Lesson video

Hello and welcome to this lesson on scale diagrams for convex or converging lenses.

This is from the unit called electromagnetic waves.

And my name's Mr. Norris.

So convex or converging lenses are everywhere.

They're in magnifying glasses.

They're in phone cameras.

There's one here that I'm talking into right now.

However, these are not convex lenses.

Glasses only have convex lenses for people who have long sight, I have short sight.

More on that in a future lesson.

But what we need to do is for convex lenses, they always form an image.

And if you want to make a device like a camera or a projector using a convex lens, you need to know what type of lens to use to create the size and type of image that you want.

Do you want an image that is magnified or do you want an image that is diminished and inverted? So a scale diagram can help us predict where the image will appear and how big and what way up for a specific object at a specific distance from a specific lens.

So let's look at how we do that.

The outcome of this lesson is that by the end of the lesson, hopefully you'll be able to describe uses of convex lenses and also, you'll be able to draw scale ray diagrams to determine the position and magnification of an image.

Here are some keywords that will come up this lesson.

Object distance, image distance, principal focus, focal length and principal axis.

Each word will be explained as it comes up in the lesson.

The lesson has two sections.

We're gonna firstly look at uses of lenses to form images on a screen.

And we're secondly gonna look at scale ray diagrams for convex lenses.

Let's get going with the first section.

So we know that a convex lens can focus light to produce an inverted image of an object on a screen and have a look at the diagram.

And you might have had the opportunity to do that yourself using a convex lens.

And we know that the distance between the object and the lens is called the object distance.

And we also know that to produce a sharp image that's not blurry, the screen's got to be placed at exactly the right distance from the lens, which is called the image distance.

So for that specific object, however far away it is from the lens and that specific lens, the rays will come together at a point which is at a distance called the image distance from the lens.

And that's where you've got to put a screen to get a sharp image on your screen.

If an object is distant, then the rays of light from that distant object that pass through the lens are effectively parallel.

So there's an object and if it's a very large distance away, then kind of that middle ray, we can think of all of the rays kind of around that middle ray are effectively hitting a very distant lens parallel coming from that point.

And parallel rays are focused to a point called the principal focus of the lens, which is labelled there.

The image that's produced is diminished and inverted like we saw in that first slide.

The focal length of a lens is the distance from the centre of the lens to the principal focus.

So that's that distance there.

Let's just check those terms. Give the correct terms for labels X and Y, which is what we just went through.

Pause the video if you need to.

Okay, here are the answers.

So point X is called the principal focus of the lens.

And it's the principal focus 'cause it's where parallel rays of light are converged to.

If the rays of light incident on the lens are not parallel, then they'll meet at a focus, but it's not the principal focus unless the rays of light hitting the lens are parallel.

And Y is the focal length of the lens, the distance between the centre of the lens and the principal focus.

Well done if you got both of those.

So the image distance, the point where rays of light will be converged to, where you need to put a screen to get a clear image of the object the other side of the lens, that image distance and the image size or magnification depend on two factors, the power of the lens and the object distance.

So the shorter the object distance, the greater the image distance and the greater the image size.

And that is a pattern that goes on until the object distance reaches the focal length or closer.

So this animation shows that.

We start with a distant object, which gets closer to the lens.

And you could see that image distance becomes greater as the object distance becomes shorter.

We can also see the image size increasing as the object distance decreases.

And that happens all the way till the object reaches the focal length or closer because when the object is too close to the lens, i.

e.

at the focal length or closer, then no image can be formed on a screen.

So at the focal length is here, or closer is here.

Okay? And at those points, no image can be formed on a screen because the lens can't refract the rays enough to make them converge when the object's too close because from closer objects, the rays are diverging, spreading out more the closer the object.

So if the object is too close, the rays are too diverging and the lens can't refract the rays enough to make them actually come together and converge.

So the rays then leave the lens parallel or diverging and they never meet at a point because they're parallel or spreading apart, diverging.

So they never form a clear image or a focus.

Let's do a quick check about what we've just gone through.

Fill in the gaps on this slide, pause the video and have a go at working out what goes in each gap.

Off you go.

Right, I'll give you some feedback now.

As the object distance decreases towards the focal length, the image distance and the image size both increase.

The image goes from being diminished to being magnified and the image is always inverted.

So here's the animation that shows that, again, as the object distance decreases, image distance and image size, both increase, the image goes from being diminished to being magnified.

Now it's magnified, but it started diminished, smaller than life size.

But the image is always inverted.

Let's look at a specific use of this kind of lens now.

So in a camera, a lens focuses lights onto a small light sensitive CCD, which stands for charge capture device.

It's just the tiny light-sensitive screen that detects the light and creates electrical signals, which are sent to the digital camera or phone's memory to store the image as digital information.

In older-style cameras, the CCD would've been photographic film.

So in a phone camera, and in fact, in all cameras, the CCD is very close to the lens.

We're talking perhaps like five millimetres between where the lens is and where the CCD is within the phone casing.

So when light is focused on the CCD, the object distance is far greater than the image distance for almost anything you photograph.

And phone cameras now sometimes have very specialist, like separate lenses, like called macro lenses for photographing something that's very, very close to the phone because it needs to switch to a different lens.

Otherwise, the normal lens for taking pictures of objects normal distances away won't be able to focus the light on the CCD successfully.

And the human eye works in a similar way to how cameras work.

But instead of a CCD or photographic film, the back of the eye is called the retina, which is covered in light-sensitive cells.

It's the screen where light rays must be focused to form an image.

And the depth of an eyeball is only about two centimetres.

So the retina is only about two centimetres behind the lens.

So again, just like for cameras, the object distance for most objects we look at it will be much greater than the image distance.

The image distance will only be about two centimetres.

And just like cameras, the image on the retina is inverted.

But with the human eye, our brain actually flips this image kind of in our mind so that we see things the right way up even though the image that forms on the back of our eyes is upside down compared to how the objects are in real life.

And we know that moving an object closer to a lens will increase the image distance, okay? When you bring the object closer to a lens, that's gonna push the image further away.

But for a camera or the human eye, the image must always form in the same place, which is on the CCD or photographic film or on the retina.

So if an object's too close, then the rays are too diverging, the image distance is too far, and the image on the CCD would be blurry.

So in cameras, adjustments can be made to the lens position to ensure that the image always forms on the CCD.

And cameras can also switch between multiple lenses, like I mentioned earlier, and that will make sure that the image always forms on the CCD.

A clear image always forms on the CCD.

Now, the human eye does a similar job, but in a different way.

If an object's too close, then the image on the retina would be blurry 'cause the light from a closer object is too diverging.

Now, in the human eye, the ciliary muscles, that's these, they actually change the power of the lens by kind of pulling it to change its shape to change the lens curvature.

And that changes the lens power so that light from closer objects can be focused on in the right place.

So keeping the image distance constant by changing the power of the lens.

Obviously, with both cameras and the human eye, there is still a distance, which the lens cannot focus the image onto the right place.

So there is a minimum distance that objects have to be away for you to be able to focus them.

And if you bring your finger close to your eye, you'll notice there's a point where your eye can't focus on your finger anymore and it becomes blurry and you've reached that minimum distance.

Okay, here's a different use of this kind of lens.

A projector.

So in a projector, the object is actually a small bright screen behind the lens that's giving out light.

And then the lens focuses that light and it focuses on a projection screen, which is a large distance away.

Now, with a projector, you want to create a much bigger image of whatever's on that small bright screen within the projector.

So how do you produce a magnified image on a screen? Well, the small bright screen must be placed between one and two focal lengths behind the lens so that the image distance is bigger than the object distance because the image needs to form on the screen, which is much further away from the lens than the small bright screen is.

And that also means you get a magnified image of whatever's on that small bright screen.

So you get a magnified inverted image on the projection screen.

And what that means is the small bright screen must display an inverted version of the image that you want to be projected so that when whatever's on the small, bright screen is inverted, so it's projected on the screen, it then appears the correct way up that you want.

And this is how classroom projectors work and also cinema projectors.

Let's do a check on what we just talked through.

Which diagram best represents a lens forming an image in a camera and the human eye and a projector? So which one, A or B, links with a camera and the human eye and which one links with a projector? It should be fairly quick to do.

Five seconds.

I'll give you some feedback now.

A links to a camera and the human eye because the image distance is much shorter than the object distance because the camera and the human eye have to focus the light on the CCD or photographic film or the retina at the back of the eye, which is only a very tiny distance from the lens.

So that's much more like A, whereas a projector is much more like B where the object is a tiny screen, a tiny distance from the lens and the lens of the projector then projects a magnified image onto a screen, which is a much further distance away from the lens than the object was.

So well done if you got those the right way round.

Okay, time for a task now on the uses of lenses to form images on a screen.

So for each pair of statements, identify which applies to cameras and which applies to projectors.

So put C for cameras, P for projectors, both if the statement applies to both or neither if the statement applies to neither.

Pause the video now and have a go at that task.

Okay, I'll give you some feedback now.

So the object is a small bright screen placed between f and 2f.

That is projectors.

And there's a minimum object distance, but no maximum.

That's cameras because the object can be as far away as you like because the further away light is, the rays will be parallel and they'll still be able to be projected on the camera's CCD.

But there is a minimum object distance because if the object is too close, then a camera can't focus it.

So well done if you got those the right way around.

The image needs to form close to the lens, that's a camera.

The image needs to form much further from the lens.

That's a projector.

And then the next set of statements basically say the same thing.

The object distance has to be bigger than the image distance basically in a camera.

And the object distance has to be much smaller than the image distance in a projector because the screen that you're projecting onto is much further away than the tiny screen that you're projecting in a projector.

The image is diminished in a camera and the image is magnified in a projector.

And the image is inverted in both and the image is upright in neither.

So well done if you got most of those right? Okay, it's time for the second section of the lesson.

So how do we draw these scale diagrams to predict where the image is going to appear for a given object and a given lens and the size of the image and whether it's magnified or inverted? So let's look at how we draw these scale diagrams. So to engineer an optical device like a camera or projector, you need to know exactly where the image will form and what size it's gonna be.

And this can be predicted by drawing a scaled ray diagram.

The lens used can be represented using the correct symbol, and we're gonna be focusing on convex lenses which have that symbol.

So just be aware if you ever see this symbol, that is a concave lens and that's not the focus of today's lesson.

So lights from the surroundings reflects in every direction from every point on the object, the tree in this case.

And many rays of light pass through the lens from each point.

So they're the rays of light that pass through the lens from that point on the object.

And they're gonna be refracted to a point here by the lens on the screen.

And here are the rays of light that come from that point on the object.

And these are the ones that pass through the lens and they're gonna be refracted to a point here on the screen.

And these are the rays that reflect from this point on the object.

And these are the ones that pass through the lens from that point.

And they're gonna be refracted to a point here on the screen.

And here are the rays of light that come from that point on the object.

And here are the ones from that point that go through the lens.

And they are going to be refracted to a point here on the screen.

So all of the rays that pass through the lens from the same point on the object meet at a single point a set distance from the lens, the image distance.

That's where you need to put the screen.

That's where I put the screen in this diagram.

And that creates a bright spot of light from that point on the object.

So the image's created from the different bright spots made by all such rays from the different points on the object, like this.

Okay, let's do a quick check on that understanding of how lenses form an image.

So Sndeep creates an image on a screen using a convex lens, the top one that's pictured on the right of the screen there.

Andeep then swaps the lens for one with the same power, but half the diameter.

So the second lens down, it's smaller, it's half the diameter.

And then Andeep swaps back to the original lens but with half of the lens covered and no other conditions change.

So which statements are true about the three images Andeep creates with those three lenses? Take a moment to pause the video, read through each option and decide which statements are true about the three images created.

Pause the video now, have a go, Right, I'm gonna give you some feedback now.

Okay, statement A, the focal length of the lens, it's always the same.

Well, that's true 'cause the lens is always the same power.

It says that in the question.

So the image distance, and magnification is always the same.

Well, that's true because it's the same power of lens and the same object at the same object distance.

So that's true.

Statement B, the amount of light creating the image is always the same.

Now, that's not gonna be true 'cause in the second lens with half a diameter, only half as many rays or less actually will go through that lens and be focused onto the screen than with the larger lens.

And with the third lens, which was half covered, only half as much light is gonna go through as the first lens.

So the image brightness won't always be the same because the amount of light creating the image is less when Andeep swaps to the smaller lens and the half covered lens.

So B was not true.

Well done if you worked that out.

And then statement C, light can always pass through the lens from every point on the object to the screen.

And that's true.

So the full image does appear in all three.

And that's a big misconception sometimes with lenses.

Sometimes pupils think that covering up half a lens means only half the image appears.

But that's not how lenses work because light can still pass through that half covered lens from every point in the object to the screen.

So the full image will appear in all three.

Now, that was quite a difficult kind of test of your understanding of how lenses form images.

So extremely well done if you got all three of those right.

So to predict the image distance and the image size, you only have to draw two or three of those rays that form the image from a single point on the object.

So that's great because there's a huge number of rays that form the image, but we only have to draw two or three of them.

That simplifies our job massively.

So rays from every other point on the object, we know they're gonna meet at the same distance, the image distance.

That's why we only need to bother drawing two or three.

And in theory, it doesn't matter which point on the object you choose to draw the rays from.

However, it's a good idea to choose the highest point on the object 'cause that's gonna lead to the highest point on the image and so give you the clearest indication of the image size or the magnification.

And we should say that the object and the image are usually represented using arrows.

So instead of forming an image of a tree, we're gonna be forming an image of an arrow on the screen.

Okay, let's make sure we've got this set up properly.

Which of the following shows an object, a lens, and the principal axis in the correct arrangement to form an image to the right of the lens? Pause the video now and choose the right option.

Okay, I'll give you some feedback now.

The correct option that shows an object, so the arrow, a lens, the lens symbol, and the principal axis, which is the horizontal line in the correct arrangement to form an image to the right of the lens, it's arrangement B.

Arrangement A could form an image on the left of the lens 'cause light would come from the object through the lens.

C and D are drawn completely incorrectly because the principal axis should pass through the middle of the lens and the base of the arrow should be on the principal axis and pointing vertically upwards.

So here is the beginnings of a scale diagram.

It's done on graph paper and it's done to scale and I've put scales on the axes as well.

Scales might not be included in all examples of this kind of task, but I've included them so you're really clear that it's a scale diagram and everything's drawn to scale.

So this object has a height of 10 centimetres.

That's the green arrow.

It's 40 centimetres from the lens.

So look at the object distance on the horizontal axis and the focal length of the lens is 10 centimetres.

Now, if that's not marked for you, you have to mark in the principal focus at the correct distance, which is the focal length.

So a little cross at 10 centimetres to show the position of the principal focus of this lens.

And then the rays that you need to draw.

Remember, we're only gonna draw two or three.

They're called the principal rays and here are the rules for drawing them.

So for the first principal ray, a ray that's parallel to the principal axis is refracted to the principal focus.

So parallel to the principal axis is refracted to the principal focus.

So it goes through the principal focus like that.

And we know that parallel arrays get refracted to the principal focus.

So that's just a rule that actually we knew already.

Okay, the second principal ray, a ray that passes through the centre of the lens, it turns out it's not refracted.

And that's all to do with the symmetry of a lens.

So you just draw it passing through the centre of the lens and it's not refracted.

So straight through.

And actually, that's enough to give the position of the image.

So we could draw it in there.

Okay, it looks like the image distance is 13 centimetres from the lens if you look at the image distance scale and the image height, it looks like it's about four boxes.

So the image height is four centimetres, so that's a diminished image, smaller than the object is in real life.

Now, you can draw the third principal ray, which is this, a ray arriving at the lens from the principal focus.

So that's from the principal focus on the left of the lens.

Can you see how it's drawn from the top of the arrow and goes through the principal focus that was on the left of the lens and then it hits the lens? And that's gonna become parallel.

And you can see all three of those rays meet at the same point, which is the top of the image.

So you don't have to draw the third ray, but it can help you confirm that you've drawn the other two correctly if all three of the rays meet at the same point.

And in this case, we know that the image is inverted 'cause it's upside down, the arrow is upside down compared to the object and it's diminished because it's smaller compared to the actual object size.

Okay, let's check you've got the idea of doing these ray diagrams. So make a quick copy of this diagram first and then add rays and the image to show what a ray diagram should look like.

Use a ruler to do this so it looks neat.

And your principal focus of the same distance in front of the lens and behind the lens, those two crosses.

But this doesn't need to be to scale.

So this is just getting the hang of what these diagrams should look like.

Pause the video now and have a go at this check.

Right, I'll give you some feedback now.

So you should have just started by copying what's on the screen now.

And then you should have added the three principal rays for this kind of lens.

So a ray that's parallel to the principal axis is refracted through the principal focus.

So it should look like that.

And then a ray that passes through the dead middle of the lens isn't refracted.

So use a ruler just goes straight through the middle of the lens and then a ray that goes to the other principal focus and hits the lens doesn't change direction at any of those points.

It only changes direction at the lens when it's refracted.

So it becomes parallel.

So make sure that your diagram looks like that and it's worth spending two or three minutes now, pause the video to just practise drawing that out maybe two or three times just so you get the hang of and get the feel for how to draw those rays to predict where the image should be.

And of course, you should have added the image in.

And in this case, it's an inverted image.

So following that process, we'll find the location of the image for any object distance that will produce an image on the screen.

So any object distance from very, very far away up to the focal length will produce an image on the screen and you just follow that process and it will work.

So here's another example, but the object is now much closer to the lens.

It's only 15 centimetres from the same lens.

So follow the same process.

A parallel ray is refracted through the principal focus.

A ray that goes through the centre of the lens is not refracted.

That's actually enough to form the image but I'm gonna do the third ray anyway, which goes through the other focus and becomes parallel.

They should all meet.

And that this time forms a magnified image, which is 28 centimetres from the lens, reading the image distance scale there.

So this works to find the location of the image for any object distance that's gonna produce an image on a screen.

Let's do a check then, a final check before you do some of these yourself.

Match each principal ray to the correct description of where it is refracted by this kind of lens.

Pause the video now and have a go at that task.

Okay, time for some feedback now.

So a ray parallel to the principal axis is refracted to pass through the principal focus the other side of the lens.

A ray that passes through the centre of the lens is not refracted, that's the easy one.

And that leaves a ray that arrives at the lens from the principal focus that's refracted to become parallel.

Well done if you've got all three.

So a final thing you can do with your scale ray diagram is calculate the magnification of the image using magnification equals the image height divided by the actual height of the object.

So here is the ray diagram we just drew.

So for this image, the magnification is gonna be 19 centimetres divided by 10 centimetres.

So the image height is 19 centimetres.

That's the height of the arrow, of the image arrow, the inverted arrow, and the original object, which is the arrow on the left, that's only 10 centimetres tall.

So magnification is image height of actual height, 19 centimetres over 10 centimetres, the units cancel.

So magnification has no units, it's just a ratio.

And in this case, it's just 1.

9.

So if magnification is above one, then the object is magnified.

If the magnification is less than one, then it is diminished 'cause magnification really is just the scale factor ratio.

How much is the image being scaled up or down by? In this case, it's a factor of 1.

9.

Let's do a check of magnification.

Calculate the magnification of this image, and I've given you an enlarged version so you can accurately measure the image height.

Pause the video now, have a go at doing that and show all your working.

Okay, here's some feedback now.

You should have written the equation that you're gonna use.

Magnification is image height divided by the actual height.

The image height in this case is four centimetres because it's four little boxes, that arrow on the right and the actual height is 10 centimetres.

Centimetres cancel, gives you an answer with no units, 0.

4.

So this image is diminished.

And we'll just note at this point that sometimes people can get different answers from the same diagram for the magnification because there might be slight differences in how the diagram's being drawn.

And there might be slight differences in how the diagram's being interpreted, especially if the rays are drawn too thickly.

So use a sharp pencil when you're drawing these diagrams, draw them as precisely as you can, and then hopefully your answer will match the expected answer.

Okay, so you're now ready to do a task.

There are gonna be three scale ray diagrams to complete.

And then for each one, describe the image and calculate the magnification after you've drawn the diagram.

So here is the first ray diagram to draw.

And this task is really designed to be done on the sheet that accompanies the lesson.

However, if you don't have that sheet, you can just do them on graph paper or square paper using a ruler.

You'll just need to draw out the diagram first before adding the rays.

So the first one is this 15-centimeter high object, which is 25 centimetres from a lens of focal length that's eight centimetres.

The second one you need to do is a 12 centimetre high object, which is 30 centimetres from a lens of focal length 15 centimetres.

Where does the image appear for that object? Close to that lens.

And then the third one you need to do is for a six centimetre high object, 21 centimetres from a lens of focal length 15 centimetres.

So what I suggest you do is have a go at that now, pause the video and go back to the previous slides if you're working from the video if you need to see the questions again.

And I will see you in a few moments when you've had a good go at all three diagrams. And remember, once you've drawn each diagram, remember to describe the image and find the magnification.

Off you go.

Okay, here's some feedback on that task.

So here is the first diagram.

It should look like this.

There's the first principal ray, the second principal ray through the centre of the lens, and the third principal ray.

And that should give you that image there.

That is about 12 centimetres from the lens and about seven centimetres tall.

So the image is inverted and diminished.

Magnification is the image height divided by the actual height.

Image height is seven centimetres, object height is 15 centimetres, giving you a magnification of 0.

466.

So about 0.

47, the magnification.

Well done if you got that.

Okay, here's the second one.

Here's the first ray refracted through the principal focus from parallel, the second ray through the centre of the lens and the third ray through the focus becomes parallel.

There's the image.

This image is inverted and it's the same size as the object.

And that's because the object has been placed at exactly double the focal length from the lens and that's the point where the image turns out life size, so not magnified or inverted.

So the magnification should come out to one, which it does.

So well done if you got that.

And here is the third ray diagram to do.

So a ray that's parallel is refracted through the focus, straight through the middle, and then through the focus and then straight.

There's the image.

This image is inverted but magnified.

And what's the magnification? Well, the image height is 15 centimetres.

The original object height was six centimetres.

The units cancel, giving you a magnification of about 2.

5 for this one.

Very well done if you got that.

So here's a summary of the lesson.

In cameras and the human eye, a convex lens needs to focus the image on the CCD or the retina.

And the object distance is much bigger than the image distance.

So the image is diminished and inverted in cameras and the human eye.

Whereas in a projector, the image needs to be focused on the screen, which is a large distance away.

So the object distance is much smaller than the image distance and the image is magnified and also inverted.

And we've seen how the image distance and height can be predicted by drawing a scale ray diagram from a single point on the object.

And the ramification is calculated by doing the image height divided by the actual height.