In this video David solves a few exmaple problems involving concave and convex mirrors using the mirror equation and magnification equation.
Want to join the conversation?
- Now I'm confused... Sal said the real image can be seen only if you put a screen there... In this video, they say the real image, in front of the concave, can be seen anyway, and it feels like you can reach out and touch it.... Which one is it?(6 votes)
- Sal didn't mean so.. In a real image, the light rays come to a point from a point of object and this happens to all other points of that object..So, if we put a screen at that distance, we can see a clear image and as the image is really created for every point it is called real image .
This image will still create if you don't put a screen there and if you go to mirror at that distance, you'll be able to see an image of that thing.. So you'll feel that there was something before you and you could touch it....
I hope you understand the matter..... :)(13 votes)
- when I did the problem where the object was 3cm from the mirror I got symmetrical results di=-12 but the object height was 12cm I think that is because the object height was quadrupled, Is this right and what would you see?(3 votes)
- At7:08, David said "focal length" when I think he meant "focal point".(1 vote)
- Both are correct. If the focal point is behind the mirror then the focal length will also be behind the mirror.(2 votes)
- what are sign conventions of every mirrors?(1 vote)
- Concave Mirror +F -Do ±Di May be real and inverted or virtual and right side up
Convex Mirror -F +Do -Di Virtual and right side up(1 vote)
- Can distances behind the mirror be negative?(1 vote)
- David explains, that with a convex mirror and the eye and object on the left, the image is formed on the right (negative image length). I'm confused about why light forms an image through the mirror, is it that the mirror is only partially reflective? Thanks!(1 vote)
- Nice question. The mirror is not partially reflective, which is generally the case unless otherwise mentioned.
If you construct the ray diagram for this case, you will see that the light rays diverge after reflection from the mirror, they do not move on a path that would make them intersect. And we know that image is formed when light rays intersect, so no image formation should take place, and yes, no real image is formed.
However, when you extend the rays to behind the mirror you will observe that the lines intersect at a single point, i.e. the rays appear to converge (/emanate from) a single point, so the rays are virtually intersecting, and yes, a virtual image is, in fact, formed! So in this case, a virtual image forms instead of a real image.
The image you see when you look into the mirror while brushing your teeth is also a virtual image, see how it appears to be behind the mirror, in a mirror world? This is why. Also, check out this video for more on virtual images:
- I have a problem where I need to calculate the image distance if the image is between the focus and the object and the distance between the image ad the object is 15cm. Focus of the concave mirror is 20cm. If I try to do this problem even with this sign convention the results come out as complex numbers.(1 vote)
- In our textbooks we have magnification formula like Г = H/h = f/d
My question is why we don't have minus before f/d which we have in video or just the concept is different and it is understood to be like that(1 vote)
- Is radius of curvature of any curved mirror double its focal length?
And is there any equation for plane mirrors?(0 votes)
- Yes, the radius of curvature of any curved mirror is double of its focal length. No, there are no equations for plane mirrors.(2 votes)
- In a convex mirror, why is the image considered to be behind the mirror?(0 votes)
- Actually the image formed in this case was virtual .. Which means that the image is formed without the convergence of rays but instead our brain extrapolates those rays which appear to be coming from a point.( and since we are intelligent beings our brain tries to make sense and tries to visualise things ..so it extrapolates those rays) and well coming back to your question the virtual image is not actually formed behind the mirror but instead it's formed inside the mirror itself ...and also you may have noticed that your mirror image cannot be caught on a screen and that is what virtual images are..the can't be caught on a screen..is it fynn??(2 votes)
- [Instructor] Mirror equation problems can be intimidating when you first deal with them. And that's not because the mirror equation's all that difficult. It's kinda easy. It's just a few fractions added together. The place where it gets tricky is deciding whether these should be positive or negative. So there's a bunch of positive and negative sign decisions that you have to make. And if you make even one of those incorrectly, you can get the wrong answer. So let's do a few mirror equation problems, and you can see how the signs work. Now I have to warn you, everyone has their own sign convention. There's a lot of different sign conventions when you deal with optics. The one I'm gonna use is the one that I feel like most textbooks are using these days, and it's the one where anything on the front side of the mirror, so, by front side, your eyes should be over here somewhere. So let's say your eyes are over here, right? You're looking at some object. Maybe it's an arrow or a crayon right here, a blue crayon, you're holding in front of this mirror right here. So this is the mirror right here. And the convention I'm using is that anything on this side of the mirror is gonna be counted as positive. So if it's a focal point on that side, positive. If it's an object distance on this side, it's positive. And if the image distance comes out positive, you'll know that it's also on this front side of the mirror. Anything that comes out negative, so if we get a negative image distance after we do our calculation, we'll know that that thing is behind the mirror. That kinda makes sense, negative like behind, positive like in front. So this side over here is in front of the mirror, positive. And back here would be behind the mirror, and that would be negative. So I've got some numbers in here already. Let's just solve this one. So what do we do? We're gonna use this mirror equation. We're gonna say that one over the focal length, and already we have to decide on a positive or negative sign. So this mirror, the way it's shaped right here, based on how we're looking at it, is a concave mirror. And with the sign conventions we just discussed and the signs I'm using in this formula, concave mirrors always have a positive focal length. So, in other words, since this focal point is four centimeters from the center of the mirror, I'm gonna have to plug in the focal length as positive four centimeters. Notice I'm not converting. That's okay. If you leave everything in centimeters, you'll just get an answer in centimeters. So it's okay, you don't have to convert as long as everything's in the same units. We'll set that equal to one over the object distance. Just one other warning, sometimes instead of d o, you'll see this as s o for object distance, or you might see d i as s i for image distance. It's the same thing. It's just a different letter. So the object distance, again, is on this side in front of the mirror. So it's gonna be positive 12 centimeters since it's located 12 centimeters from the center of the mirror. So this is also gonna be positive 12 centimeters. And now we're gonna add to that one over the image distance. The image distance, we don't know. I haven't drawn the image on here. It's gonna be a surprise. We don't know what this is gonna be, but we can solve for it. So we can solve for image distance. I'll subtract one over 12 from both sides, which will give us one over four centimeters minus one over 12 centimeters, and that's gonna have to equal one over the image distance. So 1/4 you could rewrite as 3/12. So 3/12 minus 1/12 is just gonna be 2/12, and that's gonna equal one over the image distance. But 2/12 is just 1/6, so one over the image distance is just gonna be 1/6. But that's what one over the image distance is. Sometimes people forget to flip this at the end. We don't want one over the image distance. We want the image distance. So finally, you take one over each side. And we solve, and we get that the image distance is gonna be six centimeters. And it came out positive. That's important. This came out to be positive six centimeters. So where's our image gonna be? So since this image distance came out to be positive, our image is gonna be in front of the mirror. So it's gonna be over here. It's gonna be six centimeters from the mirror, somewhere around here. So at this point right here is where our image is gonna be. But we don't know how big it's gonna be or whether it's right-side up or upside down. To figure that out, we have to use a different equation, and that equation's called the magnification equation. It says that the magnification is equal to the height of the image divided by the height of the object. That's equal to negative the image distance divided by the object distance. So it turns out this ratio of negative image distance over object distance is always equal to the ratio of the height of the image over the height of the object. So what's the height of our image gonna be? Let's just solve for it. If we solve for the height of our image, we get that the height of the image is gonna be, multiply both sides by h o, we'll have the negative sign's already here, so negative height of the object times this ratio of the image distance over the object distance. And now we can just plug in numbers. The height of the object, it says it's three centimeters tall right here. So it's three centimeters, so negative three centimeters times the ratio of, the image distance was six, the object distance was 12. And so if you solve this, you'll get one half of negative three, which is negative 1.5 centimeters. The negative sign means that this image got inverted. So it got flipped over. It's gonna be upside down compared to what it was before. And the 1.5 is how tall it's gonna be. So what we end up with is an image six centimeters from the mirror, and it's gonna be, have a height half as tall, so 1.5 centimeters. And it's gonna be upside down because of this negative sign. So it's gonna be 1.5 centimeters tall and upside down. That's what you're gonna see if you look into this mirror. It's like a funhouse mirror. It's a weird, curved mirror. You'd see an upside-down image right here. It might look like you could reach out and grab it, but it's gonna be an optical illusion. There's gonna be no object there. It's just gonna be the image of this object here. So that's an example using a concave mirror. What would change, what if we did this? What if we took our object, so say we don't put it here anymore, we move it inside here. So instead of being at 12 centimeters, we moved it to like three centimeters. What would we do differently? Everything would be the same. We just would plug in, instead of positive 12 down here, we'd plug in positive three. So the mirror equation works exactly the same. You still plug in whatever that object distance is. You solve for your image distance. We're, of course, gonna get a different image distance. But whatever you get, that would tell you where the image is. And then you would take that, plug it into the magnification equation if you wanted to decide how big the image is gonna be and whether it's gonna be right-side up or upside down. So these numbers are gonna be different, but you would use this equation the exact same way. Now what would happen if instead of using a concave mirror, we used a convex mirror? Let's say we used a mirror shaped like this. So imagine our eye, again, is over here, looking at this object inside of the mirror, and we're gonna see an image of the object. We're gonna see the object right here, but we're also gonna see the image of the object. This mirror, this time instead of concave, this is a convex mirror. So its focal length is behind the mirror. So what do we do now to figure out where the image is? We, again, use the mirror equation. We're gonna use the same equation. We're gonna have one over the focal length. And, again, I immediately have to make a decision on the sign. With the convention that I'm using, this focal length is behind the mirror. So this focal length for a convex mirror is gonna be negative. So this would be negative four centimeters. And that's gonna equal one over the object distance. Well, again, the object is in front of the mirror 12 centimeters. So this object distance is gonna be positive 12 centimeters. And then we add to that the image distance, which we don't know. This is what we're gonna solve for. So this time if we solve, we're gonna have one over negative four. And, again, we have to subtract one over 12, and that's all gonna have to be equal to one over the image distance. So now on the left-hand side, we have negative 1/4, but that's the same as negative 3/12. So negative 3/12 minus 1/12 is the same as negative 4/12, and that's got to equal one over the image distance. But negative 4/12 is the same as negative 1/3. So one over the image distance has to be equal to negative a third. And if we flip that over, we get that the image distance finally is gonna be negative three centimeters. So, in other words, this here is negative 1/3. So when you flip that over, you get that the image distance is negative three centimeters. Where is this image gonna be? Well, since it came out negative, that means it's behind the mirror 'cause that's the sign convention we're using. So it's gonna be three centimeters behind the mirror. So it's gonna be like over here about at this point right around here somewhere, three centimeters behind this mirror. And, again, if we want to figure out how tall it's gonna be, whether it's right-side up or upside down, we're gonna have to use the magnification equation. And that magnification equation looked like this. It said that the height of the image over the height of the object had to be negative image distance over object distance. So let's just see what this right-hand side's gonna be. If we just set this equal to, it's gonna be negative of, negative three is the image distance, so I have to plug in the negative three. And you keep that negative in there. This negative out here comes along always, but now we have another negative inside of this image distance of negative three. And the object distance, again, was 12 centimeters. So what are we left with? We get negative of negative three, which is positive three over 12, which is positive 1/4. And the centimeters cancel. So what this ratio tells you, which is the magnification, is that the image is not gonna be inverted. This positive means it's gonna be right-side up. And the 1/4 ratio means it's gonna be 1/4 as large, the image, I should say, is gonna be 1/4 as large as the object is gonna be. And the reason is this ratio is equal to height of the image over the height of the object. So, in other words, if I can multiply, if I want to multiply both sides, I can say that height of the image is equal to positive 1/4 times the height of the object. Well, the height of our object was three centimeters in this case. But whatever your height of the object is, you multiply it by this ratio of negative d r over d o, and you get what the height of the image is gonna be. So we're gonna get 1/4 three. So we get positive 3/4 of a centimeter. So it's gonna be tiny. This is gonna be little, little image that's gonna be right-side up. So it's gonna be right here, but it's gonna be teeny. It's only gonna be like this, 3/4 of a centimeter tall. That's what our image is gonna look like. So recapping, you can use the mirror equation to figure out where the images are gonna be located. The sign convention we're using is that objects, images, and focal lengths in front of the mirror are gonna be positive. Anything behind the mirror is gonna be negative. And you could use the magnification equation to figure out how tall the image is gonna be relative to the object by taking negative the image distance over the object distance.