AP Physics 1 & 2 Wave Phenomena

Hi everyone and welcome back to Educator.com.0000

I am Dan Fullerton and we are going to continue our study of waves today as we talk about wave phenomena.0002

Our objectives are going to be to understand qualitatively the Doppler Effect in order to explain why there is a frequency shift for both moving sources and moving observers, explain the origin of diffraction patterns resulting from single slits, double slits, and diffraction gratings...0008

...calculate locations of maxima and minima for these situations, and explain qualitatively why diffraction grating is better than a two slit grading for making accurate wavelength measurements.0023

Let us start with the Doppler Effect.0036

The shift in a waves observed frequency due to relative motion between the source and observer is known as the Doppler Effect.0039

When the source or observer are moving toward each other, the observer is going to perceive a shift to a higher frequency.0045

If the source and the observer are moving away from each other, there is a perceived shift at a lower frequency.0052

Practically, you guys have probably all experienced this before.0061

Have you ever heard a car go by you? As it comes toward you, you hear the higher frequency and as it goes away you hear a lower frequency.0065

That is where that shift in frequency comes from; as it is coming toward you, it is emitting waves at a specific frequency, but because it is coming toward you, you are picking up those frequency shifts, each of those wave fronts, a little bit more quickly; it is coming toward you as it is emitting them.0074

The same thing happens as it goes away from you because it is emitting them at a constant frequency -- but they are spreading out due to the added velocity of the object moving away from you -- you actually perceive them as being further apart having a lower frequency.0092

Higher frequency, or a pitch for sound, is perceived as it approaches and a lower frequency or pitch is perceived as it moves away.0108

Now the applications of this are all over the place.0116

Radar guns, for example -- As your car is traveling down the highway, a radar signal is sent from the radar gun, bounces off your car, comes back, and the shift in its frequency can tell people how fast you are moving or measuring the speed of a baseball for example.0119

Meteorology -- you have probably heard of Doppler radar, super Doppler radar, super-secret extra wonderful Doppler radar, depending on the station that you are listening to for your weather report.0135

Well, now what they are doing is they are sending the waves up into the atmosphere, looking for reflections back from the wave and that is also telling them then the speed of the winds at different altitudes.0145

It is used in medical imaging and even in astronomy.0156

We will talk a little bit more about astronomy and Doppler Effect on our next slide.0159

We use Doppler Effect in astronomy to analyze radiation from distant stars and galaxies, but the observed frequencies that we pick up are slightly lower than what scientists would predict based on knowing what the radiation is coming from.0164

It is usually shifted toward the red end of the spectrum, toward slightly lower frequencies, which is why it is sometimes called a 'red shift'.0179

What that means is that those objects must be moving away from the earth because they are shifted to lower frequencies.0189

The more distant the object, the greater that red shift happens to be.0196

What does that mean? The more distant objects in space are moving away from Earth faster than nearer objects.0200

The universe must be expanding; a big, big, big point in astronomy and studies of space.0208

Let us take a couple of examples around the Doppler Effect.0217

A car's horn produces a sound wave of constant frequency.0220

As the car speeds up going away from a stationary spectator, the sound wave detected by the spectator does what?0224

Well, of course it is going to go down in amplitude.0232

As things get further away, they get quieter, so we can get rid of (C) and (D).0234

What happens to the frequency as it is going away?0239

It drops and you get a lower frequency, so we have a decrease in amplitude and a decrease in frequency.0244

Another example -- The vertical lines in this diagram represent compressions in a sound wave of constant frequency propagating or traveling to the right from a speaker toward an observer at point (A).0254

Determine the wavelength of the sound wave.0266

The wavelength of a longitudinal wave is the distance between compressions, so this is going to be the wavelength if these lines here represent compressions, so from 1-2 1/2 meters, our wavelength must be 1.5 m.0270

Now the speaker is moved at constant speed toward the observer at point (A).0288

Compare the wavelength of the sound wave received by the observer while the speaker is moving to the wavelength observed while the speaker was at rest.0293

If the speaker is moving toward (A), you are going to have a shift to a higher frequency.0301

If frequency is going up -- V = F(λ) -- you have frequency going up, then we must have a wavelength going down, so frequency increases, our wavelength must be decreasing.0308

The wavelength appears shorter.0322

All right, let us talk a little bit about diffraction.0334

Diffraction is the bending of waves around obstacles or the spreading of waves as they pass through an opening.0337

You have probably observed this before as well.0342

Close the door to your room at night when all the lights are off in your room, but the hall light is on and you can see some of the light coming under the door, but yet you can see other things in the room besides just what is right at the crack at the door...0345

...the light is bending around there and there is also some reflection involved too, in all honesty.0356

Or you can hear sound around the corner of a building; that is what is happening as a wave bends around a corner.0362

Now it is most apparent when the wavelength is on the same order of magnitude as the obstacle or the opening.0370

Let us talk about single slit diffraction first.0376

If we have something like a monochromatic light source or a single wavelength, a single color on this side of our very small opening...0380

...we are going to put a screen some distance away and we will call this distance (l), so that we can see the effects of our light going through this opening.0397

We will say that the dimensions of our opening will be called (d), the width of our opening.0406

Let us call this amount of diffraction, how much it bends toward one side -- we will call this angle here theta.0415

What we are going to observe when we put this monochromatic light source behind a very small opening is on a screen some distance away, we are going to see a graph of intensity that is going to look kind of like this at the same point...0422

...the same spot directly behind the opening, we are going to have a fairly large spike in intensity and then as we go out, we are going to have some fringes of intensity...0435

...where we are going to call this order 1, order 2, order 3, and so on for the minima.0454

Now with single slit diffraction, we can relate all these together by saying that (d), the opening width, times the sine of that angle theta is going to be equal to (m) times the wavelength and that is going to give you your minima.0464

You can use this Geometry to figure out how far apart these minima are -- single slit diffraction.0485

Even more interesting is double slit diffraction.0493

This is an overview of Young's double slit experiment -- evidence that light is a wave because you can see it interfere with each other.0496

What we are going to do is we are going to take a monochromatic source again, but we are going to put it behind two slits that are situated some distance (d) apart from each other and again we will put a screen some distance (l) from our two slits.0504

When we do this, the pattern that we are going to see again is we are going to get that central bright spot again, but now we are going to have these fringes again and I will not draw them all the way out, but you get the idea.0523

We will call this our order 0, and at the maxima, we are going to call that order 1, the first order maximum, the second order maximum, and so on and so on in both directions.0537

If this is (l) and this is (d) -- what we were going to get our first order maximum -- let us draw this in here.0558

If we draw a line from here to the spot of our first order maximum, it looks like it is about here on the screen -- something like that -- we could call this angle θ.0565

Now d sin(θ), again, is going to equal m(λ), but we are going to manipulate this just a little bit further using what we know.0579

We are going to have to use a small angle approximation.0588

If we assume that we have an angle θ that is relatively small, then for small levels of θ, the sin(θ) is also equal to θ, but it is also equal to the tan(θ).0590

They are all very close to each other when θ is small, so if the sin(θ) is equal to the tan(θ) -- well tan(θ) is the opposite side -- we will call this distance (x) divided by the adjacent side (l).0602

This is our small angle approximation.0619

If that is the case then, we could write that d × x/l (sin(θ)), must equal m(λ).0622

Or then we can solve for (x), the distance between our central maximum and our first order maxima as x = m(λ)l/d.0634

Again, what is that? That is our distance between our bright spots.0652

What happens then, if instead of just looking at two slits, we take on a lot of slits?0667

Well, now we are going to have (d) as the separation between our slits again, but we have a lot of them and this is called a diffraction grating.0673

We still have our monochromatic source over here and we are going to get the same basic pattern as we had with double slit diffraction.0683

We still have our distance (l) to our screen; we still have our first order maxima in the spot, which is still (x); that is still θ, and d sin(θ) equals m(λ)...0690

...so all the same math works out, but what you are going to see over here on intensity is instead of those different peaks that are not quite as clear, we are going to see some very, very sharp and bright maxima and minima.0704

It is going to be very easy to see these and the closer those slits are to each other, the more expression, the clearer this is going to be.0718

This is very useful for determining wavelengths accurately.0727

If you do not know the wavelength of this light -- well you do not know wavelength -- you can find (m) -- set that to 1 -- you can find (x), you can find (l), and you can find (d).0730

If you know all of those other things you can calculate wavelength using a tool like this, which is much, much much clearer than using that double slit diffraction.0740

Let us take a look at this in an example.0750

We have red light of wavelength (700 nanometers) or λ = 700 × 10^-9 m is incident upon a double slit of separation (.0005 m).0753

That is going to be our (d), d = 0.0005.0767

What is the largest possible distance between a first order and third order bright spot on a screen that is 1 m from the slit? So l = 1 m.0771

Let us draw out what we have a little bit, what we expect to see on the screen.0782

If I had a graph of intensity, we are going to have our central order, our bright line, and we are also going to have our maxima that is something like that.0786

If we want the maximum distance between our first and third order of bright spots -- well, here we have 0; here is our first or our 1's are right there, so that is a first order and that is a first order; second, second; third and third.0804

If we want the maximum distance between our first and third order bright spots, we are going to need to go from third order, all the way here to first order, so we need to find that distance.0820

We know how to find that, the distance to the first order bright spot and added to the distance to our third order bright spot.0834

We will call this x3, the distance from our central max to the third order bright spot and we will call this x1.0841

That will give us the maximum possible distance between them.0848

If we wanted the minimum possible distance, well then we would have just gone with that, but that is not what it is asking in this problem.0852

Let us see if we cannot calculate that. Let us start with that x3.0860

x3 = m(λ)l/d and if we want the third order maximum...0864

...that is going to be 3 × 700 × 10^-9 m (wavelength) × 1 m (l)/.0005 m = x3 = 0.0042 m.0876

Now let us do that for x1.0896

The same idea gives us m(λ)/d and that is going to be 1 × 700 × 10^-9 × 1 m (length)/.0005, which implies then that x1 = 0.0014 m.0898

So the maximum distance between these two -- you have to add x3 to x1 to get this total distance, so the distance between them then is going to be 0.0042 + 0.0014 = 0.0056 m.0917

Part B -- What would happen to your answer if you replaced the double slit apparatus with the diffraction grating having the same slit separation?0939

Well the math stays the same, but remember with a diffraction grating, we are going to have much more defined bright spots, maxima and minima, so it is going to get sharper and brighter.0947

Let us do a problem where we are trying to determine the wavelength.0966

Monochromatic, single wavelength radiation is incident upon a diffraction grating with 2,000 lines/cm.0968

If the distance between the central and first order bright spot on a screen 1 m away is 11 cm, determine the wavelength of the radiation.0977

Well the first thing we need to do is let us figure out the distance between those slits.0986

If we have 2,000 lines/cm, then that implies that we have 0.01 m/2,000 lines.0992

Or when I do that division, I am going to find that d = 5 × 10^-6 m or 5 microns is going to be our separation between slits.1009

We also know that (l) is 1 m and our (x) here is 0.11 m or 11 cm, so now we will go down and start our calculations.1021

x = m(λ)l/d and in this case if we are solving for wavelength, λ then is going to be equal to dx/ml...1032

...where our d = 5 × 10^-6 m, x = 11 cm or 0.11 m/1 m (m- first order in our length).1045

I get a value of λ (wavelength) = 5.5 × 10^-7 m or about 550 nanometers.1058

We are using that diffraction grating to determine wavelength.1079

All right, let us just review with a couple of problems here.1084

A radar gun can determine the speed of a moving vehicle by measuring the difference in frequency between emitted and reflected radar waves.1087

This process illustrates which of the following: resonance, the Doppler Effect, diffraction, or refraction?1094

Of course, that is the Doppler Effect.1103

One more -- When observed from Earth, the wavelength of light emitted by a star is shifted toward the red end of the electromagnetic spectrum.1110

This red shift occurs because the star is -- well if it is shifting toward the red end of the spectrum, it is shifting to a lower frequency, so it must be moving away from Earth, which is another Doppler Effect problem.1119

Hopefully that gets you a good start on wave phenomena.1133

Thank you for visiting us here at Educator.com and make it a great day everyone. We will talk to you soon.1137