together and say, okay fine. It just makes it look more beautiful. Because they have an opposite angle and they have right angles. For make‐up and shaving mirrors the most common ratings are 3X, 5X, 7X and 10X. We know u, we know v, we know f. How can we relate them? No matter what it is, right is positive and up is positive. Because before this there was a pretty difficult convention where you measured this way in a particular manner, that way, no. Whatever we learn in math is going to get connected to physics and vice versa, it keeps happening that way. Example Problem #1 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. That’s how it will be stored in your mind. So right and up – positive. What do we do? So once we’re done with these 3 we’ll see how they are related for mirrors. And what is the origin? So I am going to do something pretty sneaky here. Anything down is negative, and anything up is positive. Why do we need it? Just giving them some names. What did we see till now? formula gives the relationship between image distance, object distance and the Aah there. Yes, that’s right. Why are we doing this by the way? The image distance, the object distance and the focal length. A plain flat mirror would be rated at 1X and one that makes an object 3 times larger would be rated at 3X. The distance from the image to the mirror is called v. Yeah. We will all write A this way and not in some other way. No matter how you want to rearrange it make it look that way. Mirror formula gives the relationship between image distance, object distance and the focal length of the mirror and this formula is valid in all situations for all spherical mirrors for all positions of the object. Second Formula for Magnification There is another formula of magnification Note : - If magnification (m) is positive , It means image formed is virtual and erect If magnification (m) is negative, It means image formed is real and inverted Questions Example 10.1 - A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. And now those two being equal, just cross multiply them and see what happens. We don’t want to use even pencil, scale, nothing. perpendicular to principle axis and then its areal magnification is: \({{m}_{s}}=\frac{Area\,of\,image\left( {{A}_{i}} What struck us? Because we know we have to connect some quantities. We’re going to drive something that’s very favorite you know, it’s pretty weird to most of the people who’ll ask you a few questions. Right? height of the image to the height of the object. Right? What’s a convention? \(m=\frac{I}{O}=-\frac{v}{u}=\frac{f}{f-u}=\frac{f-v}{f}$$. It’s very simple right now. 1) Mirror Formula: $$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$$. And we saw that in different cases the image forms in different places. T he linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance, with a minus sign Practice Questions Find the size, nature and position of image formed when an object of size 1 cm is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. Two pairs of similar triangles equate. Left’s negative and down is negative. There’s one more there. d. All the distances parallel to the principal axis are measured from the pole (p) of the mirror. Makes it so that our old method is too long. All their angles are equal, but also all their sides are proportional. All we can do is just take 2 rays, extend them, reflect them back and find out where the object image forms, from where the object forms. Now look at that. Of course you’re going to use math and physics right? Range of Expressions using Trigonometric Substitution, AYUSH Counselling Schedule for NEET AIQ GOVT./ GOVT. e. All the distances … So that for you is the mirror formula. Left and down – negative. 1 by v, plus 1 by u equals 1 by f. Right? Right? 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