Fisika untuk Universitas

Fisika untuk Universitas

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

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All physics of the 19th century and earlier is called classical physics.

 Examples are Newtonian mechanics, which we dealt with this whole term, and electricity and magnetism, which you will encounter the next term. In the early part of this century, when we learned about the composition of atoms, it became clear that classical physics did not work on the very small scale of the atoms. The size of an atom is only ten to the minus ten meters.

 If you take 250 million of them and you line them up, that's only one inch. In 1911, the English physicist Rutherford demonstrated that almost all the mass of an atom is concentrated in an extreme small volume at the center of the atom. 

We call that the nucleus, it's positively charged. And there are electrons which are negatively charged, which are in orbits around the nucleus, and the typical distances from the nucleus to the electrons is about 100,000 times larger than the size of the nucleus itself. 

As early as 1920, Rutherford named the proton, and Chadwick discovered the neutron in 1932, for which he received the Nobel Prize. Now, let us imagine that this lecture hall is an atom. 

And the size of an atom is defined by the orbits, the outer orbits of the electrons. If I scale it properly, now, in this ratio 100,000 to 1, then the size of the nucleus would be even smaller than a grain of sand.

 And it just so happens that yesterday I went to Plum Island, I walked for three hours on the beach and I ended up with some sand in my pockets. And so I will donate to you one proton; make sure you hold onto it... Ooh, this is two protons, that's too generous. So keep it there this is one proton. 

And there would be an electron, then, anywhere there, near the walls, going around like mad in orbit and that would then be a hydrogen atom. Just think about what an atom is. 

 An atom is all vacuum. You and I are all vacuum. You think of yourself as being something, but we are nothing. You can ask yourself the question, If you are all vacuum, why is it, then, that I can move my hand not through the other hand, like a ghost can walk through a wall? That's not so easy to answer, and in fact, you cannot answer it with classical physics and I will not return to that today.

 But you are all vacuum. According to Maxwell's equations, Maxwell's law of electricity and magnetism, an electron, because of the attractive force of the proton, would spiral into the proton in a minute fraction of a second, and so atoms could not exist. Now, we know that's not true. 

We know that atoms do exist. And so that created a problem for physics and it was the Danish physicist Niels Bohr who in 1913 postulated that electrons move around the nucleus in well-defined orbits which are distinctly separated from each other, and that the spiraling-in of the electrons into the nucleus does not occur, for the reason that an electron cannot exist in between these allowed orbits. It can jump from one orbit to another, but it cannot exist in between. 

Now, Bohr's suggestion was earth-shaking, because it would also imply that a planet that goes around the sun cannot orbit the sun just at any distance. You couldn't move it just a trifle in or a trifle farther out. It would also require discrete orbits. 

It would also mean that if you had a tennis ball and you would bounce the tennis ball up and down, that the tennis ball could not reach just any level above the ground, but it would only be discrete levels, and that is very much against our intuition. We'd like to think that when you bounce a tennis ball, that it can reach any level that you want to. You give it just a little bit more energy and it will go a little higher. That, according to quantum mechanics, would not be possible. 

Now, all this seems rather bizarre, as it goes against our daily experiences, but before we dismiss the idea of quantization see, the quantization comes in when you talk about discrete orbits-- you have to realize that the differences in the allowed heights of the tennis ball and the differences between the allowed orbits of the planets around the sun would be so infinitesimally small that we may never be able to measure it. 

 In other words, quantum mechanics really plays no role in our macroscopic world. Now, atoms are very, very small compared to tennis balls, and the quantization effects are much larger in the sub-microscopic world of electrons and atoms than in our familiar world of baseballs, pots and pans, and planets. So before we continue, I would like to repeat to you one of the cornerstones of quantum mechanics. And it says that the electrons in atoms can only exist at well-defined energy levels think of them as being orbits around the nucleus, and they cannot exist in between. 

Now, when I heat a substance, the electrons in the atoms can jump from inner orbits to allowed outer orbits, and when they do so, they can leave a hole, an opening, an empty space in the inner orbits. But later on, they can fall back to fill that opening. They can occupy that place again.

 And when I keep heating this substance, there is some kind of a musical chair game going on. The electrons will go to outer orbits, they may spend there some time and then they may fall to lower orbits, to inner orbits. You see here a vase, a very precious vase, and when I pick up this vase, I have to do work. I bring it further away from the center of the Earth.

 Now, is that energy lost? No. I could drop the vase, and it would pick up kinetic energy. I will get that energy back. Gravitational potential energy will be converted to kinetic energy. It will crash to pieces, and it will generate some heat.

 In fact, the breaking itself of this vase would take some energy. In a similar way, the energy that you put into electrons when you bring them to outer orbits is retrieved when the electrons fall back. 

So there is a parallel-- dropping this vase and getting your work back that I put in. It wouldn't be a nice thing to do to this 500-year-old vase, but as far as I'm concerned, perfectly reasonable to do it with Ohanian, so we can let that go, and the energy will come out in the form of heat and also in the form of, perhaps, some noise. 

When electrons fall from an outer orbit back to an inner orbit, it's not kinetic energy that is released, but it comes out often in the form of light, electromagnetic radiation. Light has energy.

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:

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:

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

Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

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

  » Download this transcript (PDF)

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

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

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

  I have N, capital N.

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

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

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

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

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

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

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

  And that's only in the small angle approximation.

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

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

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

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

  We had exactly the same equation.

  There was no difference.

  And D now is the separation between two sources here.

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

  But now comes the big surprise.

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

  We haven't discussed the destructive interference.

  And to derive that properly is very tricky.

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

  But I will give you the results.

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

  And minima means complete destructive interference.

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

  We had only one zero in between the two maxima.

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

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

  I take the small angle approximation.

  So this angle is now in radians.

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

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

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

  And the same is the case here.

  And this can be huge.

  N can be a few hundred.

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

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

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

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

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

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

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

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

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

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

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

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

  So you get nine times more light.

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

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

  Of course you gain a factor of three.

  You have three times more sources.

  You get three times more light.

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

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

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

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

  A micron is 10 to the -6 meters.

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

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

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

  Which is about 10 meters.

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

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

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

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

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

  That is 0 degrees.

  That's immediately obvious.

  I use that equation.

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

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

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

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

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

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

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

  Well, my grating is like so.

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

  There are 2500 of those lines per inch.

  And my laser beam is roughly 2 millimeters in size.

  So this is about 2 millimeters.

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

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

  Because this factor of N enters into it here.

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

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

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

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

  So this is what I want to show you first.

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

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

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

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

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

b. Prof. Bernd Surrow, Ph.D.

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

Staff

Visualizations:
Prof. John Belcher

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

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

Technical Instructors:
Andy Neely
Matthew Strafuss

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

Acknowledgements

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

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Double-Slit Interference
Interferometers

Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

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

  » Download this transcript (PDF)

  I'm very proud of you.

  You did very well on the last exam.

  Class average is a little bit above 70.

  Congratulations.

  There were 22 students who scored 100.

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

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

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

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

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

  But today I will focus on the wave character only.

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

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

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

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

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

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

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

  So the mountains become higher and the valleys become lower.

  We call that constructive interference.

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

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

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

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

  So the waves go out on a sphere.

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

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

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

  Because you're talking about 3-D waves.

  You're talking about spherical waves.

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

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

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

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

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

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

  And we call that destructive interference.

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

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

  That's when you would have destructive interference.

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

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

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

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

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

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

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

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

  If the two are in phase.

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

  So this line would be always a maximum.

  Constructive interference.

  It's this 0, substitute there.

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

  Going perpendicular to the blackboard right through the middle.

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

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

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

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

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

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

  And so this is what we call an interference pattern.

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

  Figure that out at home.

  That's very easy.

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

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

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

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

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

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

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

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

  The mountains and the valleys arrive at the same time.

  The water is never moving at all.

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

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

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

  Starts tapping already.

  I can see the nodal lines very well.

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

  At all moments in time it's standing still.

  Here's one.

  Here is one.

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

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

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

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

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

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

  You would not be bobbing up and down at all.

  And that is very characteristic for waves.

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

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

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

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

  And you see that butterfly here.

  Tom, perhaps you can turn off that overhead.

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

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

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

  So no other butterfly was sacrificed.

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

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

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

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

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

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

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

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

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

  This is theta.

  And so this is theta.

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

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

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

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

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

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

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

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

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

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

  And then I have the whole family of destructive interference.

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

  Just as we had it on the blackboard there.

  We discussed that earlier.

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

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

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

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

  It's a different scale.

  And here is a screen.

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

  And here is one of those direction theta.

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

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

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

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

  That is simple geometry.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

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

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

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

b. Prof. Bernd Surrow, Ph.D.

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

Staff

Visualizations:
Prof. John Belcher

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

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

Technical Instructors:
Andy Neely
Matthew Strafuss

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

Acknowledgements

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

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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.

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

Free Downloads

Video

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

  » Download this transcript (PDF)

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

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

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

  I can only touch upon a few of them.

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

  Let's first look at magnetic materials.

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

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

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

  Has to do with quantum mechanics.

  It's not part of 8.02.

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

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

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

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

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

  And the current I is flowing through the solenoid.

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

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

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

  The magnetic permeability, and this can be huge.

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

  So you get an enormous increase in magnetic field strength.

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

  That's just the definition of self-inductance.

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

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

  We have an interesting problem.

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

  Let's now turn to transformers.

  A transformer often comes in this shape.

  Let me move it a little bit to the right.

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

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

  N1 windings, index, self-inductance L1.

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

  And this is the secondary side.

  N2 windings.

  Self-inductance L2.

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

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

  That's a good approximation.

  Depends on how well the coupling goes.

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

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

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

  That is very, very special.

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

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

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

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

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

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

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

  It was enough to melt that iron nail.

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

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

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

  So let's now go to RLC circuits there.

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

  AC.

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

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

  There is nothing special about cosine in life.

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

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

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

  We call this the reactance.

  The upstairs.

  For which we give often the symbol X.

  And so this is also X then divided by R.

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

  The units are ohms.

  And we call that Z.

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

  Then the cosine term is either plus or minus 1.

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

  So here is frequency and here is I max.

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

  I max is 0.

  There's no current flowing at all.

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

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

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

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

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

  Because when the reactance is zero this part vanishes.

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

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

  When I am at resonance, phi becomes 0.

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

  They are in phase with each other.

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

  Because the impedance itself becomes R.

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

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

  I have here some numbers which you have seen before.

  During my lectures.

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

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

  And I calculate for you here the resonance frequency.

  I calculate the frequency also in terms of kilohertz.

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

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

  So your reactance in this case becomes -86 ohms.

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

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

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

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

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

  They annihilate each other and so the reactance becomes 0.

  So Z now is just pure R.

  X is 0.

  And so the maximum current in this case is 1.

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

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

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

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

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

  That power comes out in the form of heat.

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

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

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

  Either one will be fine.

  I've decided to take this one.

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

  This is the power at any moment in time.

  I will do the time averaging a little later.

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

  That was drilled into my memory here.

  I will never forget that, I think.

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

  So this is this term.

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

  So this term vanishes.

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

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

  So I get a two here.

  And then I still have my cosine phi there.

  And I'm done.

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

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

  You still see it there.

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

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

  That's this part.

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

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

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

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

  So now we can look at resonance.

  It's always a very special situation.

  When we are at resonance, Z equals R.

  So you replace this capital Z by R.

  And then you find V0 squared divided by 2R.

  That's utterly trivial.

  You could have predicted that.

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

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

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

  And you find immediately then that answer.

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

  You see that immediately here.

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

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

  All right.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

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

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

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

b. Prof. Bernd Surrow, Ph.D.

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

Staff

Visualizations:
Prof. John Belcher

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

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

Technical Instructors:
Andy Neely
Matthew Strafuss

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

Acknowledgements

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

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.