Hello and welcome to this lesson called "Refraction Through a Semicircular Block".

My name is Mr. Norris.

This lesson is from the unit on electromagnetic waves.

And in this lesson we're going to look at another visual effect caused by the refraction of light.

So I want you to start by thinking about the surface of water.

Is it clear or transparent? Can you see through it? Does it transmit light? Does it always? Well, actually the answer is it doesn't always.

So I've got some water here, which I'll just adjust the camera so I can show you.

And what I want you to look at is I want you to look up through the surface of this water.

There it is.

And it might look like it's clear, but when I dip this pencil in the water, there's a moment when you can't see it coming towards the surface of the water anymore.

What's going on there, okay? And then the pencil gets closer to the surface of the water and then breaks through the surface.

And now you can probably see that actually the inside surface of that water is acting like a mirror 'cause there's a reflection of the pencil there.

And just to prove that the inside surface of the water is acting like a mirror, here's a coin that I'll put up here.

And you can see the reflection of the coin and my hand in the surface of the water.

Now interestingly, the surface of water doesn't always act like a mirror because if I just look up through the surface of the water, I can see through it.

So it must be something to do with the angle that the light is passing through the water-air boundary.

That means it behaves differently.

So that's what we're going to investigate in this lesson using a semicircular block.

So let's get going looking at the lesson.

The outcome of this lesson is, hopefully by the end of the lesson you'll be able to describe and explain when refraction happens and when total internal reflection happens at boundaries between different materials, different media.

Some key words that will come up this lesson are refraction, medium, angle of incidence, and total internal reflection.

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

This lesson has two sections.

The first section looks at refraction in a semicircular block.

And then we will look at this idea of total internal reflection.

Let's get going with the first section.

So the bottom edge of these rulers are at the same level, But when viewed through the water, the image is distorted.

Look at the scale on that ruler on the left.

They're at the bottom of those rulers are at the same level as in the photo on the left.

But when viewed through the water, look how distorted the image is.

So that water is seven centimetres deep, but from this angle it only looks four centimetres deep because when we measure it using the ruler on the right go, the bottom of the water looks like it's at three centimetres on the scale, when in fact it's at zero centimetres on that scale.

And the top of the water is at seven centimetres.

That's the depth of four centimetres.

So this is caused by refraction as light ray reflecting off the tank floor move out of the water into air.

And this is why water really is often deeper than it looks.

That's not just a saying.

Water really is often deeper than it looks because of this effect of refraction.

So refraction is when waves enter a new medium, a new material, and change direction because they move with a different speed in the new medium.

So when waves speed up, they turn away from the normal.

For example, when light moves from water where it's a bit slower back into air where light is faster, of course light is always incredibly fast, 300 million metres per second in air and only about 225 million metres per second in water.

So still really fast, but slower in water than in air.

And remember that the normal is an imaginary line drawn at 90 degrees to the surface between the two materials, between the two media.

And it's drawn at the point where the incident ray, that's the ray coming in towards the surface.

The normal is drawn at 90 degrees to the surface at the point where the incident ray hits the surface.

So when wave speed up, they turn away from the normal.

Whereas when waves slow down, if they were going into the water from air instead of out of the water, then the opposite happens, they turn towards the normal.

And that's shown in that diagram.

So let's do a check on those basic ideas of refraction and which way does array of light turn when it enters new medium.

Which is the correct refracted ray for each of the following.

So these are rays of light and the media are air and glass.

So we've got diagram one and diagram two.

So for each diagram you've got to choose which is the correct refracted ray.

Pause video now, make sure you've chosen between A, B, C, and D for each of these.

Right, I'll go through the answers now.

For diagram one, the incident ray is going into glass, so slowing down so it turns towards the normal.

So the answer must be D.

Well done if you've got that.

And for diagram two, the incident ray is moving out of glass into air.

So it's speeding up.

So it's gonna turn away from the normal as it comes out of the glass, it's refracted out away from the normal.

So the answer is C in that diagram.

Well done if you've got both of those.

Here's a reminder that if the angle of incidence is zero, the incident ray travels along the normal, the change in speed still occurs but refraction which is the change in direction, does not occur.

So you can have transmission without refraction because refraction is the change of direction and there's no change in direction if the angle of incidence is zero, the ray just continues along the normal 90 degrees to the surface.

And this is why semi-circular transparent blocks are useful.

So let's do a check on what we've just said but looking at semi-circular transparent blocks.

So in each diagram, which is the correct transmitted ray? Can you work it out? Five seconds to decide, pause video if you need to.

I'll tell you the answers now.

So actually for all three of these diagrams, the incident ray is entering the boundary at 90 degrees to the surface.

So along the normal, each of these incident rays is travelling along the normal.

So that's an angle of incidence of zero.

What that means is each ray isn't refractive, there's no change in direction, they're just transmitted straight into the block without a change in direction.

So the answer is B in all three diagrams for that reason.

Well done if you spotted that.

So when entering a semicircular transparent block, any incident ray that's directed towards the centre of the circle has an angle of incidence of zero when it enters the block.

So these rays are not refracted of that air-glass boundary.

'Cause at that boundary the light's travelling along the normal to that boundary.

And what this means is that using a semicircular transparent block, we can direct rays of light to arrive at the same point on the internal or inside surface of the block, which is the centre of the circle, but with different angle of incidence when rays of light arrive at that same point.

So we're gonna make use of that in the investigation that we're gonna do.

So we are gonna investigate refraction at that internal surface of a semi-circular transparent block.

Could be glass, could be plastic, it won't really matter.

So a array box and slit can produce a narrow beam of light and take care 'cause these can get hot.

And what we're gonna do is we're gonna change that angle of incidence each time and trace the paths taken by the rays after they change direction at the boundary from glass going into air.

And we're gonna measure the relevant angles using a protractor.

So first we'll need to place the block on paper and draw around it.

And then use a ruler to mark the centre of the straight side.

You might want to add a baseline as well.

And then mark the centre of the straight side.

You probably need to measure how long that straight side is and halve it.

And then mark a point that's exactly at that halfway point.

And then we will remove the block.

There it goes and mark out the angle of incidence.

So you need to pop the protractor on, aligned with the baseline and your marked central point and mark out the angle of incidence.

So different angle of incidence which you can then trace in.

Make sure that your normal goes through the baseline and then you've got different angle of incidence.

Then we can replace the block in position with the straight edge against the baseline again and shine each ray through one at a time.

So here's the first ray.

What you need to do is mark the centre of each refracted ray in two places with a sharp pencil.

So take your time to do this, make sure your pencil's really sharp so you can make a really precise mark in the centre of each ray in two places, one, two.

And the more precisely you can do that, hopefully the more accurate your results will be.

And then you can do that for each incident ray in turn.

Okay, so mark each refracted ray in two places as it leaves the block.

Then you can remove the block to draw in the rays and then measure the angles.

Okay, time for a check for understanding on this experiment.

Which of the following would reduce random errors in the results? Which are the angle measurements in this experiment? A, "Making the beam of light as narrow as possible." B, "Aiming to mark the centre of the beam every time." C, "Taking care not to accidentally nudge the block out of position between tests." D, "Doing repeats and calculating the mean if there's time to do so." Which would reduce random errors in the results? Pause video now and decide which ones would do that.

Well the answer is they all would.

All of these are ways of making your results more accurate in this experiment.

So well done if you identify that actually all of these would.

Okay, so it's time to do the investigation now.

So there's different steps to this task.

So start with step one.

Collect a semicircular transparent block and a array box with a slit and a power supply if needed.

Then place the block on paper and draw around it like I showed you.

And use a ruler to find a mark the centre of the straight side.

With a sharp pencil, ruler and protractor, mark up the angle of incidence again like I just showed you.

And then direct array into the block along each line.

Mark the centre of each array that leaves the block in two places with a sharp pencil.

Then step four, remove the block and use a ruler to draw each ray's path, including any path the ray takes back through the block.

Bit of a clue there that something different is gonna happen at some point.

And then step five, describe what happens to each ray in the table and use a protractor to measure the angle of refraction.

And the angle of refraction that you need to measure is shown there in the diagram.

So those are all the instructions to follow and so you should pause the video now and complete that investigation.

Off you go.

I'm going to go through some sample results of the experiment now.

So the key thing actually is the second column.

What happens to the ray? 'Cause for smaller angle of incidence up to about 40 degrees, the ray is mainly transmitted out of the block and turns away from the normals.

There might be a partial reflection from the inside surface, but the ray is mainly gonna be transmitted out of the block and it's gonna turn away from the normal because it's going from glass to air so speeding up.

And there's also that faint reflection back in the block.

But you might have spotted that as the angle of incidence increased, the partial reflection got a bit brighter up to 40 degrees or so because beyond about 40, 41, 42 degrees, the ray, larger angle of incidence, the ray became totally reflected off the inside of the block and it was very bright because it was all of the light was being reflected and none of the ray was being transmitted out of the block.

So there hopefully should have been no angle of refraction to measure for angle of incidence bigger than about 40, 41, 42 degrees.

So the angles of refraction you should have measured hopefully are in that third column, but only for the smaller angle of incidence.

So in this part of the lesson, we'll look at explaining why we get this result of refraction at some angles, the smaller angles of incidence, but then total internal reflection where all of the ray is reflected from the internal surface of the glass block at other larger angle of incidence.

So a conclusion states what you found out in an experiment and how you know this.

But in this experiment, the conclusion is more complex than just a single pattern because there's a pattern, but then something changes once a certain angle is reached.

So here are some conclusions from this experiment.

So in the sample results which are shown in the table for angle of incidence from zero degrees to 40 degrees, we can say this, as the angle of incidence increase, the angle of refraction also increased, but at an increasing rate.

So in other words, the angle of refraction increased faster than the angle of incidence.

And that rate of increase actually sped up for the same increase in angle of incidence, which was 10 degrees every time.

And we can also say that the angle of refraction was always bigger than the angle of incidence, and that's because refraction occurred away from the normal.

When light refracts, when light moves from glass into air, it speeds up and refracts away from the normal.

So the angle of refraction is always bigger than the angle of incidence.

And that's shown in this animation here.

As we increase the angle of incidence, which is the ray on the left from zero to around 40, 42 degrees, you can see that angle of refraction also increases.

We've also got that partial reflection from the inside surface and that partial reflection got stronger and stronger and stronger as the angle of incidence increased, and actually the refracted ray got weaker and weaker.

Yeah, so there we go.

The partially internally reflected ray is internally reflected 'cause it's reflected from the inside surface, internal means inside, and it's a partially reflective.

So we've got partial transmission out through the boundary and also partial internal reflection.

And the intensity of that reflected rate increased as the angle of incidence increased.

However, for larger angle of incidence, refraction did not occur because the ray was totally reflected from the inside surface.

This is called total internal reflection.

So the app that shows this now, starting from this angle, there's no refraction.

So all of those angles marked as large angle of incidence, large i with the yellow arrow, there's no refraction out across the boundary, it is total internal reflection.

And you hopefully notice that the reflected rays obey the law of reflection.

I didn't ask you to measure those in the task, but you might have done just to see if they were equal, they looked equal or they should look equal.

'Cause the angle of incidence should be equal to the angle of reflection when you get total internal reflection from the inside surface of a boundary.

So time for a check of the results of the experiment.

Which light rays in each of the following show the path taken by the light when it emerges from the block? So we've got diagram one, diagram two, diagram three, what happened in the experiments.

Pause the video now make sure you choose the right options for each diagram.

Off you go, I'll tell you the answers now.

So in diagram one, the incident ray is incident on the inside surface of the glass along the normal, so there's no refraction and ray B is the correct option.

In diagram two, we've got a smaller angle of incidence.

So I would think that that ray is going to be refracted out of the, across the boundary, out of the glass block and into the air.

So that would be ray C, but there might also be a partial reflection, which would be ray D.

So well done if you identified both of those as paths that the ray takes after interacting with the boundary between the glass and air.

And in diagram three, that looks like a large angle of incidence to me.

So we're gonna get total internal reflection and only ray D as the path taken by the ray.

So well done if you've got all of those right.

So for large angle of incidence, there's no refraction, and instead all the light is reflected.

There's total internal reflection.

So refraction stops and total internal reflection starts at the angle of incidence where the angle of refraction approaches 90 degrees.

And the angle of refraction cannot increase beyond 90 degrees.

As in this diagram that's impossible, because if it did, then the ray would not leave the glass.

So there'd be no changing wave speed to cause any refraction in the first place.

And that's why refraction stops at the angle of incidence where the angle of refraction approaches 90 degrees.

So at a glass-air boundary, if the angle of incidence is greater than a certain angle, which we could just call C, the light will be totally internally reflected.

So if the angle of incidence is smaller than a certain angle for that boundary, then you get refraction out of the glass block as you might expect.

But if the angle of incidence is bigger than a certain angle for that specific boundary, then you get total internal reflection instead.

And that certain angle is the angle of incidence where the angle of reflection is 90 degrees.

And you should be aware that the partial reflections are often not shown in diagrams like this.

So total internal reflection can only happen at a boundary across which the wave speed increases.

For example, when light goes from glass into air because light is faster in air.

Total internal reflection can happen at the opposite kind of battery, at the opposite kind of boundary, for example, from air into glass.

And that's because light has been bending away from the normal, so light has been moving into a faster medium for the angle of refraction to reach 90 degrees before the angle of incidence does.

And here are some example angles at which total internal reflection starts to happen for some different boundaries.

And you can see that the boundary with the smallest critical angle is the boundary between diamond and air.

So what that means is that when light is incident on the internal surface of a diamond, at almost all angles of incidence from 24 degrees to 90 degrees, the light is totally internally reflected back into the diamond.

And that's why diamonds are so kind of shiny and sparkly.

So in which of the following situations could total internal reflection occur? When light is incident on an air-glass boundary at 20 degrees? When light is incident on an air-glass boundary at 60 degrees? When light is incident on a glass-air boundary at 20 degrees? When light is incident on a glass-air boundary at 60 degrees? And only one of those situations, could total internal reflection occur.

Can you work out which one? Five seconds, off you go.

Okay, so we're looking for a large angle of incidence.

So that's only gonna be B or D.

And the light has to be hitting the internal surface of the glass.

So that is light incident on a glass to air boundary in D, not the air to glass boundary 'cause total internal reflection can't happen going into the glass.

It can only happen when the light's trying to come out of the glass.

So that's a glass-air boundary, that's a large angle of incidence.

So the answer was D.

Well done if you've got that.

Okay, so it's time to do the final task of this lesson.

This task has two parts, here's the first.

So complete these diagrams to illustrate what can happen when light arrives to glass-air boundary with different angle of incidence that are labelled i.

So you can see in the first diagram, there's a small angle of incidence.

In the second diagram there's a larger angle of incidence, and in the third diagram there is a very large angle of incidence.

So the idea here is to show the three different possibilities of what can happen when light arrives at a glass-air boundary, with a small angle of incidence with a large angle of incidence.

And then there's a certain angle in the middle, which you should probably illustrate as well the crossover point.

So you should add brief notes to each diagram to explain what's happening.

So that's the first part of the task.

And the second part of the task is about this decorative lamp.

It's made of many clear tubes called optical fibres.

I'm sure you'll have heard of fibre optic, like fibre optic broadband.

That's where instead of like fibre optic cables send flashes of light down a cable instead of pulses of electricity.

Anyway, in this light, the optical fibres are lit from the bottom, the bottom edge and light travels along each fibre making the ends glow brightly.

So these are nice decorative lamps.

So the two parts of this part of the task.

Part A, you, I would like you to complete the diagram to show how a light array could be totally internally reflected along the fibre, and exit at the top end.

And then part B is a written explanation.

Explain why most light rays will reach the end of the optical fibre.

So you should pause the video now and have a good go at both parts of this task.

So part one, part two, which has 2a and 2b.

Pause the video now, off you go.

Right, I'll give you some feedback now, we'll start with part one of the task.

So it's probably a good idea to pause the video in a moment and just compare what you've done to these example diagrams and then make any improvements.

But hopefully you've got something along the lines of the first diagram showing the angle of incidence, a small angle of incidence, which is less than a certain angle.

The middle diagram shows the angle of incidence exactly at a certain angle where the angle of refraction is 90 degrees, after which you get to the third diagram with large angles of incident where you get total internal reflection because that angle of incidence is larger than a certain angle for that boundary.

So I'll pause the video now and make any improvements to your diagrams. Time's of feedback for part two of the task 2a, complete the diagram to show how a light ray could be totally internally reflected along the fibre.

So what you need to make sure is that your, your reflections look like they obey the law of reflection.

So the angle of incidence should be equal to the angle of reflection for each reflection along that clear internal edge of the fibre.

So it's important to realise that these fibres are completely transparent, they're not mirrored on the inside, it is just that you get mirror-like total internal reflection when the angle of incidence is large, which it is in this case.

And that's the reason really why most rays will reach the end of the optical fibre.

So here's an example explanation, check that yours is along the right lines.

The light ray (indistinct), in fact, any light ray, which enters the bottom of the fibre will almost always hit the sides of the fibre at large angle of incidence.

So it's almost always gonna be totally internally reflected and therefore reach the end of the optical fibre, which is why the end of the optical fibre looks very bright, because that's where lots and lots of light rays are coming from the ends of the fibre because they're all totally internally reflected along the length of the fibre.

And the light ray reflects at the same angle as the angle of incidence.

And this keeps happening all along the fibre.

So very well done if your answer was along those lines.

Here's a summary of the lesson.

At a boundary to a medium where wave speed increases, waves are refracted away from the normal.

But if the angle of incidence is too big, bigger than a certain angle for that boundary, then the light ray will undergo total internal reflection instead.

And that's shown in the diagrams with the middle diagram showing the certain angle at which that angle of reflection reaches its maximum, which is 90 degrees.

And for a glass-air boundary, that certain angle is about 42 degrees.

So for angle of incidence on a glass-air boundary above 42 degrees, you'll get total internal reflection.