side, we have negative 1/4, but that's the same as negative 3/12. That's okay. It's gonna be upside down So it's gonna be like over here about at this point right around here somewhere, three centimeters behind this mirror. f = -14 cm . in front of the mirror. The ray diagram for the given specifications that will show the image location is shown in Figure 1. sometimes instead of d o, you'll see this as s We're gonna use the same equation. the object distance was 12. What is the focal length of the mirror? The one I'm gonna use is the one that I feel like most Example Problem #1 A 4.0-cm tall light bulb is placed a distance of 35.5 cm from a convex mirror having a focal length of -12.2 cm. So we get positive 3/4 of a centimeter. Sometimes people forget is equal to height of the image over the height of the object. is gonna be negative. It's the same thing. image of the object. sign's already here, so negative height of the object times this ratio of the image distance is that anything on this side of the mirror is gonna over the focal length. So one over the image distance has to be equal to negative a third. And, again, if we want to figure get that the image distance finally is gonna be up or upside down. centimeters from the mirror, somewhere around here. So negative 3/12 minus 1/12 Everything would be the same. So it turns out this ratio positive six centimeters. The sign convention we're using is that objects, images, and focal lengths in front of the mirror If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Place the mirror at the right side, place C at the center and F halfway between the mirror and C . And then we add to that Maybe it's an arrow or a crayon one over the image distance. We, again, use the mirror equation. But whatever your height of the object is, you multiply it by this ratio are over here, right? It says that the magnification is equal to the height of the image divided on that side, positive. of positive 12 down here, we'd plug in positive three. images are gonna be located. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. same as negative 1/3. So 1/4 you could rewrite as 3/12. So this is the mirror right here. on this side, it's positive. So recapping, you can There's a lot of Thus, the ray diagram is shown in Figure 1. We're gonna say that one So these numbers are gonna be different, but you would use this So finally, you take one over each side. So it's gonna be tiny. So what's the height of those incorrectly, you can get the wrong answer. To find: Position and height of using mirror equation. Sketch a vertical line at the mirror point to represent the mirror. again, was 12 centimeters. So 3/12 minus 1/12 is just gonna be 2/12, and that's gonna equal one And the 1/4 ratio means That's what our image is gonna look like. We're, of course, gonna get tall, so 1.5 centimeters. could reach out and grab it, but it's gonna be an optical illusion. And back here would be behind the mirror, and that would be negative. It passes through F , reflects the principal plane, and is reflected parallel to the principal axis. So that's an example this front side of the mirror. And, again, we have to So let's do a few mirror Here, hi is the height of the image and ho is the height of the object. always equal to the ratio of the height of the image And it's gonna be upside down because of this negative sign. over the object distance. Problem: a.) It's just a few fractions added together. where our image is gonna be. Now I have to warn you, everyone has their own sign convention. If we solve for the height of our image, we get that the height You solve for your image distance. gonna be right-side up. Donate or volunteer today! So we're gonna get 1/4 three. So there's a bunch of positive So let's say your eyes Here, do is the position of the object, di is the position of the image, and f is the focal length. out how tall it's gonna be, whether it's right-side up or upside down, we're gonna have to use The vertical scale of 1 block is 4.0 mm . This came out to be So the object distance, again, is on this side in front of the mirror. Provide the procedure to draw the ray diagram as follows: The horizontal scale of 1 block is 1.0 cm . There's gonna be no object there. image is equal to positive 1/4 times the height of the object. So where's our image gonna be? Determine the image distance and the image size. Notice I'm not converting. What would change, what if we did this? So, in other words, this So at this point right here is We're gonna use this mirror equation. Let's just solve for it. Well, again, the object is in front of the mirror 12 centimeters. You'd see an upside-down image right here. And you could use the The expression for the mirror equation is. right here, a blue crayon, you're holding in front comes out positive, you'll know that it's also on problems can be intimidating when you first deal with them. The expression for the magnification equation is. We want the image distance. The equation … Where is this image gonna be? equation problems, and you can see how the signs work. of our image gonna be? We don't want one over the image distance. is behind the mirror. image of this object here. of this mirror right here. conventions we just discussed and the signs I'm using in this formula, concave mirrors always have The image is located where rays 1 and 2 cross after reflection. So it's gonna be three You still plug in whatever That's equal to negative from the mirror. right-hand side's gonna be. Let's just solve this one. right-side up or upside down. And so if you solve this, you'll get one half of negative three, which is negative 1.5 centimeters. subtract one over 12, and that's all gonna have to be equal to one over the image distance. Wanted : The distance between the patient’s teeth and the mirror. of the image is gonna be, multiply both sides by h o, we'll have the negative to use a different equation, and that equation's called it's gonna be 1/4 as large, the image, I should say, With the convention that I'm using, this focal length is behind the mirror. This mirror, this time instead of concave, this is a convex mirror. And if you make even one negative three centimeters. To draw:The ray diagram for the given specifications that will show the image location. As a demonstration of the effectiveness of the Mirror equation and Magnification equation, consider the following example problem and its solution.

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