字幕表 動画を再生する 英語字幕をプリント The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. MICHAEL SHORT: OK. I think things have been getting pretty derivy lately, so I wanted to shift gears to something a little bit more practical. So I started alluding to this hypothetical radiation source I might have right here, and things like if you have a source of known activity, which we calculated yesterday, and you have a detector of unknown efficiency, how do you know what the efficiency is? How do you know what, let's say, your dose distance relationship is? And how do you calculate all this stuff? So let's take the general situation that we're starting to work out. Let's say we have a Geiger counter right here. That's our GM tube. And we have a point source that's emitting things in all directions. Let's go with the stuff from yesterday. Let's say it's a cobalt 60 source. It's now 0.52 microcurie. The question is, how many counts do you expect in this detector when it's a certain distance away? So I've actually laser-cut out a little Geiger counter jig from a previous class. And you guys can all do this too. Who here has been to the IDC before? A couple. The international design center-- so they've got a laser cutter that you can sign up to use, which is where I did this. And it's set to just take a Geiger counter and put your sources at some fixed distance away so you can discover the dose distance relationship with things. Speaking of, does anybody know what the relationship is between dose and distance or measured activity and distance? Yeah, Luke. AUDIENCE: [INAUDIBLE] r cubed. MICHAEL SHORT: Close. It's, let's say, the measured activity would be proportional to 1 over r squared. Who knows where this comes from? I'll move the source a bit away to lessen the beeping. Yeah. AUDIENCE: Well, the flux of particles coming out is just [INAUDIBLE] over the surface area of [INAUDIBLE] and the [INAUDIBLE] is 4 pi r squared. MICHAEL SHORT: Yeah, exactly. If you were to draw a hypothetical sphere around the source right here, then you've got, let's say, a detector that's roughly rectangular with a fixed area. Let's say it's got a half length L and a half width W. Then the area-- I'm sorry, let's just say length L, width W-- would be just L times W. And actually, what Chris mentioned as the solid angle subtended by this detector right here-- in other words, at a certain distance r away, how much of this sphere-- how much does the area of this sphere-- does this detector take up? In other words, how many of these gamma rays are going to go in a different direction than the detector, versus how many we'll actually enter the detector? And a simple formula for the solid angle is just the surface area of whatever you've got over r squared. It's a pretty good approximation to the solid angle of something for very long distances, and it's probably the one that you'll see in the reading. But I wanted to show you the actual formula, in this case, for a rectangle-- solid angle comparison. Good, that's up there. So let's say on the x-axis, right here, this would be distance from the source to the detector in meters. And I've said that we've got some sort of a detector that is 2.5 by 10 meters in size. That's an enormous detector. Let's actually switch it to the units right here. So this is roughly 10 centimeters long. So let's change our length to 0.1. And what do you think the width of this Geiger counter is in meters? AUDIENCE: A centimeter MICHAEL SHORT: A centimeter. 0.01. We're going to have to change our axes so we can actually see the graph. So instead of looking all the way out to 15 meters away, let's look one meter away, maybe less. This whole thing is probably 50 centimeters. And we'll take a look there. And what we notice is that except for extremely short distances, this approximate formula for the solid angle-- or in other words, if I were to draw a sphere around the source that's the radius of the distance between the source and the detector, how much of that sphere's area does the detector take up? This approximate formula-- the blue curve-- is a pretty good approximation of the red curve until you get really, really close to 5 centimeters away, or about this distance right here. Does anyone know why this formula would break down? What happens as r goes to 0? What happens to our solid angle or our approximation for our solid angle? AUDIENCE: Goes to Infinity MICHAEL SHORT: It goes to infinity, right? Can a detector actually take up infinity area on, well, anything? Never mind that unit sphere. Not quite. If you were to take this detector and bring the radius down to 0 so that the source and the detector, if not counting for the thickness of the plastic, were right upside each other, if that solid angle went to, well, infinity , then the count should go to infinity, and it does not compute. Does anyone know how many-- first of all, who here has heard of solid angle before? So a little more than half of you. That's getting clicky. I'm going to turn that off. Solid angle is kind of the analog to regular old angle, except in 3D. So instead of looking at things in radians, this has the unit of what's called steradians-- steradians-- with a full sphere taking up 4pi steradians. Interestingly enough, 4pi is also the surface area of a unit sphere with radius of 1. So that's where this comes from. If something were to completely cover a unit sphere-- like, if you were to, let's say, encase a light source in tin foil completely, and say, how much of that solid angle does the tin foil encase? It would be 4pi steradians, regardless of the size of the sphere or how much tin foil you had to use. So this pretty simple formula isn't the best approximation for it. And I'm not going to go through the derivation, because like I said, today is going to be a more practical nature. There is a more complex and rigorous formula for the solid angle of something, let's say, in this case, a rectangle of length L and with W, from a certain distance r, or, in this case, on our graph, x away from the sphere. And you can actually see that red curve right there. Once you get to a few centimeters away, it's pretty close. Anyone want to guess what the maximum value of the red curve is? If I take this source and slam it right up next to the detector, how much of sphere is the detector subtending? AUDIENCE: 2pi MICHAEL SHORT: 2pi-- half the sphere. Because let's say this whole side of the source is completely obscured by the detector and this whole side is free to move. And if you look really closely, yep, at 0, the correct formula does give you 2pi steradians. Which is to say that half the gamma rays leaving the source would enter the detector.