## Rabu, 29 Juni 2011

### 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

### Video

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

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.

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.

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.

## Senin, 20 Juni 2011

### Himpunan Mahasiswa Pendidikan Fisika Indonesia

Himpunan Mahasiswa Pendidikan Fisika Indonesia

(Indonesian Physics Student Association)

Visi

International Quality in Physics Education

Misi

Is to enhance the understanding and appreciation of physics through teaching

Program

HMPFI bertujuan :

1. Mewujudkan cita-cita Proklamasi Negara Kesatuan Republik Indonesia, dan mempertahankan, mengamankan, serta mengamalkan pancasila dan Undang-undang Dasar 1945
2. Berperan aktif mencapai tujuan nasional dalam mencerdaskan bangsa dan membentuk manusia Indonesia seutuhnya
3. Berperan serta mmengembangkan system dan pelaksanaan pendidikan nasional
4. Mempertinggi kesadaran dan sikap Mahasiswa Pendidikan Fisika, meningkatkan mutu dan kemampuan profesi guru di masa depan dan tenaga kependidikan lainnya
5. Menjaga, memelihara, membela, serta meningkatkan harkat dan martabat Mahasiswa Fisika melalui peningkatan kesejahteraan anggota serta kesetiakawanan organisasi.

Fokus

Inovasi, Riset dan Pengembangan Pendidikan Fisika

PENDIDIKAN DI INDONESIA : MASALAH DAN SOLUSINYA

OLEH : M. SHIDDIQ AL-JAWI

2. Masalah Mendasar : Sekularisme Sebagai Paradigma Pendidikan

Jarang ada orang mau mengakui dengan jujur, sistem pendidikan kita adalah sistem yang sekular-materialistik. Biasanya yang dijadikan argumentasi, adalah UU Sisdiknas No. 20 tahun 2003 pasal 4 ayat 1 yang berbunyi, "Pendidikan nasional bertujuan membentuk manusia yang beriman dan bertakwa kepada Tuhan Yang Maha Esa, berakhlak dan berbudi mulia, sehat, berilmu, cakap, serta menjadi warga negara yang demokratis dan bertanggungjawab terhadap
kesejahteraan masyarakat dan tanah air."

Tapi perlu diingat, sekularisme itu tidak otomatis selalu anti agama. Tidak selalu anti "iman" dan anti "taqwa". Sekularisme itu hanya menolak peran agama untuk mengatur kehidupan publik, termasuk aspek pendidikan. Jadi, selama agama hanya menjadi masalah privat dan tidak dijadikan asas untuk menata kehidupan publik seperti sebuah sistem pendidikan, maka sistem pendidikan itu tetap sistem pendidikan sekular, walaupun para individu pelaksana sistem itu beriman dan bertaqwa (sebagai perilaku individu).

Sesungguhnya diakui atau tidak, sistem pendidikan kita adalah sistem pendidikan yang sekular-materialistik. Hal ini dapat dibuktikan antara lain pada UU Sisdiknas No. 20 tahun 2003 Bab VI tentang jalur, jenjang dan jenis pendidikan bagian kesatu (umum) pasal 15 yang berbunyi: Jenis pendidikan mencakup pendidikan umum, kejuruan, akademik, profesi, advokasi, keagaman, dan khusus.

Dari pasal ini tampak jelas adanya dikotomi pendidikan, yaitu pendidikan agama dan pendidikan umum. Sistem pendidikan dikotomis semacam ini terbukti telah gagal melahirkan manusia salih yang berkepribadian Islam sekaligus mampu menjawab tantangan perkembangan melalui penguasaan sains dan teknologi.

Secara kelembagaan, sekularisasi pendidikan tampak pada pendidikan agama melalui madrasah, institut agama, dan pesantren yang dikelola oleh Departemen Agama; sementara pendidikan umum melalui sekolah dasar, sekolah menengah, kejuruan serta perguruan tinggi umum dikelola oleh Departemen Pendidikan Nasional. Terdapat kesan yang sangat kuat bahwa pengembangan ilmu-ilmu kehidupan (iptek) dilakukan oleh Depdiknas dan dipandang sebagai tidak berhubungan dengan agama. Pembentukan karakter siswa yang merupakan bagian terpenting dari proses pendidikan justru kurang tergarap secara serius. Agama ditempatkan sekadar sebagai salah satu aspek yang perannya sangat minimal, bukan menjadi landasan dari seluruh aspek kehidupan.

Hal ini juga tampak pada BAB X pasal 37 UU Sisdiknas tentang ketentuan kurikulum pendidikan dasar dan menengah yang mewajibkan memuat sepuluh bidang mata pelajaran dengan pendidikan agama yang tidak proposional dan tidak dijadikan landasan bagi bidang pelajaran yang lainnya.

Ini jelas tidak akan mampu mewujudkan anak didik yang sesuai dengan tujuan dari pendidikan nasional sendiri, yaitu mewujudkan suasana belajar dan proses pembelajaran agar peserta didik secara aktif mengembangkan potensi dirinya untuk memiliki kekuatan spiritual keagamaan, pengendalian diri, kepribadian, kecerdasan, akhlak mulia, serta keterampilan yang diperlukan dirinya, masyarakat, bangsa dan negara. Kacaunya kurikulum ini tentu saja berawal dari asasnya yang sekular, yang kemudian mempengaruhi penyusunan struktur kurikulum yang tidak memberikan ruang semestinya bagi proses penguasaan tsaqafah Islam dan pembentukan kepribadian Islam.

## Minggu, 19 Juni 2011

### 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

### Video

• iTunes U (MP4 - 108MB)
• Internet Archive (MP4 - 211MB)

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.

## Sabtu, 18 Juni 2011

### Fisika untuk Universitas

Fisika untuk Universitas

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

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--

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Rabu, 15 Juni 2011

### 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

### Video

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

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?

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?

Two?

One?

And who knows the answer to zero?

Most of you, right?

I haven't seen a lot of hands though.

All right.

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.

## Jumat, 10 Juni 2011

### 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

### Video

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

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.