Now let's proceed with a few spherical mirrors. We'll start with the case of a converging mirror with the object at its focus. The first thing we do is draw a line that is parallel to the horizon. Once it strikes the mirror it will converge through the focus. So this is parallel to this line. Now the second line we draw is straight through the focus.

And then, we look and see what our eye sees. As our eye looks at these light rays, it draws back the point of convergence right here. So essentially, if a mirror is reflecting an object that's at its focus, we get no image. Now we can also use the thin spherical mirror equation to determine this, 1/o + 1/i = 1/f.

Now with mirrors, we can essentially define one side to be positive and the other side to be negative, just as a reference. Now for this problem, let's say that our focal length is 1 and our object is also the same 1. If we plug in these values, 1/1 + 1/i =1/1, we find that 1/i = 0.

Meaning, that our image would have to be at some infinite location, or another which it doesn't exist. Now, let's look at the case of a converging mirror, outside the focus. Here, we again draw a straight line, parallel to the horizon that bounces back and converges through the focus. The second line we draw through the focus, and when it hits the mirror surface, it will bounce back parallel.

So this line, this line, and this line are all parallel. Now our eye sees these light rays cross and fills in the image. Is this image real or virtual? Well, since it's actual light passing through this point, this is a real image. Again remember, images that are on the same side of the mirror as the object are real.

Now we could also use the thin spherical mirror equation to determine this problem, that is 1/o + 1/i = 1/f. Now let's take our focal length to be 2 and our object distance to be 3. What is the image distance? We then plug these in, 1/3 + 1/i = 1/2. Or 1/i = 1/2- 1/3 is 1/6.

So our image is at a distant 6, which would be further from the plane of the mirror than the object in the focus. What's the magnification? m = -i/o, we can plug in our values, m = -6/3 or -2, which means that this image is in the opposite direction of the object or down, and it is larger.

Now let's put our object inside the focus and see what happens. Now it works the same way, we first draw a line parallel to the horizon. When it bounces off the mirror, it will reflect back through the focus. Next, we draw a line as if it had originated through the focus to hit our object. Once this one hits the mirror, it bounces back parallel.

So again, just like last time, this line, this line, and this line are parallel. Now your eye sees these two light rays, and since your brain assumes that light travels in straight lines, it extrapolates this backwards, to convergence point and then we get our image. Now is this image, real or virtual? Since this image, is the convergence of extrapolated light and is behind the plane of a mirror, this is a virtual image.

Now let's also use the thin spherical mirror equation to explore converging mirrors inside the focus. We have 1/o + 1/i = 1/f. For this problem, let's have our object distance be 4/7, and our focal length be 1. You can then plug these values in.

1/4/7 + 1/i = 1/1, we can subtract this from this side and say 1/i = 1/1- 7/4. We can change 1/1 to 4/4, it's the same thing = -3/4. So our i= -4/3. This negative sign, indicates that it's on the opposite side, making it a virtual image.

So we have a distance of -4/3. What's our magnification? Magnification is simply going to be, m = -i/o =- (-4/3) / 4/7. We then simplify this to be-- is + + 4/3*7/4.

These cancel to get a magnification = 7/3. Now, the last mirror we'll look at is the diverging mirror. Now the converging mirror, when we drew the light ray parallel to the horizon it would converge, bend towards and go through the focus. This doesn't happen in the diverging mirror. When we draw the line to the horizontal, instead of bending towards focus it reflects away as if it had come from the focus.

When we draw our second light ray and we try to get it to go through the focus, but what happens instead, is that the light ray hits the mirror and reflects away parallel. Now, our eye sees this and extrapolates backwards to a point of convergence and we get our image. Now, is this image real or virtual?

It's behind the plane of the mirror and it involves extrapolated light, so this is a virtual image. And now we'll use the thin spherical mirror equation to confirm what we've just drawn, 1/o + 1/i = 1/f. So for this problem we'll use an object distance of 2 and the focal length of -1, since it's on the opposite side of the mirror's plane.

We can plug these in, 1/2 + 1/i = 1/-1, or 1/i = -1/1- 1/2. Multiplying both sides by 2 we get -2/2 = -3/2 or -2- 1 = -3. The i = -2/3, which makes sense, that should be on the opposite side of the mirror and it's virtual.

Now the m = -i/o or -(-2/3) / 2 = 1/3, meaning that the image is in the same direction as the object, but it's smaller.

Read full transcriptAnd then, we look and see what our eye sees. As our eye looks at these light rays, it draws back the point of convergence right here. So essentially, if a mirror is reflecting an object that's at its focus, we get no image. Now we can also use the thin spherical mirror equation to determine this, 1/o + 1/i = 1/f.

Now with mirrors, we can essentially define one side to be positive and the other side to be negative, just as a reference. Now for this problem, let's say that our focal length is 1 and our object is also the same 1. If we plug in these values, 1/1 + 1/i =1/1, we find that 1/i = 0.

Meaning, that our image would have to be at some infinite location, or another which it doesn't exist. Now, let's look at the case of a converging mirror, outside the focus. Here, we again draw a straight line, parallel to the horizon that bounces back and converges through the focus. The second line we draw through the focus, and when it hits the mirror surface, it will bounce back parallel.

So this line, this line, and this line are all parallel. Now our eye sees these light rays cross and fills in the image. Is this image real or virtual? Well, since it's actual light passing through this point, this is a real image. Again remember, images that are on the same side of the mirror as the object are real.

Now we could also use the thin spherical mirror equation to determine this problem, that is 1/o + 1/i = 1/f. Now let's take our focal length to be 2 and our object distance to be 3. What is the image distance? We then plug these in, 1/3 + 1/i = 1/2. Or 1/i = 1/2- 1/3 is 1/6.

So our image is at a distant 6, which would be further from the plane of the mirror than the object in the focus. What's the magnification? m = -i/o, we can plug in our values, m = -6/3 or -2, which means that this image is in the opposite direction of the object or down, and it is larger.

Now let's put our object inside the focus and see what happens. Now it works the same way, we first draw a line parallel to the horizon. When it bounces off the mirror, it will reflect back through the focus. Next, we draw a line as if it had originated through the focus to hit our object. Once this one hits the mirror, it bounces back parallel.

So again, just like last time, this line, this line, and this line are parallel. Now your eye sees these two light rays, and since your brain assumes that light travels in straight lines, it extrapolates this backwards, to convergence point and then we get our image. Now is this image, real or virtual? Since this image, is the convergence of extrapolated light and is behind the plane of a mirror, this is a virtual image.

Now let's also use the thin spherical mirror equation to explore converging mirrors inside the focus. We have 1/o + 1/i = 1/f. For this problem, let's have our object distance be 4/7, and our focal length be 1. You can then plug these values in.

1/4/7 + 1/i = 1/1, we can subtract this from this side and say 1/i = 1/1- 7/4. We can change 1/1 to 4/4, it's the same thing = -3/4. So our i= -4/3. This negative sign, indicates that it's on the opposite side, making it a virtual image.

So we have a distance of -4/3. What's our magnification? Magnification is simply going to be, m = -i/o =- (-4/3) / 4/7. We then simplify this to be-- is + + 4/3*7/4.

These cancel to get a magnification = 7/3. Now, the last mirror we'll look at is the diverging mirror. Now the converging mirror, when we drew the light ray parallel to the horizon it would converge, bend towards and go through the focus. This doesn't happen in the diverging mirror. When we draw the line to the horizontal, instead of bending towards focus it reflects away as if it had come from the focus.

When we draw our second light ray and we try to get it to go through the focus, but what happens instead, is that the light ray hits the mirror and reflects away parallel. Now, our eye sees this and extrapolates backwards to a point of convergence and we get our image. Now, is this image real or virtual?

It's behind the plane of the mirror and it involves extrapolated light, so this is a virtual image. And now we'll use the thin spherical mirror equation to confirm what we've just drawn, 1/o + 1/i = 1/f. So for this problem we'll use an object distance of 2 and the focal length of -1, since it's on the opposite side of the mirror's plane.

We can plug these in, 1/2 + 1/i = 1/-1, or 1/i = -1/1- 1/2. Multiplying both sides by 2 we get -2/2 = -3/2 or -2- 1 = -3. The i = -2/3, which makes sense, that should be on the opposite side of the mirror and it's virtual.

Now the m = -i/o or -(-2/3) / 2 = 1/3, meaning that the image is in the same direction as the object, but it's smaller.