Jumat, 01 Juli 2011

Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Resolving Power
Single-Slit Diffraction
Angular Resolution
Human Eye

Instructor/speaker: Prof. Walter Lewin

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    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.

b. Prof. Bernd Surrow, Ph.D.


Prof. John Belcher

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


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.

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