Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:

Farewell Special
Bring a Friend!

Instructor/speaker: Prof. Walter Lewin

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• iTunes U (MP4 - 103MB)
• Internet Archive (MP4 - 203MB)

  » Download this transcript (PDF)

  Today I would like to talk to you about some of the research that I did during my early days at MIT.

  It's a long time ago.

  I got my Ph.

  D in the Netherlands on nuclear physics and I came over to MIT in 1966.

  I was supposed to be here only for one year.

  I had a one-year postdoc position.

  But I loved it so much I never left.

  I changed fields.

  I joined the research group of Professor Bruno Rossi here at MIT, I changed from nuclear physics to x-ray astronomy.

  X-ray astronomy speaks for itself.

  You're trying to do astronomy in x-rays.

  You cannot see any x-rays from the ground because the earth atmosphere absorbs them completely.

  So you have to go outside the atmosphere unlike optical astronomy and radio astronomy which you can do from the ground.

  When I use the word x-rays I'm thinking of the kind of x-rays that your dentist would be using, medical purposes, about 1 to 50 kilo electron-volts.

  And since all of you took 8.02 you should know by now what a kilo electron volt is.

  Uh, optical light is 2 electron volts, where x-rays, way more energetic than optical light.

  Uh, during the Second World War Wernher von Braun in Peenemunde developed under Hitler Germany destructive rockets.

  They were used to destroy the Allies.

  To destroy you and me.

  And after the war, around 1948, the Americans used these rockets to do science.

  They also got Werner von Braun over to this country and for reasons unknown to me he became a hero.

  They tried to observe x-rays from the solar system.

  And they found indeed that the sun emits x-rays.

  The sun is very close.

  So you may say well that's not a surprise.

  But it's really very unusual because to create x-rays you need extremely high temperatures, which we didn't think existed on the sun.

  And if you take the power that the sun puts out in x-rays, this is joules per second now, and you divide that by the power in the optical and the infrared light of the sun, this symbol stands for sun, that ratio is about 10 to the -7.

  So you must conclude that the sun emits largely optical light and infrared and that the amount of x-rays is a modest byproduct.

  Interesting as it is all by itself.

  In 1962 several scientists here in Cambridge, Massachusetts, among them Bruno Rossi and Riccardo Giacconi and Herb Gursky, made an attempt to observe x-rays from stars outside our solar system.

  The odds were strongly against them.

  The detectors were not sensitive enough.

  If you take the sun and you bring the sun to the nearest stars, which is a distance say of 10 light-years, then there would be no hope that you would be able to detect x-rays from an object like the sun.

  In fact the detectors were too insensitive by about a factor of one billion.

  They tried anyhow and they were successful.

  They did indeed find to everyone's surprise and joy, they found x-rays from at least one object outside the solar system.

  This object was later called SCO X-1.

  SCO stands for the constellation Scorpio in the sky and X for x-rays and 1 for the first source observed in that constellation.

  We now know that this object is a faint blue star.

  And what is extremely special about the object SCO X-1 is that if you take the ratio x-ray power over optical power then that ratio is about 10 to the 3rd.

  Compare that with the sun.

  This object, we had no clue what it was in those days, primarily emit x-rays, and the optical emission is a byproduct.

  Whereas with the sun it is reversed.

  And so the burning question was in those days, what kind of animal is this?

  It must be a totally different beast.

  Something very different from our sun.

  And when I came to MIT in 1966, there were six sources known outside our solar system.

  And they were all discovered with rockets.

  The rockets in those days could spend about five minutes above the earth atmosphere and they would quickly make a scan over the sky, five minutes, that's all they had.

  And I joined the group of uh George Clark, who is still at MIT, uh he was doing x-ray astronomy from very high-flying balloons, very close to the top of the atmosphere, and the advantage of balloons was that you could observe the sky for many, many hours, if you're lucky sometimes even a day or more.

  But on the other hand, since there is always a little bit of atmosphere left above you, even though there's very little, there is still some left, the x-rays are absorbed, almost all x-rays below 20 kilo electron-volts would be absorbed, and we would not be able to see them.

  But of course the compensation was that we could look at the sky for many, many, many hours.

  Nowadays no one is doing these balloon observations anymore.

  No more rocket observations.

  Everything is done of course from satellites.

  So when I came to MIT, together with George Clark, I developed new x-ray detectors for these balloon observations.

  Many graduate students were involved.

  Many undergraduates.

  It would take about two years to build a telescope.

  To give you a rough idea it would take a million dollars in terms of 1966 dollars, and the weight of such a telescope would be roughly 1000 kilograms.

  The balloons in those days would cost about $100,000 to get us up to these high altitudes, and we would need about $80,000 of helium, and you will see some slides of that.

  We have to go to altitudes of about 140000 feet.

  We had huge balloons for that.

  You will see one.

  They have diameters of about 600 feet.

  And the material was polyethylene.

  Extremely thin to make them light-weight so that they can go high.

  The thickness of that polyethylene was about half of one-thousandth of an inch.

  Which is thinner than the saran wrap that you have in the kitchen.

  It is thinner than cigarette paper.

  A very risky business to fly these balloons.

  No guarantee of course that they would work.

  You pay your money.

  If they work that's great.

  If they don't work that's just tough luck.

  There is a good chance that you have a failure when you launch the balloon.

  They're very fragile.

  There could be damage right at the launch.

  But even if they make it up in the atmosphere they have to go through the tropopause, near about a hundred thousand feet where it's very, very cold, the balloons get brittle, and then they can burst.

  And that of course would be the end then of that balloon flight.

  And that could also be the end of a Ph.D thesis.

  Because that all these flights of course were connected with research and therefore with Ph.

  D work and so the tension during these early phases of the launch were always extremely high.

  Sometimes even unbearable.

  So now I would like to show you some slides, which will give you a good idea of what these expeditions were like.

  Oh, yeah, a classic problem.

  This is nice that these -- ah, now they work.

  All right, so if I can have the first slide, you see here Jim and Pat who at the time were undergraduates, they are now both Ph.Ds, and they are working there, very tedious work, trying to put the electronics together.

  You may think that science is not very romantic.

  But I can assure you it is.

  They fell in love.

  They married.

  They have kids.

  And that's the way it sometimes goes in life.

  And so here you see the plant in Texas where these huge balloons were made.

  Uh, balloons are put together sort of like the -- the way that the tangerine is put together.

  At the surface you see these gores of the balloon.

  And the sealing of these gores to make up the balloon were -- was only done by women.

  Only women were allowed to work there.

  Has nothing to do with sex discrimination of any kind.

  It just turned out that women were more patient.

  They did the work better.

  They make way fewer mistakes than men did.

  That's the way it goes sometimes in life.

  Here you see balloon coming out of the box.

  Nicely protected in a plastic cover.

  And we also have here cloth on the -- on the grass because the balloon is so thin that it would certainly get damaged if it touches the grass, it's enormously thin.

  This was not my balloon.

  Uh we were worried that there was something wrong with it.

  You can see here the concern.

  They thought it was a -- there was a hole in the balloon.

  And that if there is a hole in the balloon there's just nothing you can do about it anymore.

  You can't patch it because the hole is almost always through many many layers.

  What you're looking at here is hundreds of layers of balloon that are folded together.

  But as I said, since it wasn't my balloon I wasn't too worried, but of course it's never nice if you see a failure of your colleagues.

  Now I bring you to the desert town Alice Springs in Australia.

  Right at the heart of Australia.

  And now you get a pretty good idea of what it's like.

  Here you see the launch truck.

  The telescope is there.

  And then you see this enormously big balloon.

  All of it is empty now and most of this will stay empty.

  This is the roller arm which holds this part down.

  This is the only part that will be inflated.

  And here you see the helium truck.

  And here you see the inflation tubes.

  And we will let helium in from both sides which will then gradually begin to fill this top part of the balloon.

  And you see here the roller arm in detail.

  The roller arm is very important because when this part of the balloon is being filled it wants to lift, it wants to go up, and of course you have to keep it down, you have to keep it under control.

  And so this roller arm and this -- this car is loaded down with concrete.

  It's very heavy.

  And then just before the launch this roller arm by command is [fweet] flipped over, and then as you will see later then the balloon will make it up.

  And here you see the early part of the inflation.

  Helium comes in from both sides.

  And so we -- we fly these balloons almost always early morning because then the winds are very calm.

  You need extremely reliable winds.

  You need to know the direction very well.

  And the winds should be no more than something like three or four miles per hour.

  If they are stronger you would lose the balloon.

  You see here these gores that I mentioned to you earlier.

  Where the sun is behind the balloon.

  Here the bubble is nearly fully inflated now.

  Here it's still going on.

  Still going on, the inflation.

  But we are very close to the end of the inflation.

  Here is the roller arm and then in this direction here, 500 feet or so down is the payload with the truck.

  We're now very close to a launch.

  We're still in Alice Springs.

  This is -- was my graduate student Jeff McClintock at the time.

  He's now Dr.

  McClintock.

  Here you see radar reflectors which allows us to follow the balloon -- a radar.

  Here you see the telescope hanging on the launch truck.

  Here is the roller arm.

  All this is empty.

  And here you see the parachute.

  We have here a connection between the parachute and the bottom of the balloon.

  And we can control that on radio command.

  We can separate that so that the telescope safely comes back to earth.

  At least that's the idea on paper.

  And so now you see the release of the bubble.

  So the roller arm is up and this bubble now takes off.

  This is an incredibly fantastic moment.

  This is really butterflies in your stomach and ants in your pants.

  This is the moment that balloon can easily fail.

  Very thin material, the helium goes up, reflects against the top, is pushed back again, you get this peculiar mushroom shape, it makes an enormous sound like a storm.

  The idea now is that this balloon will go higher and higher in the sky.

  Will pick up all this empty part.

  This is not inflated.

  As the balloon goes up in the atmosphere the atmospheric pressure will go down.

  And the helium will expand and will fill the balloon.

  And the -- the trick now is for this truck to manipulate, to maneuver itself under what we call the bubble.

  And therefore the wind has to be in this direction so that the balloon comes to the truck.

  And then the truck tries to get straight under the balloon.

  And then the payload will be released here.

  Here you see a close-up of this mushroom.

  You can actually see this reflection of the helium going up and coming back.

  You can also see these gores very clearly.

  It's tedious work.

  By these women who have to seal these balloons.

  Enormous amount of labor goes into it.

  Amount of helium as I said earlier is about $80,000.

  About the same price of the balloon.

  And here it goes higher.

  We're in Alice Springs.

  The cover is falling off.

  Balloon is going up.

  See the engine is already running.

  The truck cannot move yet because if it started to move this part of the balloon would slide over the cloth.

  There would be friction and there would be holes in the balloon.

  So this truck has to wait until all of this is off the ground.

  Going higher.

  And I'm now so close to the balloon that I couldn't continue my picture-taking from Alice Springs.

  So I will jump back to an earlier flight in the United States.

  We flew these balloons in the United States from a town called Palestine, Texas.

  And so you will see then the remaining part of the flight from Palestine, Texas.

  So the balloon is now completely off the ground.

  See a little bit of gas, well it's not so little, but it looks very little compared to the size of this balloon.

  You see the parachute here and here then is the connection which on radio command we can separate.

  So now this is a very crucial moment.

  The person in charge on this launch truck has probably driven the truck to get straight under the balloon.

  And when it's straight under there they will allow the payload to go free.

  The payload is attached to this truck.

  If the balloon is too far ahead and the payload is released it will pendulum into the ground.

  And if you release it too early then of course the payload will pendulum back into the launch truck.

  Both would be a disaster.

  If the pull of the balloon is not enough, for instance if a hole developed during the launch, so if the tension is not strong enough, you would release the payload, it would go bang, back to the ground.

  So all these factors have to be taken into account, and then finally the person in charge commits to a launch.

  And then there it goes.

  All the way empty.

  Here you see the helium.

  The parachute.

  And you see the -- the payload.

  And here you see the balloon at 150000 feet, 45 kilometers high in the sky.

  The helium has now expanded.

  The balloon is fully inflated.

  And you can look straight through it.

  It's only half of one-thousandth of an inch of polyethylene.

  And these are huge ducts which have openings of about ten meters each.

  And they are there because the balloon cannot stand any over-pressure.

  If there is any over-pressure the balloon would pop and so when the balloon keeps rising and rising and rising when it reaches -- reaches its maximum volume the helium would escape at the bottom.

  That's the idea of these ducts.

  Here you see George Ricker who was graduate student, my graduate student at the time.

  This is in Australia.

  He is now Dr.

  Ricker.

  He's still at MIT.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher

Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Course Co-Administrators:
Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao

Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:
Doppler Effect
The Big Bang
Cosmology
Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

• iTunes U (MP4 - 103MB)
• Internet Archive (MP4 - 203MB)» Download this transcript (PDF)
  Today I want to talk with you about Doppler effect, and I will start with the Doppler effect of sound which many of you perhaps remember from your high school physics.
  If a source of sound moves towards you or if you move towards a source of sound, you hear an increase in the pitch.
  And if you move away from each other you hear a decrease of a pitch.
  Let this be the transmitter of sounds and this is the receiver of sound, it could be you, your ears.
  And suppose this is the velocity of the transmitter and this is the velocity of the receiver.
  And V should be larger than 0 if the velocity is in the direction.
  And in the equations what follow, smaller than zero it is in this direction.
  The frequency that the receiver will experience, will hear if you like that word, that frequency I call F prime.
  And F is the frequency as it is transmitted by the transmitter.
  And that F prime is F times the speed of sound minus V receiver divided by the speed of sound minus V of the transmitter.
  So this is known as the Doppler shift equation.
  If you have volume one of Giancoli you can look it up there as well.
  Suppose you are not moving at all.
  You are sitting still.
  So V receiver is 0.
  But I move towards you with 1 meter per second.
  If I move towards you then F prime will be larger than F.
  If I move away from you with 1 meter per second then F prime will be smaller than F.
  The speed of sound is 340 meters per second.
  So if F, which is the frequency that I will produce, is 4000 hertz, then if I move to you with 1 meter per second, which I'm going to try to do, then the frequency that you will experience is about 4012 hertz.
  It's up by 0.3 percent.
  Which is that ratio one divided by 340.
  And if I move away from you with 1 meter per second, then the frequency that you will hear is about 12 hertz lower.
  So you hear a lower pitch.
  About 0.3 percent lower.
  I have here a tuning fork.
  Tuning fork is 4000 hertz.
  I will bang it and I will try to move my hand towards you one meter per second roughly.
  That's what I calculated it roughly is.
  Move it away from you, towards you, away from you, as long as the sound lasts.
  You will hear the pitch change from 4012 to 3988.
  Very noticeable.
  Have you heard it?
  Who has heard clearly the Doppler shift, raise your hands, please?
  OK.
  Chee chee chee chee it's very clear.
  Increased fre- frequency and then when I move my hands, away a lower pitch.
  Now you may think that it makes no difference whether I move towards you or whether you move towards me.
  And that is indeed true if the speeds are very small compared to the speed of sound.
  But it is not true anymore when we approach the speed of sound.
  As an example, if you move away from me with the speed of sound, you will never hear me.
  Because the sound will never catch up with you, and so F prime is 0.
  And you can indeed confirm that with this equation.
  But if I moved away from you with the speed of sound, for sure the sound will reach with you.
  And the frequency that you will hear is only half of the one that I produce.
  So there's a huge asymmetry.
  Big difference whether I move or whether you move.
  So I now want to turn towards electromagnetic radiation.
  There is also a Doppler shift in electromagnetic radiation.
  If you see a traffic light red and you approach it with high enough speed you will experience a higher frequency and then you will see the wavelengths shorter than red and you may even think it's green.
  You may even go through that traffic light.
  To calculate the proper relation between F prime and F requires special relativity.
  And so I will give you the final result.
  F prime is the one that you receive.
  F is the one that is emitted by the transmitter.
  And we get here then 1 - beta divided by 1 + beta to the power one-half.
  And beta is V over C, C being the speed of light, and V being the s- speed, the relative speed between the transmitter and you.
  If beta is larger than 0, you are receding from each other in this equation.
  If beta is smaller than 0, you are approaching each other.
  You may wonder why we don't make a distinction now between the transmitter on the one hand, the velocity, and the receiver on the other hand.
  There's only one beta.
  Well, that is typical for special relativity.
  What counts is only relative motion.
  There is no such thing as absolute motion.
  The question are you moving relative to me or I relative to you is an illegal question in special relativity.
  What counts is only relative motion.
  If we are in vacuum, then lambda = C / F and so lambda prime = C / F prime.
  Lambda prime is now the wavelength that you receive and lambda is the wavelength that was emitted by the -- by the source.
  So I can substitute in here, in this F, C / lambda which is more commonly done.
  So this Doppler shift equation for electromagnetic radiation is more common given in terms of lambda.
  But of course the two are identical.
  And then you get now 1+ beta upstairs divided by 1- beta to the power one-half.
  The velocity, there if I'm completely honest with you, is the radial velocity.
  If you are here and here is the source of emission and if the relative velocity between the two of you were this, then it is this component, this angle is theta, this component which is V cosine theta, which we call the radial velocity, that is really the velocity which is in that equation.
  Police cars measure your speed with radar.
  They reflect the radar off your car and they measure the change in frequency as the radar is reflected.
  That gives a Doppler shift because of your speed and that's the way they determine the speed of your car to a very high degree of accuracy.
  You can imagine that in astronomy Doppler shift plays a key role.
  Because we can measure the radial velocities of stars relative to us.
  Most stellar spectra show discrete frequencies, discrete wavelength, which result from atoms and molecules in the atmosphere of the stars.
  Last lecture I showed you with your own gratings a neon light source and I convinced you that there were discrete frequencies and discrete wavelengths emitted by the neon.
  If a particular discrete wavelength, for instance in our own laboratory, would be 5000 Angstrom, I look at the star, and I see that that wavelength is longer, lambda prime is larger than lambda, then I conclude -- lambda prime is larger than lambda, that means the wavelength the way I observe it is shifted towards longer wavelength, is shifted in the direction of the red, and we call that redshift.
  It means that we are receding from each other.
  If however I measure lambda prime to be smaller than lambda, so lambda prime smaller than lambda, we call that blueshift in astronomy, and it means that we are approaching each other.
  And so we make reference to the direction in the spectrum where the lines are moving.
  I can give you a simple example.
  I looked up for the star Delta Leporis what the redshift is.
  There is a line that most stars show in their spectrum which is due to calcium, it even has a particular name, I think it's called the calcium K line, but that's not so important, the name.
  In our own laboratory, lambda is known to a high degree of accuracy, is 3933.664 Angstroms.
  We look at the star and we recognize without a doubt that that's due to calcium in the atmosphere of the star and we find that lambda prime is 1.298 Angstroms higher than lambda.
  So lambda prime is larger than lambda.
  So there is redshift and so we are receding from each other.
  I go to that equation.
  I substitute lambda prime and lambda in there and I find that beta equals +3.3 times 10 to the -4.
  The + for beta indeed confirms that we are receding, that our relative velocity is away from each other, and I find therefore that the radial velocity -- I stress it is the radial component of our velocity is then beta times C and that turns out to be approximately 99 kilometers per second.
  So I have measured now the relative velocity, radial velocity, between the star and me, and the question whether the star is moving away from me or I move away from the star is an irrelevant question, it is always the relative velocity that matters.
  How can I measure the wavelength shifts so accurately that we can see the difference of 1.3 angstroms out of 4000?
  The way that it's done is that you observe the starlight and you make a spectrum and at the same time you make a spectrum of light sources in the laboratory with well-known and well-calibrated wavelength.
  Suppose there were some neon in the atmosphere of a star.
  Then you could compare the neon light the way we looked at it last lecture.
  You could compare it with the wavelength that you see from the star and you can see very, very small shifts.
  You make a relative measurement.
  So you need spectrometers with very high spectral resolution.
  So there was a big industry in the early twentieth century to measure these relative velocities of stars.
  And their speeds were typically 100, 200 kilometers per second.
  Not unlike the star that I just calculated for you.
  Some of those stars relative to us are approaching.
  Other stars are receding in our galaxy.
  But it was Slipher in the 1920s who observed the redshift of some nebulae which were believed at the time to be in our own galaxy and he found that they were -- had a very high velocity of up to 1500 kilometers per second, and they were always moving away from us.
  And it was found shortly after that that these nebulae were not in our own galaxy but that they were galaxies in their own right.
  So they were collections of about 10 billion stars just like our own galaxy.
  And so when you take a spectrum of those galaxies, then of course you get the average of millions and millions of stars, but that still would allow you then to calculate the redshift, the average red shift, of the galaxy, and therefore its velocity.
  And Hubble, the famous astronomer after which the Hubble space telescope is named, and Humason made a very courageous attempt to measure also the distance to these galaxies.
  They knew the velocities.
  That was easy because they knew the redshifts.
  The distance determinations in astronomy is a can of worms.
  And I will spare you the details about the distance determinations.
  But Hubble made a spectacular discovery.
  He found a linear relation between the velocity and the distances.
  And we know this as Hubble's law.
  And Hubble's law is that the velocity is a constant which is now named after Hubble, capital H, times D.
  And the modern value for H, the modern value for H is 72 kilometers per second per megaparsec.
  What is a megaparsec?
  A megaparsec is a distance.
  In astronomy we don't deal with inches, we don't deal with kilometers, that is just not big enough, we deal with parsecs and megaparsecs.
  And one megaparsec is 3.26 times 10 to the 6 light-years.
  And if you want that in kilometers, it's not unreasonable question, it's about 3.1 times 10 to the 19 kilometers.
  So I could calculate for a specific galaxy that I have in mind, I can calculate the distance if I know the red shift.
  I have a particular galaxy in mind for which lambda prime -- for which lambda prime is 1.0033 times lambda.
  So notice again that the wavelength that I receive is indeed longer than lambda, so there is a redshift.
  I go to my Doppler shift equation which is this one.
  I calculate beta.
  One equation with one unknown, can solve for beta.
  And I find now that V is 5000 kilometers per second.
  Very straightforward, nothing special, very easy calculation.
  But now with Hubble's law I can calculate what D is.
  Because D now is the velocity which is 5000 kilometers per second divided by that 72 and that then is approximately 69 megaparsec.
  Again we have the distance if we do it in these units in megaparsecs.
  That's about 225 million light-years.
  And so the object is about 225 million light-years away from us.
  So it took the light 225 million years to reach us.
  So when you see light from this object you're looking back in time.
  And if you have a galaxy which is twice as far away as this one, then the velocity would be twice as high.
  And they're always receding relative to us.
  I'd like to show you now some spectra of three galaxies.
  Can I have the first slide, John?
  All right, you see here a galaxy and here you see the spectrum of that galaxy.
  That may not be very impressive to you.
  The lines that are being recognized to be due to calcium K and calcium H are these two dark lines.
  Some of you may not even be able to see them.
  And this is the comparison spectra taken in the laboratory.
  These lines are seen as dark lines, not as bright lines.
  We call them absorption lines.
  They are formed in the atmosphere of the star.
  Why they show up as dark lines and not as bright lines is not important now.
  I don't want to go into that.
  That's too much astronomy.
  But they are lines and that's what counts.
  And these lines are shifted towards the red part of the spectrum by a teeny weeny little bit.
  You see here this little arrow.
  And the conclusion then is that in this case the velocity of that galaxy is t- 720 miles per second which translates into 1150 kilometers per second, and so that brings this object if you believe the modern value for Hubble constant at about 16 megaparsec.
  This galaxy is substantially farther away.
  No surprise that it therefore also looks smaller in size, and notice that here the lines have shifted.
  These lines have shifted substantially further.
  And if I did my homework, using the velocity that they claim, which they can do with high degree of accuracy because you can calculate lambda prime divided by lambda, those measurements can be made with enorm- accuracy, I find that this object is about 305 megaparsecs away from us, so that's about 20 times further away than this object.
  So the speed is also about 20 times higher of course because there's a linear relationship.
  And if you look at this one which is even further away, then notice that these lines have shifted even more.
  The next slide shows you what I would call Hubble diagram.
  It was kindly sent to me by Wendy Freedman and her coworkers.
  Wendy is the leader of a large team of scientists who are making observations with the Hubble space telescope.
  You see here distance and you see here velocity in the units that we used in class, kilometers per second.
  Forget this part.
  That's not so important.
  But you see the incredible linear relationship.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher
Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Course Co-Administrators:
Dr. Peter Dourmashkin
Prof. Robert Redwine
Technical Instructors:
Andy Neely
Matthew Strafuss
Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao

Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:

Gratings
Resolving Power
Single-Slit Diffraction
Angular Resolution
Human Eye
Telescopes

Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

• iTunes U (MP4 - 107MB)
• Internet Archive (MP4 - 211MB)

  » Download this transcript (PDF)

  So last time we discussed the interference patterns due to two coherent light sources.

  Today I will expand on this by exploring many, many light sources.

  Suppose instead of having two slits through which I allow the light to go I have many.

  I have N, capital N.

  And let the separation between two adjacent ones be D, and so plane parallel waves come in and each one of these light sources is going to be a Huygens source, is going to produce spherical waves.

  And so now we can ask ourselves the same question that we did before, and that is look at a long distance far away at certain angle theta.

  Where will we see maxima and where will we see minima?

  And then we can put up here a screen at a distance L and we will call this X equals 0, and then we can even ask the question where exactly on that screen will we see these maxima?

  You will have constructive interference, exactly the same situation that we had with the double-slit interference pattern, when the sine of theta of N equals N lambda divided by D.

  And if you're dealing with very small angle theta, you should all remember that the sine of an angle is the same as the angle itself, provided that you work in radians.

  So for small angles, you can always use this approximation, if you remember that it is in radians.

  And that's only in the small angle approximation.

  And so the conclusion then is if we work in radians for now that theta of N for the maxima is then at N lambda divided by D, N being 0 right here, N being 1 right here, N being 2 right there.

  And if you want to express that in terms of a linear displacement from 0, then X of N again for small angles is L times that number.

  And so now you get displacement here in terms of centimeters or in terms of millimeters.

  So you will say well big deal, it's the same result that we had for the double-slit interferometer.

  We had exactly the same equation.

  There was no difference.

  And D now is the separation between two sources here.

  It is obvious that it is the same because if these two are constructively interfering then these two will too and these two will too and these two will too so all of them will, so it's not too surprising that you get exactly the same result.

  But now comes the big surprise.

  We haven't discussed yet the issue where the locations are where light plus light gives darkness.

  We haven't discussed the destructive interference.

  And to derive that properly is very tricky.

  In fact if you take 8.03 you will see a perfect derivation.

  But I will give you the results.

  What is not so intuitive, that if you have N sources, that between two major maxima, that means between this maximum at N equals 0 and a maximum at N equals 1, there are now N, capital N, minus 1 minima.

  And minima means complete destructive interference.

  So if capital N is 2, which we did last time, 2-1=1, exactly, that was correct.

  We had only one zero in between the two maxima.

  But that's not the case anymore when capital N is much larger than 2.

  And so let me now make you a -- a sketch whereby I plot the intensity of the light as a function of angle theta and this is the intensity, so that's in watts per square meter, remember that's the Poynting vector, and let this be 0, and let the angle theta 1 be here, and for small angles then that's lambda divided by D, and here you have theta 2, which is 2 lambda divided by D, and so on.

  I take the small angle approximation.

  So this angle is now in radians.

  What you're going to see now is the following intensity, as a function of theta.

  You see here a peak, and you're going to see here a peak, and you're going to see here one, and so on, and the same of course is true on the other side.

  And here in between you're going to see now N-1 locations whereby you have total destructive interference.

  And the same is the case here.

  And this can be huge.

  N can be a few hundred.

  So we have many many locations where you have 100% destructive interference.

  Now this point, this first location, where we hit the zero, that now is at the position lambda divided by D divided by capital N.

  And I will call that angle from the maximum to that zero, from this maximum to this zero, I will call that angle for now delta theta.

  Because that delta theta is a measure for the width of the line, here is at maximum, here it is zero, and so that angle delta theta in terms of radians is lambda divided by D times N, which then is approximately theta 1 divided by N, because theta 1 itself is lambda divided by D.

  And so you see that it is N times smaller than this distance.

  And so if N is large, these lines become extremely narrow, and that's the big difference between two-slit interference and multiple-slit interference.

  And the larger N is, the higher these peaks will be.

  The height of these peaks, the intensity here, is proportional to N squared.

  And you may say, "gee, why -- why not -- is why is it not linearly proportional to N?" Well that's easy to see.

  Suppose I increase capital N, the number of sources, by a factor of three.

  Then the electric field vector where there are maxima is three times larger.

  But if the electric field vector is three times larger the Poynting vector is nine times larger.

  So you get nine times more light.

  Now you may say, "gee, that's a violation of the conservation of energy.

  Three times more sources, nine times more light, how can that be?" Well, you overlook then that if you make N go up by a factor of three that the lines get narrower by a factor of three, because of this N here, and so they get higher by a factor of nine, and they get narrower by a factor of three, and so you gain a factor of three in light.

  Of course you gain a factor of three.

  You have three times more sources.

  You get three times more light.

  So you see there's no violation of the conservation of energy here.

  And I want to demonstrate this to you using a -- a red laser which we have used before.

  And I will use what we call a grating, a grating is a plate which is specially prepared, a transparent plate, which has grooves in it, and the one that I will use has tw- 2500 grooves, we call them lines, per inch.

  That means the separation D between two adjacent grooves in my case is about 2.16 microns.

  A micron is 10 to the -6 meters.

  And the wavelength that I'm going to use is our red laser, which is about 6.3 times 10 to the -7 meters.

  And I'm going to put the whole thing there.

  I'm going to make you see it there at a distance L.

  Which is about 10 meters.

  And so this allows me now to calculate where the zero order will fall, where the first order and where the second order will fall.

  We call when N is 0, we call that zero order, so this is zero order, when N is 1 we call that first order, and when N is 2 we call that second order.

  And you have of course the first order also on this side and the second order also on this side.

  Everything that you have here you have to also think of it as being on the other side.

  So I can predict now where the zero order will be when N is 0.

  That is 0 degrees.

  That's immediately obvious.

  I use that equation.

  If N is 0 the zero order is always right at the center, provided that all these sources are in phase.

  And they will be in phase because I use plane waves.

  So Huygens will tell you that they're going to oscillate exactly at the same time, they produce the same frequency, they produce the same wavelength, and they're all in phase with each other.

  So there will be a maximum at theta 1 equals 0.

  And then there will be a maximum which I calculated to be at 3.55 degrees.

  I calculated that from this equation and then theta 2 will be at roughly 7.1 degrees.

  If you want to know how wide the width of this peak is going to be, then you have to know how many lines of my grating I will be using.

  Well, my grating is like so.

  Here I have these lines not unlike the grating that you have in your optics kit.

  There are 2500 of those lines per inch.

  And my laser beam is roughly 2 millimeters in size.

  So this is about 2 millimeters.

  And that tells me then that I cover about 200 lines.

  And if I have 200 lines I can now calculate how wide that line is going to be.

  Because this factor of N enters into it here.

  And if I express that in terms of that angle delta theta, then the angle delta theta, going back here, so delta theta is then the 3.55 degrees divided by 200, and that's an extremely small angle, that angle is approximately one arc minute, which is 60 times smaller than one degree.

  And if you want to translate that in terms of how wide that spot will be, if I see it on the screen 10 meters away from me, and if you want to call that delta X, then you would naively predict that delta X is something like 3 millimeters, and the reason why I say naively because you will not see that it is 3 millimeters, it will be extremely narrow, but it will be more than 3 millimeters, because the limiting factor is always the divergence of my laser beam.

  And so the divergence of my laser beam is more than one arc minute, and so I don't get down to the one arc minute narrow beam.

  I'm not too far away from it, though.

  So this is what I want to show you first.

  I will turn on the laser first, and then make it very dark because we do need darkness -- or this has to come off because that would obviously -- oh, I turned off the wrong laser, but that -- I turned on the wrong laser, but that's OK.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher

Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Course Co-Administrators:
Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao

Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:

Double-Slit Interference
Interferometers

Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

• iTunes U (MP4 - 108MB)
• Internet Archive (MP4 - 212MB)

  » Download this transcript (PDF)

  I'm very proud of you.

  You did very well on the last exam.

  Class average is a little bit above 70.

  Congratulations.

  There were 22 students who scored 100.

  Many of you are interested in where the dividing line is between C and D.

  If I take only the three exams into account, forget the quizzes, forget the homework, forget the motor, and you add up the three grades of your three exams, the dividing line between C and D will be somewhere in the region 135 to 138.

  So you can use that for your calibration where you stand.

  The controversy between Newton and Huygens about the nature of light was settled in 1801 when Young demonstrated convincingly that light shows all the characteristic of waves.

  Now in the early twentieth century, the particle character of light surfaced again and this mysterious and very fascinating duality of being waves and particles at the same time is now beautifully merged in quantum mechanics.

  But today I will focus on the wave character only.

  Very characteristic for waves are interference patterns which are produced by two sources, which simultaneously produce traveling waves at exactly the same frequency.

  Let this be source number one and let this be source number two.

  And they each produce waves with the same frequency, therefore the same wavelength, and they go out let's say in all directions.

  They could be spherical, in the case of water surface, going out like rings.

  And suppose you were here at position P in space at a distance R1 from source number one and at a distance R2 from source number two.

  Then it is possible that at the point P the two waves that arrive are in phase with each other.

  That means the mountain from two arrives at the same time as a mountain from one, and the valley from two arrives at the same time as the valley from one.

  So the mountains become higher and the valleys become lower.

  We call that constructive interference.

  It is also possible that the waves as they arrive at point P are exactly 180 degrees out of phase, so that means that the mountain from two arrives at the same time as the valley from one.

  In which case they can kill each other, and that we call destructive interference.

  You can have this with water waves, so it's on a two-dimensional surface.

  You can also have it with sound, which would be three-dimensional.

  So the waves go out on a sphere.

  And you can have it with electromagnetic radiation as we will also see today, which is of course also three dimensions.

  If particles oscillate then their energy is proportional to the square of their amplitudes.

  So therefore since energy must be conserved, the amplitude of sound oscillations and also of the electric vector in the case of electromagnetic radiation, the amplitude must fall off as one over the distance, 1 / R.

  Because you're talking about 3-D waves.

  You're talking about spherical waves.

  And the surface area of a sphere grows with R squared.

  And so the amplitude must fall off as 1 / R.

  Now if we look at the superposition of two waves, in this case at point P and we make the distance large, so that R1 and R2 are much, much larger than the separation between these two points, then this fact that the amplitude of the wave from two is slightly smaller than the amplitude from the wave from one can then be pretty much ignored.

  Imagine that the path from here to here is one-half of a wavelength longer than the path from here to here.

  That means that this wave from here to here will have traveled half a period of an oscillation longer than this one.

  And that means they are exactly 180 degrees out of phase and so the two can kill each other.

  And we call that destructive interference.

  And so we're going to have destructive interference when R2 - R1 is for instance plus or minus one-half lambda, but it could also be plus or minus 3/2 lambda, 5/2 lambda, and so on.

  And so in general you would have destructive interference if the difference between R2 and R1 is 2N + 1 times lambda divided by 2 whereby N is an integer, could be 0, or plus or minus 1, or plus or minus 2, and so on.

  That's when you would have destructive interference.

  We would have constructive interference if R2 - R1 is simply N times lambda.

  So then the waves at point P are in phase and N is again, could be 0, plus or minus 1, plus or minus 2, and so on.

  If the sum of the distance to two points is a constant you get an ellipse in mathematics.

  If the difference is a constant, which is the case here, the difference to two points is a constant value, for instance one-half lambda, then the curve is a hyperbola.

  It would be a hyperbola if we deal with a two-dimensional surface.

  But if we think of this as three-dimensional, so you can rotate the whole thing about this axis, then you get hyperboloids, you get bowl-shaped surfaces.

  And so if I'm now trying to tighten the nuts a little bit, suppose I have here two of these sources that produce waves and the separation between them is D, then it is obvious that the line right through the middle of them and perpendicular to them is always a maximum if the two sources are oscillating in phase.

  So this line is immediately clear that R2 - R1 is 0 here.

  If the two are in phase.

  And they always have to generate the same frequency, of course.

  So this line would be always a maximum.

  Constructive interference.

  It's this 0, substitute there.

  And in case that we're talking about three-dimensional, this is of course a plane.

  Going perpendicular to the blackboard right through the middle.

  The different R2 - R1 equals lambda would again give me constructive interference.

  That would be a hyperbola then, R2 - R1 equals lambda, that would again be a maximum, and you can draw the same line on this side, and then R2 - R1 being 2 lambda again would be a maximum.

  And again, if this is three-dimensional, you can rotate it about this line and you get bowls.

  And so in between you're obviously going to get the minima, the destructive interference, lambda divided by two, and then here you would have R2 - R1 is 3/2 lambda.

  We call these lines where you kill each other, destructive interference, we call them nodal lines or in case you have a surface it's a nodal surface.

  And the maxima are sometimes also called antinodes, but I may also refer to them simply as maxima.

  And so this is what we call an interference pattern.

  If you look right here between -- on the line between the two points, then you should be able to convince yourself that the linear separation here between two lines of maxima is one-half lambda.

  Figure that out at home.

  That's very easy.

  Also the distance between these two yellow lines here right in between is one-half lambda.

  And so that tells you then that the number of lines or surfaces which are maxima is very roughly 2D divided by one-half lambda.

  So this is the number of maxima, which is also the same roughly as the number of minima, is then approximately 2D divided by lambda.

  And so if you want more maxima, if you want more of these surfaces, you have a choice, you can make D larger or you can make the wavelength shorter.

  And if you make the wavelength shorter you can do that by increasing the frequency, if you had that control.

  The first thing that I'm going to do is to make you see these nodal lines with a demonstration of water.

  We have here two sources that we can tap on the water and the distance between those two tappers, D, is 10 centimeters, so we're talking about water here.

  Uh, we will tap with a frequency of about 7 hertz and what you're going to see are very clear nodal lines, this is a two-dimensional surface, where the water doesn't move at all.

  The mountains and the valleys arrive at the same time.

  The water is never moving at all.

  So let me make sure that you can see that well.

  And so I have to change my -- my lights.

  I'll first turn it on, that may be the easiest.

  Starts tapping already.

  I can see the nodal lines very well.

  So here you see the two tappers and here you see a line whereby the water is not moving at all.

  At all moments in time it's standing still.

  Here's one.

  Here is one.

  And you even with a little bit of imagination can see that they are really not straight lines but they are hyperbolas.

  If you're very close to one tapper, the zero can never be exactly zero, because the amplitude of the wave from this one then will always be larger than the amplitude from that one, because as you go away from the source the amplitude must fall off on a two-dimensional surface as 1 / the square root of R.

  In a three-dimensional wave must fall of as 1 / R.

  But if you're far enough away then the distance is approximately the same and so the amplitudes of the individual waves are very closely the same and you can then, like you see here, the water is absolutely standing still.

  And here are then the areas whereby you see traveling waves, they are traveling waves, they're not standing waves, that here you see if you were sitting here in space the water would be up and down, bobbing up and down, and the amplitude that you would have is twice the amplitude that you get from one, because the mountains add to the mountains and the valleys add to the valleys.

  But if you were here in space you would be sitting still.

  You would not be bobbing up and down at all.

  And that is very characteristic for waves.

  If I were to tap them 180 degrees out of phase, which I didn't -- they were in phase -- then all nodal lines would become maxima and all maximum lines would become nodes, that goes without saying of course.

  It is essential that you -- that the frequencies are the same, that is an absolute must.

  They don't have to be in phase, the two tappers, if they're not in phase then the positions in space where you have maxima and minima will change but a must is that the frequency is the same.

  Now I was hiking last year in Utah when I noticed a butterfly in the water of a pond which was fighting for its life.

  And you see that butterfly here.

  Tom, perhaps you can turn off that overhead.

  You see the butterfly here, and you see here projected on the bottom the beautiful rings dark and bright, because these rings on the water act like lenses, and what you see very dramatically is indeed what I said, that the amplitude of the wave must go down with distance, because energy must be conserved of course in the wave, and since the circumference grows linearly with R, the amplitude must go down as 1 / the square root of R because the energy in the wave is proportional to the amplitude squared.

  So when I saw this it occurred to me that it would be a good idea to catch another butterfly, put it next to it, and then photograph -- make a fantastic photograph of an interference pattern.

  But I realized of course immediately, having taken 8.02, that the frequencies of the two butterflies would have to be exactly the same and so I gave up the idea and I decided not to be cruel.

  So no other butterfly was sacrificed.

  If we look at the directions where we expect the maxima as seen from the location of the sources, then I want to remind you of what a hyperbola looks like.

  If here are these two sources and here is the center, I can draw a line here, then a hyperbola would look like this.

  Let me re- remove the part on the left, doesn't look too good, but it's the same on the left of course.

  And what you remember from your high school math, that it approaches that line.

  And therefore you can define angle theta as seen from the center between these two, which are the directions where you have maxima and where you have minima.

  And that's what I am going to work out for you now on this blackboard here.

  So here are now the two sources that oscillate, there's one here and there's one here and here is the center in between them, and let this separation be D.

  And I am looking very far away so that I'm approaching this line where the hyperbolas merge, so to speak, with the straight line.

  And so I look very far away without being -- committing myself how far, I'm looking in the direction theta away.

  This is theta.

  And so this is theta.

  And I want to know in which directions of theta I expect to see maxima, and in which direction I expect to see minima.

  So this is what we called earlier R1 and we called this earlier R2, it is the distance to that point very far away.

  If I want to know what R2 - R1 is that's very easy now.

  I draw a line from here perpendicular to this line and you see immediately that this distance here is R2 - R1.

  But that distance is also -- you realize that this angle is theta -- it's the same one as that one, so that distance here is also D sine theta.

  And so now I'm in business, I can predict in what directions we will see constructive interference.

  Because all we are demanding now, requesting, that R2 - R1 is N times lambda.

  And so we need that D sine theta and I'll give it a subindex N, as in Nancy, equals N times lambda.

  In others words that the sine of theta N is simply N lambda divided by D.

  And that uniquely defines all those directions, the whole zoo of directions N equals 0, that is the center line, N equals 1, N equals 2, N equals 3, and so on.

  And then I have the whole family of destructive interference.

  Which would require that lambda R2 - R1 which is D sine theta must now be 2N +1 times lambda/2.

  Just as we had it on the blackboard there.

  We discussed that earlier.

  And so that requires then that the sine of theta N for the destructive interference is going to be 2N+1 times lambda / 2D.

  So this indicates the directions where we expect maxima and where we expect minima as seen from the center between the two sources.

  But now I would like to know what the linear distance is if I project this onto a screen which is very far away.

  And so let us have a screen at a distance capital L which has to be very far away, so here are now the two sources.

  It's a different scale.

  And here is a screen.

  And the distance b- from the two sources to the screen is capital L.

  And here is one of those direction theta.

  And you see immediately that if I call this the direction X, X being 0 here, that the tangent of theta is X/L.

  If but only if I deal with small angles, the tangent of theta is the same as the sine of theta.

  And therefore I can now tell you where the maxima will lie on that screen, away from the center line, which I call 0, that is now when X of N is L times the sine of theta, in small angle approximation.

  So this is approximately L times N lambda divided by D, and for the same reason you will get here c- destructive interference when X of N is going to be L times 2N+ 1 times lambda / 2D.

  That is simple geometry.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher

Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Course Co-Administrators:
Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao

Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered: Review for Exam 3

Instructor/speaker: Prof. Walter Lewin

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  » Download this transcript (PDF)

  These are the subjects will be covered during our third exam.

  There's no way I can cover all during this review.

  Nor can I cover all of them of course during the exam.

  I can only touch upon a few of them.

  And what I cannot cover today, what I will not cover today, can and will be on the exam.

  Let's first look at magnetic materials.

  Magnetic materials come in dia-, para- and ferromagnetic materials.

  The molecules and the atoms in para- and ferromagnetic materials have intrinsic magnetic dipole moments.

  These have always -- they're always a multiple of the Bohr magneton.

  Has to do with quantum mechanics.

  It's not part of 8.02.

  And they are going to be aligned by the external field, I call that um the vacuum field.

  And the degree of success depends on the temperature and on the strength of that external field.

  The lower the temperature, the easier it is to align them, to overcome the thermal agitation.

  And above a certain temperature which we call the Curie temperature, ferromagnet- magnetic material loses all its qualities and becomes paramagnetic and I have demonstrated that during my lectures.

  Suppose we have a solenoid and the solenoid has N windings, and the length of the solenoid is N.

  And the current I is flowing through the solenoid.

  Then the magnetic field generated by that solenoid which I have called the vacuum field, that magnetic field can be derived using Ampere's law, which you see down -- down there.

  That magnetic field is approximately mu 0 times I times N divided by L.

  If now I put in here ferromagnetic material then I have to include this factor kappa of M or K of M, whatever you want to call it.

  The magnetic permeability, and this can be huge.

  This can be 10, 100, even up to 1000 and higher.

  So you get an enormous increase in magnetic field strength.

  Self-inductance is defined as magnetic flux divided by the current I.

  That's just the definition of self-inductance.

  If the magnetic field goes up by a factor of kappa M then of course the magnetic flux will go up by the same factor and so the self-inductance will go up.

  And you may remember a demonstration that I did when I had an iron core which I moved inside the solenoid and depending upon how far I moved it in could we see that the self-inductance went up and when I pulled it out self-inductance went down again.

  We have an interesting problem.

  I think it is assignment seven, whereby we have iron core here and then we have somewhere an air gap and you may want to revisit that to refresh your memory.

  Let's now turn to transformers.

  A transformer often comes in this shape.

  Let me move it a little bit to the right.

  Often comes in this shape which is then ferromagnetic material, to give perfect coupling between the left and the right sides, also increases the magnetic field.

  This is the -- let's call this primary side.

  N1 windings, index, self-inductance L1.

  And here I put in a voltmeter to always monitor that value, I call that V1.

  And this is the secondary side.

  N2 windings.

  Self-inductance L2.

  And I put here a voltmeter which always monitors that voltage and I call that one V2.

  You can show with Faraday's law as I did in class, in lectures, that V2 divided by V1, let's not worry about plus or minus signs, is N2 divided by N1.

  That's a good approximation.

  Depends on how well the coupling goes.

  It depends on several factors, but you can come very close to this and this means then that if you make N2 larger than N1 then you can step up in voltage, we call that a step-up transformer.

  But you can also step down if you make N2 smaller than N1.

  Under very special conditions will the power generated on the primary side be all consumed for 100% or nearly 100% on the secondary side.

  That is very, very special.

  If that's the case then the time averaged power here V1 I1 is the same as V2 I2, here time averaged.

  And so as a logical consequence of that you'll find that I2 divided by I1, let's not worry about minus signs, is that N1 divided by N2.

  That, however, is not so easy as you may think.

  It only can work approximately and I mentioned that on the side in my lectures.

  If the resistance here and the resistance there is way, way smaller than the value for omega L.

  And we did try to achieve that during one of the demonstrations that I gave on this.

  I remember we had the induction oven whereby N2 was 1 and N1 was very large, I don't remember what it was anymore but it was of the order of several hundred, maybe a thousand, and we managed to get a current in the secondary which was huge, which was close to 1000 amperes.

  It was enough to melt that iron nail.

  And we made every effort then to make sure that the resistance was much, much smaller than omega L.

  I think problem 7-1 of our assignments deals with that, and very naively assumes that this is all true.

  But you should realize that it is not always so easy to achieve the conditions for that.

  So let's now go to RLC circuits there.

  Let's take an, uh, system which has a resistor R, it has a self-inductor, a pure self-inductor, L, and a capacitance, C.

  AC.

  And this driving power supply provides with a voltage, V, which is V0 cosine omega T.

  Keep in mind that this can be always be sine omega T of course.

  There is nothing special about cosine in life.

  The steady state solution, that is not when you turn the thing on but if you wait awhile, you get a steady state solution for the current.

  And the current that is going to flow now is V0 divided by the square root of R squared plus omega L minus one over omega C squared times the cosine of omega T minus phi.

  And the tangent of phi is omega L minus one / omega C divided by R.

  We call this the reactance.

  The upstairs.

  For which we give often the symbol X.

  And so this is also X then divided by R.

  And this whole square root that we have here, we call that the impedance.

  The units are ohms.

  And we call that Z.

  And so the maximum current that you can have, the current is of course oscillating with angular frequency omega, the maximum value that you can have for the current, which I call I max, is then V0 divided by Z.

  Then the cosine term is either plus or minus 1.

  I can plot now this I max as a function of frequency.

  So here is frequency and here is I max.

  If the frequency is very low or near 0, then this term here becomes infinitely high because the impedance is infinitely high and so the current is 0.

  I max is 0.

  There's no current flowing at all.

  When we go to very high frequencies it is the omega L term that goes to infinity.

  And so again Z goes to infinity so again I max goes to 0.

  And for other values of omega you get an I max which is not zero and so you get a curve like this which has the name of resonance curve.

  This I max reaches a maximum value when the system is at resonance, that's what we call resonance.

  And that's the case clearly when the reactance is 0.

  Because when the reactance is zero this part vanishes.

  And if the reactance is not 0 then the maximum current can only be lower, can never be higher.

  And so when X equals 0 you'll find that omega L is 1/omega C and so the -- the frequency for which that happens, I call that omega 0, reminds me that it is the -- the resonance, is one divided by the square root of LC.

  When I am at resonance, phi becomes 0.

  So there is no phase delay between current and the driving voltage.

  They are in phase with each other.

  And the value for I max now simply becomes V0 divided by R.

  Because the impedance itself becomes R.

  Very boring, very simple, you're looking here at Ohm's law.

  When the system is at resonance, forget the self-inductance, forget the capacitor, they are not there, they annihilate each other, and so the system behaves as if there were only a resistor, and that's exactly what you see here.

  I have here some numbers which you have seen before.

  During my lectures.

  You can download this from the Web but you have to go back to the lecture when I discussed that.

  And you see here s- some numbers for R, L and C and also for V0.

  And I calculate for you here the resonance frequency.

  I calculate the frequency also in terms of kilohertz.

  And here you see the impedance and here you see the reactance.

  If I'm 10% below resonance notice that the 1 / omega C term is always larger than omega L.

  So your reactance in this case becomes -86 ohms.

  The minus sign has of course no consequence for the current because you have an X squared here.

  But notice that Z is now almost exclusively determined by X and not by R anymore.

  Because the 10 ohms of the R here play no role, almost no role, in comparison with the 86.

  Z becomes 87 and the maximum current is one-tenth of an ampere.

  When you're on resonance, and that is characteristic for on resonance, the 2 omega L and 1/ omega C eat each other up.

  They annihilate each other and so the reactance becomes 0.

  So Z now is just pure R.

  X is 0.

  And so the maximum current in this case is 1.

  Because I chose V0 at 10 and I chose R 10.

  And then when I'm 10% above resonance then the omega L term is larger than the reactance of the capacitor and accordingly you get a lower current again, about one-eighth of the -- of an ampere.

  And so you see this curve being formed in a very natural way and that's quantitative, you see there some numbers.

  So now comes the question which of course in practice is very important.

  And that has to do with the power that is generated by the power supply.

  That power comes out in the form of heat.

  Heat in the resistor and so if you time average the power, then the time average value, you can take the -- the voltage of the power supply, multiply that by the current.

  You could also take the time average value of I squared R.

  Because all that energy will ultimately come out in the form of heat of the resistor.

  Either one will be fine.

  I've decided to take this one.

  So I will get then V0 cosine omega T -- the I becomes V0 divided by Z times the cosine (omega T - phi).

  This is the power at any moment in time.

  I will do the time averaging a little later.

  When I see cosine (omega T - phi), that reminds me of my high school days, cosine alpha minus cosine -- no, cosine alpha minus beta is cosine alpha cosine beta + sine alpha sine beta.

  That was drilled into my memory here.

  I will never forget that, I think.

  And so I will write down here -- my math teacher will be proud of me -- cosine omega T cosine phi plus sine omega T sine phi.

  So this is this term.

  If I'm going to time average it, I have a cosine omega T multiplied by sine omega T, that time average is 0.

  So this term vanishes.

  So the time average value of the power, I get a V0 squared, I get a Z here, and now I have here cosine omega T times cosine omega T.

  The time average value of cosine squared omega T is one-half.

  So I get a two here.

  And then I still have my cosine phi there.

  And I'm done.

  If you like to get rid of this cosine phi, you can do that.

  Because remember, the way that phi is defined, the tangent of phi is the reactance divided by R.

  You still see it there.

  So that means if this angle is 90 degrees that this side must be Z.

  That's the square root of X squared plus R squared.

  That's this part.

  And so the cosine of phi is also R divided by Z.

  And so if you prefer that -- there's no particular advantage but if you prefer that you can write down for cosine phi R divided by Z.

  And so you get V0 squared times R divided by 2 and now you get Z squared.

  And so there you see the power, time averaged power in an RLC circuit.

  So now we can look at resonance.

  It's always a very special situation.

  When we are at resonance, Z equals R.

  So you replace this capital Z by R.

  And then you find V0 squared divided by 2R.

  That's utterly trivial.

  You could have predicted that.

  It's really Ohm's law staring you in the face.

  There is no self-inductance and there is no capacitor at resonance.

  So you might as well have treated it as a simple system only with R.

  And you find immediately then that answer.

  At any other frequency than omega 0, Z would always be larger than R.

  You see that immediately here.

  And so that means that the average power would always be lower.

  So it's only at resonance that you generate the highest power possible.

  All right.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher

Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Course Co-Administrators:
Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao

Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.