AP Physics 1 & 2 Refraction & Lenses

Hi everyone and welcome back to Educator.com.0000

I am Dan Fullerton and today we are going to be talking about refraction and lenses as we continue our study of optics.0003

Our goals for this lesson are going to be to calculate the wavelength and velocity of an electromagnetic wave in a medium with a specific index of refraction, to utilize Snell's Law to determine the angle of refraction of an incident ray, determine the critical angle for a ray moving from a high-index material to a lower-index material...0009

...analyzing system of lenses and mirrors with respect to object and image distances, heights, magnification and image type; and analyzing the behavior of light incident upon thin, optically transparent films.0027

Refraction -- When a wave reaches a boundary between media, part of the wave is reflected and part enters the new medium.0042

As the wave enters the new medium the speed of the wave can change but frequency, again, remains constant.0049

If the speed is changing and frequency is constant, the wavelength must also change.0055

Now the front of a wave has some actual width, so if the wave enters the new medium at an angle other than 90 degrees, not all of that wave front enters the new medium at the same time.0062

When that happens, you actually get some bending of the wave as it enters the new medium, and that is known as 'refraction'.0072

I like to use a marching band analogy to help understand why waves bend when they enter a new medium where their speed is changing.0083

Imagine you are in a line in a marching band connected by all your bandmates by a short string or maybe your arms are linked.0090

You march down the field in unison until you arrive at a mud pit, so here is the line of your band, you are moving down the field, and then you are going to find this mud pit.0098

Now the band members who reach the mud pit first, they have to slow down before the band members who reach it later, but you are all tied together.0113

As this happens, the folks on this side of the line gets slowed down first, the rest of the line keeps moving at the same speed until they hit the mud pit, and you are actually going to get a shifting because this side slows down before this side.0121

Well now the direction of the band is shifted until you come back out of it again, and then maybe it shifts a little bit more depending on who comes out first and speeds up first.0136

You get this overall shifting in direction due to the speed changing at different points on the wave front, all at slightly different times.0148

Now the index of refraction is a measure of how much light slows down in a material.0158

Bigger indices of refraction (n) indicates slower velocities.0163

In vacuum and air, all electromagnetic waves travel at the same speed (c) 3 × 10⁸ m/s, which corresponds to an index of refraction of 1.0167

In other materials, the electromagnetic waves slow down by some amount.0178

Now the ratio of the speed of an electromagnetic wave in a vacuum to the speed in another material is the index of refraction or the index of refraction is the speed in the vacuum divided by that speed in the new material; and typically, you would look up the index of refraction for some material.0182

Here we have a table of indices of refraction and they are given at a specific frequency 5.09 × 10¹⁴ Hz, because to a small extent index of refraction is a little bit of a function of frequency of the electromagnetic radiation.0201

Air and vacuum are 1, corn oil (1.47) and light slows down a little more, diamond (2.42) is a very high index of refraction -- the light slows down a lot in diamond -- and you can see a bunch of other materials here, too.0217

From air (1.0) the smallest index of refraction, and the highest on the table -- the slowest material -- is diamond here.0232

Now not only this index of refraction depend upon the medium, it also has a slight variation with frequency.0242

That frequency dependence is typically relatively small but it can be very useful.0247

Dispersion is the effect that you get because you have a slightly different index of refraction based on frequency.0252

This is what is responsible for the behavior of prisms.0260

If you have white light that is coming upon a prism -- well when it comes in there and you have white light that is all the colors of the rainbow, all the frequencies of visible light, it is bent once upon entering and it is bent again upon leaving.0263

It has two areas where it is going to be refracted -- incoming and exiting point of the prism.0275

Now at each of those points you have a slightly different index of refraction, a slightly different speed in that material for the different colors, therefore they are going to be bent slightly different amounts.0281

As the light enters you get a slight shift in the paths based on color and as it leaves you get another shift in the paths due to color.0291

Therefore, because of that change in index of refraction, you actually can see the light spreading out into the spectrum.0301

You have that red, orange, yellow, green, blue, indigo, violet.0308

How do you determine index of refraction -- our first example.0314

Let us say a light ray traveling in air enters a second medium and its speed slows down to 1.71 × 10⁸ m/s.0318

Find the index of refraction of this second medium.0325

Well index of refraction is the ratio of the speed of light in a vacuum to the speed of light in that new medium, so that is going to be 3 × 10⁸ m/s over 1.71 × 10⁸ m/s.0329

So our index of refraction in this problem would be right about 1.75.0344

Let us move on to another one. We have light in diamond and crown glass.0353

In which way does blue light change as it travels from diamond -- and from our table index of refraction of diamond is 2.42 -- into crown class where n = 1.52.0357

Well you are going from a higher index to a lower index.0370

Lower indexes imply faster media, so its speed must be increasing.0374

Let us talk a little bit more about the bending of light.0385

We can analyze that analytically using Snell's Law in the quantifiable sense.0387

The amount a light wave bends as it enters a new medium is given by the Law of Refraction which we call Snell's Law, and what it says is that the first index of medium 1 times the sine of the incident angle θ1 is equal to the index of refraction of the second medium times the refracted angle sin(θ2).0392

Always measuring these angles to the normal, we have θ1 (incident angle), θ2 (refracted angle), medium 1 has some index 1, medium 2 has some index 2.0416

Now if you are going from a low to a higher index material, the light ray is going to bend toward the normal.0427

And if you go from a low index to a high index, then it is going to bend away from the normal as it enters that faster medium.0433

The key here is to always measure angles in optics to the normal.0441

What you do not want to do is take this angle to the interface, or this angle; it is always measured to the normal to that interface.0452

All right. An example with the angle of refraction.0463

A ray of light in air is incident at an angle of 40 degrees on an air crown glass interface as shown.0466

What is the angle of refraction for this light ray?0473

Well over here, we are at air so the index is 1; we know that the incident angle is 40 degrees and we are going to a higher index material -- crown glass where n2 is 1.52 -- so it is going to bend toward the normal as it enter the new medium.0476

We are going from a lower index to a higher index, and this angle we will call θ2.0496

We can use Snell's Law to figure out what is going on -- n1 sin(θ1) = n2 sin(θ2), and we want the angle of refraction or θ2 so I would say then that sin(θ2) = n1/n2 sin(θ1).0504

If the sin(θ2) is equal to this, to get just θ2, θ2 will be the inverse sine of n1/n2 sin(θ1).0527

Or as I substitute in my variables, sin(θ2) equals the inverse sine of n1 (1) over n2 -- 1.52 sin(θ1) (40 degrees), therefore θ2 equals right around 25 degrees.0541

So that angle must be 25 degrees -- an application of Snell's Law.0564

When a light wave enters a new medium and is refracted, there must be a change in the light wave's what?0572

And we are going to choose everything that applies here.0576

Color? -- No, that is a function of frequency and frequency does not change when you enter the new medium.0579

Frequency? -- We know that is not it as it does not change when you enter a new medium.0586

Period? -- No, not so much. It is 1/frequency and frequency is constant.0590

Speed? -- Yes, speed can change and if V = F(λ), if F is constant, and speed is changing, then λ has to change as well -- wavelength.0594

So D and E must be our correct answers here.0605

All right. Let us talk about total internal reflection.0612

When light passes from a high index or a slower material to a low index or faster material, the light bends away from the normal, but when that angle of refraction reaches 90 degrees, that cannot get into that new interface anymore.0615

The refracted way would travel right on the boundary between the surfaces -- that does not make any sense.0628

That is going to occur at an angle of incidence that we are going to call the critical angle θ_c.0634

When that happens for all angles of incidence greater than that, what you are going to have is instead of refraction into the new media, you are going to get that light by reflecting back into the original media.0640

We call that total internal reflection (TIR), and if we want to analyze that by the Law of Reflection, then n1 sin(θ1) = n2 sin(θ2), but this occurs when θ2 is 90 degrees.0651

That is when you are going to get our critical angle, so if I rewrite this as n2 sin(90 degrees), I can solve to find the θ1 -- n1 sin(θ1) = n2 sin(90 degrees).0672

If I want sin(θ1), that is going to be n2/n1 sin(90 degrees), but the sin(90 degrees) is 1, therefore θ1 is going to be the inverse sine of just n2/n1.0691

That is the angle, the incoming angle, at which you are going to start to see total internal reflection, so that is why that is called the critical angle (θc), which is where that formula comes from.0707

The critical angle is the inverse sine of n2/n1, and it only occurs when you are going from a higher index or a slower material to a lower index or faster material.0720

Some applications of total internal reflection are fiber-optics -- light bending through a transparent media and it keeps reflecting down that media and not going out, a great way to transfer signals.0733

Sparkly diamonds -- diamonds are actually cut in such a manner where they want to enhance total internal reflection or something like that...0748

...so that what happens is when the light ray comes in, hopefully as it comes in, it keeps reflecting internally until it comes out the top where you see that nice, bright sparkle -- that is where that sparkle comes from.0757

It is useful in binoculars, and you can even see it as the mirror effect when you are looking up from underwater from a shallow angle.0771

As you see in this image here -- you are at a very shallow angle -- the turtles here -- you see the image of the turtle in the water as you are going from a higher index of water (1.33) to air (1) -- you are getting total internal reflection; you cannot see out of the water, so all you see is the reflection there.0777

So that is total internal reflection.0793

Let us take a look at the critical angle of water.0798

Determine that critical angle for a light ray exiting from water into air.0801

Well, the critical angle occurs when n1 sin(θ1) = n2 sin(90 degrees), where the sin(90 degrees) we know is going to be 1, or 1.33 sine of our critical angle (θc) is going to be equal to our n2...0804

...which is air times sin(90), which is 1, so n1 × 1 or 1 × 1 = 1.0823

Therefore our critical angle is going to be the inverse sine of 1/1.33 or an angle of about 48 degrees; it is the critical angle for a light ray exiting from water into air.0830

Any angles greater than 48 degrees, you are going to get total internal reflection.0848

Let us talk about thin lenses for a couple of minutes.0856

Thin lenses function based on the principle of refraction, whether we have convex lenses, or converging lenses, or concave lenses which are diverging lenses.0859

The convex lenses kind of look like that and diverging lenses typically have that rough shape.0868

Now they can have some variations there but that is the general idea.0874

You can also make lenses with mirrors, though they are typically a lot larger, more complex and a lot more expensive.0877

The advantage they have is you do not have any dispersion effects; you do not have what is known as chromatic aberration, which is where you get a shift in the focal point based on the frequency of the light going through the lens because mirrors operate on reflection...0885

...so you do not have to worry about the index of refraction varying as a function of frequency.0901

Now similar rules for ray tracing applies what we had when we were talking about mirrors, and the lens equation is still applicable.0908

1/F = 1/object distance (do) + 1/image distance (di), or 'If I do, I die'.0915

All right, let us take a look at a basic convex lens.0924

Rays parallel to the principal axis are refracted through the far focal point of the lens, so if I draw a ray that is parallel to the principal axis -- something like that -- it gets refracted through the far focal point.0927

A ray drawn from the object through the center of the lens passes through the center of the lens unbent, so we could draw that one as well.0952

And finally, a ray that travels from the object through the focal point, through the near focal point gets refracted parallel to the principal axis.0969

Notice again they all meet at the same point which is where you must have your image -- something like that -- and in this case, this would be a real image; it is reduced in size and it is inverted.0986

In the example below, the diagram shows an arrow placed in front of a converging lens.1007

Determine whether the image is real or virtual, erect or inverted.1012

Well we are going to draw the image over here where we see the light rays converging, so it would look something like that, and based on that it is pretty easy to see that that is going to be a real image because we have the image where the light rays are actually converging and it is inverted.1016

Another example -- An object is located .15 m from a convex lens that has a focal length of .1 m.1039

How far from the lens is the image formed?1046

Well our object distance is .15 m, our focal length is .1 m, and we do not know our image distance.1049

So I will use our lens equation, 1/F = 1/do + 1/di or rearrange this to find that 1/di is 1/focal length - 1/do, which is going to be 1/.1 m - 1/.15 m...1060

...therefore 1/di is going to be equal to 3.33 or the image distance is 1/3.33 or about 0.3 m.1080

All right. How about a concave lens or a diverging lens?1100

Well in this case, rays from the object parallel to the principal axis are refracted away from the principal axis on a line from the near focal point through the point where the ray intercepts the center of the lens.1104

Let us draw that right away and we will do that in blue.1115

So a ray from the object parallel to the principal axis -- we will start there.1119

What happens here is it is going to be refracted but it is going to diverge, therefore in order to figure out where it is going, we have to draw a virtual line from our focal point in order to figure out the rest of the ray's path.1129

It is going to diverge; it is going to go up to that direction following the line from the focal point through there.1147

Any ray from the object through a focal point is refracted parallel to the principal axis, so if we have one that is going through the focal point, we could have through the focal point and it will get refracted back as if it were going through the principal axis.1155

In the concave lenses, any ray from the object that goes right through the center of the lens also is going to go through unfettered.1169

Let us draw the one through the center of the lens too, that is probably our easiest ray to draw here to give us our second one to figure out where they are going to converge and where we will get our image.1177

So a ray right through the center of the lens continues, and notice here that these two diverge; they are never going to meet, which means we are going to have to come back here to figure out our virtual image.1190

They are crossing right here, therefore that is where we are going to get our image.1204

It is going to be reduced, it is going to be at an image distance (di) about like that, so there is our image.1210

Concave lenses -- you usually have to draw some of those virtual lines to figure out where you are going to get your image.1222

It is upright, virtual and reduced.1229

Which ray best represents the path of light ray (R) after it passes through the lens?1237

Well this is a diverging lens, so as it comes through here, which way is it going to go?1242

If we follow this back, we draw the line that would go through the near focal point, so our correct answer must be A.1250

Now when you have systems of optical elements, you will oftentimes have more than one lens or maybe lenses and mirrors and so on.1266

In those situations, all you do is you do it piece-wise -- find the image of the first optical element first and utilize it as the object of the second optical element.1273

Use the image of the second optical element as the object for the third optical element, and so on.1282

Continue to analyze all these until you have transversed the entire optical path.1289

How does that look in practice?1295

Well we will do an example with it. This may get a little bit complex to draw, but we will see what we can do here.1297

An object sits 50 cm from a convex lens of focal length (F1 = 20 cm) and a concave mirror of focal length (F2 = 14 cm) sits 60 m behind the lens.1302

Determine the position, type and size of the image.1314

The first thing I think I am going to do is I am going to ray trace this just to get an idea of what I am looking at.1318

For our convex lens, I am going to start by drawing our ray parallel to the principal axis and that should be refracted through the far focal point -- something like that.1324

Now I can draw my second ray -- I will draw that right through the center of the lens just because that is a really easy one to do.1344

That is just a straight line right through the center of the lens, and it looks like I am going to get my image right where those two meet, so I do not need the rest of that there.1350

I will draw my image right there. There is my image from my convex or converging lens.1360

That is going to be the object for the concave mirror.1370

So now to do the concave mirror, I will switch colors here, but I know that a ray going through the focal point is going to be reflected back parallel to the principal axis, so I will try drawing that next.1374

So we have something like that, and then it should come back parallel to the axis.1390

Now the other one we could draw is if we go directly through the center of the lens, we will come back off at the exact same angle.1399

I will do my second ray like that and it should come off at the same angle which will be -- if I draw that correctly, maybe right about there.1407

That should be at least reasonably close.1427

Now I have these two rays that cross over here so I can draw my image from the mirror right there.1429

I used the image of the first lens as the object for the second.1440

When I do this -- as I go through the math, I should predict that I am probably going to get an upright and real image.1445

All right, let us analyze this with some numbers.1451

For the convex lens, we know 1/F = 1/do + 1/di, therefore our focal length is .2 m so 1/.2 = 1/.5 m = 50 cm + 1/di...1456

...and if I solve for that I get that di = .333 m, so my image distance here must be .333 m.1478

That is going to be my object distance -- well 60 cm minus that will be my object distance for the mirror.1495

That is going to be .267, so as I analyze the mirror now I have 1/F = 1/do + 1/di...1504

...where my focal length now is 1/.14 = 1/60 cm - [33.3] m (.267) + 1/di...1518

...therefore my image distance is going to be .295 m for my mirror.1535

That means that for my mirror -- let us see -- that distance there must be our .295 m.1544

We have now located our image specifically, so let us see about magnification.1554

Magnification due to our convex lens, well that is -di/do which is going to be -.333 over our object distance, which was .5 or -.666.1559

And magnification for our mirror (m2), also -di/do, but now we need to use the mirror's image and object distances, which is -.295 m for our image distance minus the object distance, which was .267 or -1.1.1575

My total magnification is going to be the magnification of the first element times the magnification of the second element.1599

That is going to be m1 × m2, which will be -.666 × -1.1 = .7326.1606

Again we have a positive number; it is an upright image; it is real and it is slightly reduced, which we expected based on our ray tracing exercise.1627

All right. Let us talk a little bit about thin film interference.1642

When light is incident upon a thin film, some light is reflected and some is transmitted into the film.1645

If the transmitted light is again reflected, it travels back out of the film and it can interfere with itself.1650

If we have light coming here into an interface, part is reflected, part is refracted or transmitted into the film, it travels some distance down and back up, and it can then interfere with itself.1656

How far has it traveled? It has traveled twice the thickness of the film, down and then back.1667

If we want to find out where we have this interference, the total distance it travels, then (2T) is going to be equal to the order (m) times the wavelength of that ray in the film -- not the original wavelength, but the wavelength in the film.1674

Now the only thing that we have to keep in mind here is that there is a little trick to these.1689

As we get a phase change, things switch -- every reflection from low index to high index materials.1693

If you have an even number of phase changes, count all your phase changes and if you have an even number you use m = 1, 2, 3, and so on.1701

If you have an odd number of phase changes, you use 1/2, 3/2, 5/2 and so on.1709

Again it sounds kind of complicated, but it is a lot easier to deal with when you actually see it in practice with an example, so let us do that.1715

Light is incident in air perpendicular to a thin film of glycerol on top of water.1723

What minimum thickness of glycerol gives the reflected light a green, 532 nanometer green color?1728

As we do this, we realize we have an incident light ray, part is going to be reflected, part is going to be transmitted and then it is going to come back out traveling this distance of 2 × T.1736

So first thing I am going to do is count the phase changes. Where do we go from low to high index?1748

And I see one spot right there where we go from 1 to 1.47, the only place where we are talking about having a phase change.1755

We have one of those, therefore when we use our (m), we are going to have to use m = 1/2.1764

Next up is let us see if we can figure out what the wavelength is going to be in the film.1774

We know the wavelength up here is 532 nanometers, but we want to know the wavelength in the glycerol.1779

To do that I am going to go back to my wave equation V = F(λ), but we are talking about light where N = C/V, so I could write that V = C/N and replace velocity with C/N = F(λ).1785

Now I am just going to rearrange this to get all the constants on the same side.1803

Frequency is a constant and so is C, so I will write this as C/F = N(λ).1807

All that is a constant so anywhere in this, these different materials, N(λ) must be the same.1817

So I am going to write that N1(λ)1 = N2(λ)2, where this is medium1 and that is medium2.1822

If I then solve for λ2, the wavelength in the glycerol, that implies that λ2 = N1(λ)1/N2 or λ2 = N1 (1) × λ1 (532 nanometers)/N2 (1.47)...1832

...or I will get 362 nanometers as the wavelength in the glycerol.1857

Well now finally, let us go figure out what that thickness has to be for the interference that will give us that green color.1864

All right. To do that, 2T = m(λ).1872

In this case we know m = 1/2, because we are looking for the minimum thickness and we have an odd number of phase changes (1), so m = 1/2 and the λ in the film, the wavelength, is 362 nanometers.1878

Therefore, T = m (1/2) × wavelength (362 nanometers) and I have to divide by 2 to get the 2 out of there, or 1/4 of 362 nanometers which is about 90.5 nanometers.1895

So what minimum thickness of glycerol gives the reflected light a green color? 90.5 nanometers of glycerol.1916

Let us take a look at a problem where we are looking for the wavelength of light in diamond.1928

A beam of monochromatic light has a wavelength of 5.89 × 10 ^-7 m in there. Find its wavelength in diamond.1932

Let us start by talking about the index of the first material in air, which is 1, and the wavelength in air is 5.89 × 10^-7 m.1941

The index of diamond (N2) -- to look up on a table, that is 2.42 and we are trying to find what it's new wavelength is.1954

We know that N = C/V and if V = F(λ), we could write this as N = C/F(λ) and again, what I am going to do is I am going to arrange these so I get all the constants on the same side for C/F = N(λ).1966

If those are constant then N(λ) must be the same anywhere in the problem and at any medium.1985

So I could write then that N1(λ)1 = N2(λ)2.1990

Let us just solve for λ2 then and our wavelength in diamond is N1/N2(λ)1.1997

N1 = 1, N2 = 2.42 and λ1 = 5.89 × 10^-7 m, which gives me an answer of about 2.43 × 10^-7 m, or 243 nanometers.2007

The wavelength has gotten considerably shorter in the diamond, which we would expect because it is going slower.2029

All right. Which diagram best represents the behavior of a ray of monochromatic light in air incident on a block of crown glass?2038

As we look at this one, we know that when light strikes a surface we are going to have part of the ray reflected and part refracted.2048

Here it is all refracted, here it looks like it is all reflected and at a goofy angle -- that is just off -- and here it is being refracted in two different directions in the wrong way, so the correct answer here must be Number 4.2056

We have reflection following the Law of Reflection and we are also bending toward the normal as we go into a higher index material, parts refracted and parts reflected, so 4 is our best answer.2071

A convex lens forms a real image that is four times larger than the object. If the image is located .16 m from the lens, what is the object distance?2085

Well we know that the magnification must be 4 and the image is located .16 m from the lens.2095

It is a real image; it is a convex lens and that is going to tell us that the image distance is going to have to be less than 0 -- there is the trick there.2103

So if it is located .16 m from the lens that means di = -.16 m and the magnification equation: m = -di/do...2111

...therefore do = -di/m which will be -.16/4 or 0.04 m.2127

All right, an example with a diverging lens.2145

In the diagram below, parallel light rays in air diverge as a result of interacting with an optical device. 2154 What could that device be?2148

Well a convex glass lens is converging -- that is not it.2156

Rectangular glass block -- that is not going to do much of anything.2161

A plain mirror -- that is not going to give us diverging on the opposite side.2165

It has to be a concave glass lens, something sort of that shape; diverging lens is another name for a concave lens.2169

Let us take a look at which glass lens in air can produce an enlarged, real image of an object.2183

Well if we want an enlarged, real image of an object, we need to have a convex lens.2190

Notice that these three are all diverging lenses, the only answer that will do that is 4.2196

Convex lenses can produce enlarged, real images.2201

The diagram here shows an object placed between 1 and 2 focal lengths from a converging lens.2209

The image of the object produced by the lenses -- well to figure this out let us do a little bit of ray tracing -- if we have an object through a converging lens we know that the ray that is coming in parallel to the principal axis right there, is going to be refracted through the far focal point.2215

We can also do the other easy one and draw the line right through the center of the lens.2236

It looks like if I extend these just a touch further, we are going to get an image somewhere way out here, so there is our image.2244

What do we know about it? Well, we are actually having the light rays converge where our image is, therefore it must be real and it has got to be inverted.2257

What is the answer? A.2266

All right, one last problem.2269

A convex lens has a focal length of 0.8 m and a light ray travels from the object to the lens parallel to the principal axis.2271

Which line best represents the path of the ray after it leaves the lens?2282

If the light ray is coming in parallel, which one is going to show its shape after it leaves the lens?2287

That has to be number 3 there -- coming in parallel you get retracted through the far focal point.2293

How far from the lens is the image formed?2302

To do that, we are going to use our lens equation (1/F = 1/do + 1/di), and that is going to imply then that 1/di = 1/F - 1/do...2305

...which our image distance is 1/.08 - 1/.1, therefore 1/di = 2.5...2322

... which gives us an image distance of 1/2.5 or 0.4 m.2340

Which one of these best explains the path of the light through the lens: diffraction, dispersion, reflection or refraction?2349

Well the path of the light through the lens is governed by refraction, the bending of the light and that is why lenses work, so our best answer there is C, refraction.2361

Hopefully that gets you started with refraction and lenses.2373

Next up, we will start talking about modern and nuclear physics.2375

Thanks so much for your time everyone and make it a great day.2379