Fisika untuk Universitas

Fisika untuk Universitas

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

» Download this transcript (PDF)

Liquids are incompressible; gases are not incompressible.

When you decrease the volume of a gas by 50%, that's no problem.

It's impossible to do that for a liquid.

In liquids, the atoms and the molecules effectively touch each other, whereas in gases, they are very far apart, so that's why you can compress the gases.

If you take air at one atmospheres, the density is a thousand times less than the density of water.

What it tells you is that the molecules are much further apart.

It is an experimental fact that there is a simple relation between the pressure that you see there, the volume of a gas, the temperature of a gas in degrees Kelvin, and the number of molecules that you have.

Now, when you see the word "molecules," I may often mean "atoms." I realize that helium and neon and krypton and argon are atomic gases, and that O2 and H2 and CO2 are molecular gases.

So I will use that word "molecules" even when I mean "atoms," and maybe vice versa, just for simplicity.

The relation that exists between these quantities, PV equals nRT: pressure, volume, n is the number of moles--

I'll get back to that--

R is the universal gas constant, which is 8.3 joules per degree Kelvin, and T must be in degrees Kelvin.

So, what is a mole? A mole has always about 6.02 times ten to the 23 molecules, or atoms, in the case that you have helium, but I will call that molecules.

And this number is called Avogadro's number.

So that's the definition of a mole.

If you take a mole of helium, or a mole of oxygen, or CO2, or N2, it doesn't matter, it always has this number of molecules, approximately.

Now, each of these substances have very different masses.

If I take, for instance, carbon, then one mole of carbon would weigh very close to 12 grams.

If I take helium, one mole of helium would weigh very close to four grams.

And if I took oxygen two, O2, then one mole would be very close to 32 grams.

So the masses are very different in a mole but not the number of molecules or the number of atoms.

When I take a neutral atom, then we have a nucleus, and the nucleus contains protons and neutrons.

It has Z protons and it has N neutrons.

The protons are positively charged, and it has Z electrons if it is a neutral atom.

There is almost no weight in the electrons; you can almost ignore that.

Everything is in the protons and in the neutrons.

N plus Z is called A, and that's called the atomic mass number.

Let's look at carbon in a little bit more detail.

If we have carbon--

and I call it carbon 12 for now, you'll see shortly why--

then carbon has always six protons in the nucleus; otherwise it isn't carbon.

And when it has six neutrons, then A is 12.

That's why we call it carbon 12.

So the atomic mass number of carbon is 12, but if you had, for instance, carbon 14--

which happens to be radioactive--

again, six protons, otherwise it wouldn't be carbon, you would have eight neutrons now, and now you would have...

atomic mass number would be 14.

A mole is this number in grams, and so you see carbon...

is the atomic mass number in grams--

you see 12 there.

If you go to helium, it has two protons and two neutrons, so A is four--

that's why you see your four grams.

If you take oxygen, it has eight protons and eight neutrons, so A is 16, but you have O2 in gas form, so now your atomic mass number has to be doubled to 32.

And so a mole of O2 is therefore 32 grams.

In fact, Avogadro's number is defined through carbon 12.

If you take 12 grams of carbon 12, and you count the number of atoms that you have, then you find exactly Avogadro's number.

That's the definition of that number, and that's very close to what we have there, 6.02 times ten to the 23rd.

The mass of the proton and the mass of the neutron are nearly equal.

I wrote down m2 for the mass of the neutron; of course, that should have been m of n.

So the mass of a molecule, or an atom, whatever the case may be, would be this number A--

because that's the sum of the protons and neutrons--

times the mass of the neutrons and the protons.

And so this is A times--

approximately, I should put a wiggle here--

1.66 times ten to the minus 27 kilograms.

So that's now an individual mass of either an atom or a molecule, and all that information, you have there and that's, of course, on the Web.

So, let's do a trivial example.

I take gases, any kind of gas--

you choose whatever you want--

and I take one atmosphere.

So that means that the pressure is 1.03 times ten to the fifth pascal.

I do it at room temperature, so T is 293 degrees Kelvin.

And I take in all cases only one mole, so n is one.

And I'm asking you now, what will be the volume of that gas? Well, you take the gas law, and it tells you that V, the volume, equals nRT divided by P.

You know n is one.

You know R, 1.03...

excuse me, you know...

[laughs]: I'm a little bit ahead of myself.

You know R, which is 8.3, you know the temperature, which is 293, and you know the pressure, which is 1.03 times ten to the fifth.

And when you calculate that, you find something very close to 24 liters, and a liter is about a thousand cubic centimeters.

And it's independent of whether it's helium or oxygen or nitrogen or CO2.

As long as you have a gas, one mole at one atmosphere pressure and room temperature always has the same volume of about 24 liters.

If a gas obeys that law exactly, we call it an ideal gas.

That's why we call that the ideal-gas law.

And many gases come very close to that.

In fact, if you took oxygen, O2, and you take one mole of oxygen at room temperature and at one atmosphere pressure and you were to calculate its volume, the actual volume that you measure is only one-tenth of a percent smaller than what you would have found with the ideal-gas law.

If you do it at 20 atmospheres, it would still be only two percent smaller, so it's a very good approximation in many cases.

What is very surprising, that in this ideal-gas law, the mass of the atoms and the molecules do not show up at all.

And that is very puzzling--

you wouldn't expect that at all.

And I'll show you why you wouldn't expect that.

Let's take two different kinds of gases with very different masses of the molecules, but we have the same number of moles, we have the same volume, we have the same temperature and therefore, we must have the same pressure, according to the ideal-gas law.

But the masses of the molecules--

very different.

So, here we have some of these molecules and the number of... density is the same, because the number of atoms is the same and the volume is the same.

Now, these molecules are flying in all directions with different speeds.

I will just now, for simplicity, take some average speed, and I assume this is going in this direction.

It's heading for the wall of the container, this area.

It hits the wall, there's an elastic collision, and it comes back in exactly the same direction.

So there is momentum transfer, and the momentum transfer for one collision is 2mv, because it comes in with mv in this direction, it comes back with mv in that direction, so the momentum transfer is 2mv.

But I'm interested in the momentum transfer per second, not just for one molecule.

And now, of course, I have to multiply by the velocity, because if the velocity is high, you have a lot of bombardments per second on here.

For each bombardment, this is the momentum transfer, but if there are many, well, you have to multiply that, of course, then, by the speed.

So the momentum transfer per second is proportional, let's say, to mv squared.

mv comes from the momentum, from one particle, and v comes from the fact that...

the number that hit it per second.

Now, momentum transfer per second is clearly...

It's a force, proportional to the force, and that is proportional to the pressure.

And yet the pressure is not affected by the mass, notice? If these are the same, the pressure must also be the same.

And so there's only one conclusion that you can draw, which is very nonintuitive--

that the pressure can only be the same if, for a given temperature, this product, mv squared, is independent of the mass of the molecule.

How can mv squared possibly be independent of the mass of the molecule? There's only one way that that's possible--

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)

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:

Rainbows
A modest rainbow will appear in the lecture hall!
Fog Bows
Supernumerary Bows
Polarization of the Bows
Halos around the Sun and the Moon
Mock Suns

Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

• iTunes U (MP4 - 104MB)
• Internet Archive (MP4 - 208MB)

  » Download this transcript (PDF)

  All of you have looked at rainbows, but very few of you have ever seen one.

  Looking at something is very different from seeing it.

  And today I will make you see the rainbow in a way that goes way beyond the beauty that we can all experience, a way that you will always remember.

  And I would like to start asking you 15 perhaps simple questions about the rainbow.

  The first question then is would any one of you remember if you see a bow whether the red color is outside or whether the -- the red color is inside?

  And then I wonder about the radius of the bow.

  If this is a bow in the sky, something like this, here is the horizon, it's clearly a perfect circle, and so the perfect circle has somewhere a center.

  And so that means there must be a radius R.

  You can measure that radius in terms of how many degrees and so what is roughly that radius.

  You've never measured it but is it 10 degrees, is it 20, 30, 50, 60?

  The length of the bow.

  Is there a difference, do you sometimes see a very long bow, sometimes a very short one?

  What is the width of the bow?

  You see colors here.

  How wide is that strip of colors in degrees?

  Perhaps some of you have noticed that there is a difference in light intensity between inside the bow and outside the bow.

  Maybe you've never seen it, and if there is a difference where is it brighter, inside the bow or outside the bow?

  What time of the day would you see bows?

  Would you see rainbows in the north, east, south or west?

  Is there perhaps a second bow in the sky?

  And if there is a second one, where should you look for the second bow?

  And if there is a second one what is the color sequence of the second bow?

  Is the red on the outside or is the red on the inside?

  And then you can ask the same question, what would be the radius of the second bow?

  And what would be the width of the second bow?

  All these first 12 questions in principle you should have been able to answer if you really have seen a rainbow.

  The last three is more difficult.

  The question is are the bows polarized?

  In what direction are they polarized?

  And are they weakly polarized or are they strongly polarized?

  Who knows the answer to 12 questions, to the first 12 questions?

  Who knows the answer to more than 10?

  Who knows the answer to nine?

  Eight?

  Seven?

  Six?

  Five?

  Four?

  Do I see a hand at four?

  Good for you.

  Five, four, three?

  Three, good, that's already good.

  Two?

  One?

  And who knows the answer to zero?

  Most of you, right?

  I haven't seen a lot of hands though.

  All right.

  So I've made my point.

  You've looked at rainbows but you've really never seen them.

  And I'm going to make you see them today.

  What you see here on the blackboard is one drop of water.

  I put the sun for simplicity at the horizon.

  Later I will put it a little bit higher in the sky.

  Light from the sun hits this raindrop.

  I've only drawn one narrow beam which hits the raindrop right there.

  And you see here the angle of incidence, which with Snell's law we call theta 1.

  I call it I here because it's nicer for me, more descriptive, it means incidence angle.

  Right at that point A some of the light will be reflected and some of the light will go into the water.

  We call that refraction.

  And Snell's law will tell me this angle R.

  Whatever goes in there reaches point B where there is a transition back to air and so some of that light will come out here and some of that light will be reflected inside.

  And then when it reaches point C again there is a transition from water to air.

  Some of that light will be reflected inside the water.

  And some of it will come out.

  And as far as the geometry is concerned, if this angle is R, then this angle is also R, this is also R, and this is also R.

  And the angle here is I.

  That follows from Snell's law, and I'll leave you with that.

  Notice that the light came in like this but it comes back like this.

  So the direction has changed over the angle delta.

  And the angle delta is very easy to calculate in terms of I and R.

  Delta is 180 degrees + 2I - 4R.

  I want you to check that at home.

  The 4 Rs come in here.

  One, two, three, four, and the 2 I's come in here and there.

  If now I think of all possible narrow beams of light that can strike this raindrop, one that would strike it here would have an I of 0 degrees.

  And then here would be 10 degrees and 20 degrees and 30 and 40.

  And the largest value for I is when the light strikes here, would be 90 degrees.

  And so I can calculate for all these values of I, which obviously all of them occur, sunlight strikes this raindrop, and all these angles for I are present.

  So I can calculate now for all these angles of I what the value is for R and then I can calculate what delta is.

  R follows from Snell's law and delta follows from this geometric relationship.

  And what you will find now very much to your surprise, that there is a minimum value for delta which is about 138 degrees.

  That means this angle phi here has a maximum value which is very roughly about 42 degrees.

  And I will show you some numbers.

  You can download this, by the way, this is on the Web, under lecture supplements.

  Here all I have done I've taken I to be from 0 to 90 degrees, all these angles are possible, with Snell's law, using an index of refraction of 1.336, that you see at the bottom, I calculate R and then in the last column using that relationship I calculate delta.

  And indeed you see that delta starts at 180 degrees when I is 0.

  And then goes to a minimum of roughly 138, after which it increases again.

  And this now is crucial, is key to an understanding of the rainbow.

  Imagine now that I have one drop of water here.

  And sunlight comes in at all angles of I, not just at one, but all angles of I.

  Whatever you see here has of course axial symmetry.

  It is a spherical drop.

  And the light comes in like this.

  So light can go this way but it can also go this way.

  And it can also go this way and this way, so there's complete axial symmetry, so this whole drawing you can rotate about this line here.

  And everything holds then in axial symmetry.

  So therefore if phi maximum, if this angle phi maximum is 42 degrees, then the light that will go back in the direction of the sun, the light that goes through the journey A B C and then comes out of the raindrop, that's all I'm talking about, now, I'm not talking about this light that sneaks out here, it is this journey, A, refraction at A, reflection at B, and then coming out at C.

  That light comes out in the form of a cone.

  And the half -- top angle of the cone must be roughly 42 degrees.

  And so I will go -- I'm going to draw that cone for you.

  Like so.

  And like so.

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:

Polarizers
Malus's Law
Brewster Angle
Polarization by Reflection and Scattering
Why is the sky blue? Why are sunsets red?
The sun will set in the lecture hall!

Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

• iTunes U (MP4 - 105MB)
• Internet Archive (MP4 - 211MB)

  » Download this transcript (PDF)

  Earlier in this course, we discussed linear polarization of electromagnetic radiation, and I demonstrate this at 75 megaHertz and at 10 gigaHertz.

  Today, I will concentrate exclusively on the polarization of light, which is at a much higher frequency.

  The light from the sun or light from light bulbs is not polarized.

  So I can ask myself the question, now, what does it mean when light is not polarized?

  Let's think of individual light photons as plane waves, with a well-defined direction of polarization.

  So each one is linearly polarized.

  A beam is coming straight out of the blackboard.

  The first photon arrives, it's linearly polarized in this direction.

  This second photon arrives, linearly polarized in this direction, so the electric field vector is oscillating like that.

  Another photon, another photon, and another photon.

  And what you see here, very clearly, that there is no preferred direction which you average over time, and that's what we call -- call unpolarized light.

  It was Edwin Land who, in 1938, developed a material that can turn this into linearly polarized light, for which he became very famous, in addition to this demonstration that I showed you last time.

  If I take one of Edwin Land's sheets, which will turn light into polarization in this direction, and I first take one photon, for instance, this one.

  That one comes in from the blackboard towards you, and so here it is.

  Oscillating the E vector like this, E0 is the maximum value of the electric field strength in that plane electromagnetic wave.

  And this is the direction of the polarizer that I have through which this photon goes.

  I can now make a simple calculation, by projecting this E-vector onto the preferred direction of polarization, and this new E-vector is now down by the cosine of theta, if this angle is theta, this E-vector is now E -- E0 times the cosine of theta.

  If you ask me now, whether the light is reduced in intensity, I would have to say, "Yes, of course," because light intensity depends on the Poynting vector, and the pointing vector is always proportional to E0 squared, because the Poynting vector is the cross-product between E and B.

  And if E is reduced, B is also reduced.

  And so we get a cosine square reduction.

  If, now, I average over all incoming photons -- so I take all of these, which represent an unpolarized beam -- so I get not only one like so, but I get one like so, and one like so, and one like so, and one like so -- then clearly, I have to calculate, now, the mean value of cosine square theta.

  And the mean value of cosine square theta is one-half, and so if the intensity of the unpolarized beam, unpolarized light was originally I0, once it comes through this polarizer that Edwin Land gave me, then I get one-half I0, but that is now 100% polarized.

  And it is 100% polarized in this direction.

  And the one-half is the result of the average value of cosine square theta.

  If this were the case, it would be an extremely ideal polarizer, we would call this an HN50 polarizer -- they don't exist, it's only in your head -- and the 50 refers to the fact that 50% get through polarized.

  In the optics kits that we hand out today that we will need throughout this course, you don't have HN50 polarizers, they don't exist.

  I don't quite know what yours is, I didn't measure it, yours may be an HN25 or maybe an HN30, which would then mean that the I0 strength of an unpolarized light of beam would not be half of I0, but maybe only .25, or .3.

  But in any case, the light that will come through your linear polarizers will be very closely to 100% polarized.

  So what I will do now, I will take unpolarized light, and I will have this light coming straight out of the blackboard perpendicular to you, with strength I0, and here is one of my polarizers, and the light that comes through here is linearly polarized in this direction.

  And so we already know that one-half I0 will come through if it is an ideal polarizer, and it is polarized in this direction.

  I take a second sheet, an identical one, I put it also in the plane of the blackboard, but I rotate it over an angle theta.

  So here is now a second sheet, which has a preferred direction of polarization in -- in this direction, and the angle is rotated over an angle theta.

  So between this one and this one is an angle theta.

  And so you can now immediately tell what the intensity of the light is that comes through this second polarizer.

  It must, of course, be polarized in this direction, because that is the allowed direction polarization for that second sheet -- and the intensity must now be one-half I0, because that's what comes in, and then I have to multiply it by the cosine square of theta.

  I don't have to average it now over all angles, because there is only one value of theta between this sheet and this sheet, so this is now the new intensity, and it's all polarized in this direction.

  And this law, whereby the light intensity is reduced by the factor cosine square theta, is known as Malus' Law.

  Malus' Law.

  If theta were 30 degrees, the light intensity here would be one-half I0 times the cosine square of 30 degrees, which is 0.75.

  If theta were 0 degrees, that means that this sheet is in the same direction as this one, if everything were ideal, one-half I0 would get through the second sheet.

  If theta is 90 degrees, then nothing will get through, because the cosine of 90 degrees is 0.

  We call that crossed polarizers.

  If you cross them like this, no light will get through.

  Now, before I demonstrate this, I have to be honest with you, because the idea of reducing the energy of individual photons by reducing their electric field strength, as I did, is a cheat.

  A light photon has a well-defined energy which depends uniquely on the frequency of the light.

  Blue light has a higher frequency than red light, so blue light has a higher energy than red light.

  And when you send blue light through a polarizer, the way I did here, it either comes through or it doesn't come through.

  But if it does come through, it is still blue light, there is no such thing as a reduction in energy.

  Whereas this reduction, by cosine theta, would imply that the energy goes down, and that moo- would imply, then, that there would be a color change, that it would no longer be blue.

  And that's not the case.

  If you want to treat this properly, you have to do it in a quantum mechanical way.

  The interesting thing is that if you use quantum mechanics, you find exactly the same law, you find also Malus' Law.

  So the law is OK, even though the derivation is not kosher.

  Now, I want you to get out of your envelope one of your green plates, which is a linear polarizer.

  This is the kind of plate that you have, you have three in there.

  Only take one out.

  These two lights shining on me, unpolarized light.

  So the light that comes to you now is unpolarized.

  I'm now going to hold in front of my face this polarizer.

  So the light that comes through is linearly polarized in this direction.

  And you are going to play the role of the second polarizer.

  Close one eye, put the polarizer in front of your eye, and rotate it.

  And you will see a huge difference in light intensity.

  If you cross-polarize with me, then you can't see me.

  That may make you very happy.

  But keep in mind, if you can't see me, then I can't see you, either.

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

» Download this transcript (PDF)

From early childhood, we can tell by touch whether an object is hot or whether it is cold.

If you want to heat an object, you bring it in contact with a hot object, for instance a flame.

If you want to cool an object, you bring it in contact with a cold object.

When objects are heated or when they're cooled, then--

and the temperature changes--

then some of their properties change, and those properties are called thermometric properties: ther-mo-metric properties.

One very characteristic thermometric property is that most substances when you heat them, they expand, and when you cool them, they shrink.

We'll talk more about it later.

If you take a gas in a closed volume and you heat it, the pressure goes up.

That's a thermometric property.

If you take an electric conductor and you heat it, in general the electric resistance will change.

If you heat an iron bar, it will expand.

And if you place it in contact with another iron bar which is cold, then the one that is hot will shrink and the one that is cold will heat up and will get longer, and this process will go on up to the point that the hot one will not get shorter and that the cold one will not get longer anymore.

And that is when the two objects, as we say, are in thermal equilibrium with each other, and that is when the temperature of the two objects are the same.

And so you can define a temperature scale by looking at the length of an object.

For instance, here is a bar, some material, I clamp it in here, has length L, and I increase the temperature by an amount delta T, and it gets longer by a certain amount delta L.

I could put the whole thing in melting ice...

melting ice...

and I could say, "Aha." The length then is L1.

Then I could put the whole thing in boiling water and I do that at one atmosphere pressure, and then I say, "Aha." I call that the length L2.

And those are my reference points for my temperature scale.

Celsius did just that.

The idea that he used melting ice, which is now called zero degrees centigrade, and he used boiling point of water, which was his 100 degrees centigrade.

He was a Swedish astronomer; in 1742 he introduced this temperature scale.

So you could make yourself a plot now of the temperature versus the length of that bar, and you could say, okay, 100 degrees centigrade...

if the length of the bar... L2, zero degrees centigrade if the length of the bar is L1.

And now you can draw a straight line--

you can always draw a straight line through two points, you have one point here, you have one point there--

and you can define temperature now by saying, if my bar has this length, L of T, then this will be the temperature.

So you can introduce a linear scale in this fashion, and the thing, in principle, could act like a thermometer.

I'll show you a demonstration of this shortly.

Centi in Greek means "one hundredth," and therefore we also call this scale often "centigrade." One degree centigrade is often called...

one degree Celsius is often called one centigrade, for the reason that it divides the scale from zero to one hundred in equal portions.

So we call them degrees centigrade, degrees Celsius.

Fahrenheit, a German scientist, invented the mercury thermometer.

We'll talk about the mercury thermometer a little later.

In 1714 he introduced a new scale.

He lived in Holland at the time, he lived there most of the time, and he used as his reference point body temperature, which he called 100 degrees Fahrenheit, and he used a mixture of salt and ice at zero degrees.

Now, neither one of these two are very reproducible.

If you pick one person, the temperature today may be a little higher than tomorrow.

A person may have fever.

In fact, the one that he picked probably did have a little bit of fever.

And so the Fahrenheit scale, in that sense, is not very reproducible, and it has been redefined now in such a way that zero degrees centigrade is 32 degrees Fahrenheit, and 100 degrees centigrade is 212 degrees Fahrenheit.

And so if you convert--

if you want to convert from Fahrenheit to centigrade or the other way around--

then the temperature in Fahrenheit is 9/5 times the temperature in Celsius plus 32.

If you take room temperature, the temperature is 20 degrees centigrade, what I was growing up with--

in Europe, everyone uses centigrade there--

then you can see that in terms of Fahrenheit, that becomes 68 degrees Fahrenheit.

9/5 times 20 gives you 36, and then you add 32.

Minus 40 degrees centigrade is the same as minus 40 degrees Fahrenheit.

Check that.

That's where the two scales cross over.

So almost the entire world uses the Celsius scale; it's part of our metric system.

United States is one of the very, very few countries who still, in a rather stubborn way, uses degrees Fahrenheit.

And it is really a pain in the neck, degrees Fahrenheit--

at least for me.

I have very little feeling for it.

I just happen to know that room temperature is 68, because that's the way I set my thermostat at my home, but that's about all.

I can't think in terms of degrees Fahrenheit.

There is no limit to high temperature, but there is a limit to the low temperatures.

There is an absolute zero.

This absolute zero below which you cannot go is about minus 273 degrees Celsius...

And if you take a system that cannot transfer energy to any other system that it is in thermal contact with, then it is at that lowest possible temperature.

This is the way we define it.

It's about minus 460 degrees Fahrenheit.

And so we now have a third scale, which was introduced by Lord Kelvin, was a British scientist.

He did a lot of research on heat, and he introduced the absolute scale whereby he uses the lowest possible temperature as zero degrees Kelvin.

But the increments in terms of increase of one degree, he uses the same as the Celsius scale.

So an increase of two or three degrees Kelvin is the same as an increase of two or three degrees centigrade.

So if we now compare the three scales--

Celsius, Fahrenheit and Kelvin--

then 20 degrees centigrade would be 68 Fahrenheit, and that would be 273.15 plus 20.

Let's round it off and make it 293, and if we take zero Kelvin, then we would have minus 273.15, but let's leave that off for now, and it is approximately minus 460.

We will almost always work with degrees Kelvin in physics and we'll discuss that in more detail Friday.

Most substances expand when you heat them, and if we start with an object which has length L and I heat it up delta T degrees, it gets longer by an amount delta L.

And that delta L can be expressed in a very simple way.

It is alpha times L times delta T, and alpha is called the linear expansion coefficient.

And the units are one over degrees centigrade, or one over degree Kelvin, which is the same, because it's the increments that matter.

The various values for alpha differ a great deal.

Give you some values for alpha.

I'll give you copper, I'll give you brass, I'll give you Pyrex, I'll give you Invar and I'll give you steel, and they are in units of ten to the minus six per degree centigrade, and we will use some of them today.

Brass is about 19.

Copper is 17.

Pyrex 3.3; Invar 0.9; and steel is roughly 12, but there are many different kinds of steel.

Invar was a great invention.

Notice it has a very low expansion coefficient.

It was very important in the 19th century, even today, to make instruments very precise, like clocks.

Clocks are affected by the expansion of the gears.

And so the invention of Invar, which is a mixture of 64% iron and 36% nickel, was invented by a physicist Guillaume in 1898, and for this discovery, he received the Nobel Prize in 1920.

It tells you something how important it was to get an alloy that has a very low expansion coefficient.

If we use these numbers, let us look at the expansion of, for instance, a railroad.

We take a railroad, and we take a piece, a stretch of rail which is, say, a thousand meters.

We take steel, iron--

so this is the expansion coefficient, roughly--

and we compare a cold day, not extremely cold, but a cold day with a hot summer day.

A cold winter day, minus 15 degrees centigrade, and a hot summer day, plus 35 degrees centigrade.

So delta T would be about 50 degrees centigrade.

So what is delta L? Well, that would be 12 times ten to the minus six times ten to the third, times 50, and that is about 0.6 meters, which is about 60 centimeters.

So what are you going to do with that now? How is that solved? If the rail wants to get longer and can't get longer, it will start to bulge either in this way, or sideways, whichever is the easiest.

But the way this is solved is actually quite simple.

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)

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:

Snell's Law
Refraction
Total Reflection
Dispersion
Prisms
Huygens's Principle
The Illusion of Color
The Weird Benham Top
Land's Famous Demo

Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

• iTunes U (MP4 - 104MB)
• Internet Archive (MP4 - 205MB)

  » Download this transcript (PDF)

  Today, I'm going to talk about light.

  Light is an electromagnetic phenomenon, and already in the sixteenth century, way before Maxwell, a lot of studies were done of the interaction of light with water and with glass.

  And the kind of experiments that were done follows -- say this is air -- I call that medium 1 -- and this is water -- call that medium 2 -- and I have a light beam that strikes this surface.

  Light comes in like so -- and I define this angle as the angle of incidence, and I call that theta 1 This is the normal to the surface, and we call that the angle of incidence.

  I will see now that some of that light is reflected -- reflected with an L, as in lion -- and some of that light goes into the water, and we call that refracted -- refracted, with an R, as in Richard -- and this angle, we'll call theta 2.

  And it was a Dutchman, Willebrord Snellius, who, in the seventeenth century, found three rules that govern the relation between these three light beams.

  The first one is that this beam, this beam, and this beam are in one plane.

  As you see, that is my plane of the blackboard.

  The second thing that he found, that this angle, theta three, which is called the angle of reflection, is the same as the angle of incidence.

  That was known before him, of course.

  And then the third one, which is the most surprising one, which is called after him, which is called Snell's Law -- although his name was Snellius -- is that the sine of theta 1 divided by the sine of theta 2, if we go from air to water, then that ratio is about 1.3.

  If you go from air to glass, it's a little higher, it's like 1.5 or so.

  He introduced the idea of index of refraction, which I will call N, as in Nancy -- index of refraction.

  For vacuum, the index of refraction, per definition, is 1, but it's very closely the same in air, we always treat it as 1 in air.

  And in water, the index of refraction is approximately 1.3, and in glass, depending upon what kind of glass you have, it's about 1.5.

  And so we can now amend this law, Snell's Law, by writing here N 2 divided by N 1, N 1 being the index of refraction of the medium where you are, your incident beam -- that's why I put a 1 here -- N 2 being the index of refraction of the medium where you're traveling to.

  You're refracted into this medium.

  And so you see, indeed, that since water is 1.3, and air is 1, that this ratio for air to water is 1.3.

  And this is called Snell's Law.

  And it is immediately obvious that if you go from air to water, or you go from air to glass, that angle theta 2 is always smaller than the angle theta 1, because this number is larger than 1.

  But if you go from water to air, then the situation is reversed, and that's what I want to address now, that's actually quite interesting.

  So now, my medium 1 is now water, and my medium 2 is now air.

  And so now, I go from here to here, and so here I have my angle of incidence theta 1 and here I have my angle of reflection, that is the theta 3, and now here, I have my angle theta 2.

  And so if I write down, now, Snell's Law, then I get the sine of theta 1 divided by the sine of theta 2 is now N 2 / N 1, but N 2 is 1, divided by 1.3, if we go from air -- from water to air.

  And what is so special here is that theta 2 can obviously never be larger than 90 degrees.

  And so if you substitute in here, theta 2 is 90 degrees, then you will find that theta 1, then, is about 50 degrees.

  And if you apply this equation, and you substitute for theta 1 an angle larger than 50 degrees, you're going to find the sine of theta 2 being larger than 1, which is nonsense.

  It cannot happen.

  And so nature ignores Snell's Law, and nature says, "Sorry, I can't do it," and what nature now does, if the angle of theta 1 is too large -- in this case, with water, larger than 50 degrees -- this is not there anymore, and all the light is now being reflected off that surface.

  And we call that total reflection.

  Total reflection.

Pengembangan Perkuliahan

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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.