Anderson Video - Resolving Power of a Telescope

Professor Anderson
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<font color="#ffffff">The nearest neighboring star to the Sun is about four light years away.</font> <font color="#ffffff">If a planet happened to be orbiting this star at an orbital radius </font> <font color="#ffffff">equal to that of the Earth-Sun distance,</font> <font color="#ffffff">what minimum diameter would an earth-based telescopes aperture have to be</font> <font color="#ffffff">in order to obtain an image that is resolved by this star planet system?</font> <font color="#ffffff">Assume the light emitted by the star and planet has a wavelength of 650 nanometers</font> <font color="#ffffff">and then it tells us about the Earth-Sun distance and one lightyear.</font> <font color="#ffffff">So, let's see if we can attack this problem.</font> <font color="#ffffff">And what we need is to understand the resolving power of a telescope.</font> <font color="#ffffff">These are new pens they don't squeak. Yeah, brand new pens don't squeak.</font> <font color="#ffffff">Okay, and so if we understand the resolving power of a telescope</font> <font color="#ffffff">then maybe we can understand how to attack this problem.</font> <font color="#ffffff">So let's see if we can draw a simple picture of what we're trying to do.</font> <font color="#ffffff">Here is a star, here is a planet.</font> <font color="#ffffff">Both of them are gonna admit light, so just pretend they're two light sources.</font> <font color="#ffffff">This one is emitting light, this one is emitting light.</font> <font color="#ffffff">You are way over here with your telescope</font> <font color="#ffffff">and you have a plane of film right there or CCD</font> <font color="#ffffff">that you're gonna record the image of the star on.</font> <font color="#ffffff">So light rays from each star come to our telescope</font> <font color="#ffffff">and when those light rays come into our lens,</font> <font color="#ffffff">they don't just make a single point on the film.</font> <font color="#ffffff">Okay? A point source here doesn't just become a point source here</font> <font color="#ffffff">because there is some diameter D of this system </font> <font color="#ffffff">that causes diffraction.</font> <font color="#ffffff">Any time light goes through a slit, no matter how big it is,</font> <font color="#ffffff">it could be a lens, it could be some aperture, it's going to diffract.</font> <font color="#ffffff">And so the light from this star in fact makes a pattern that looks like this on the film.</font> <font color="#ffffff">on the film, okay?</font> <font color="#ffffff">Light from the other star makes a similar pattern but it's shifted in space.</font> <font color="#ffffff">Let's change the color of that light just so we can distinguish them on the film.</font> <font color="#ffffff">So here's the other star.</font> <font color="#ffffff">Okay, and it is emitting light which goes to our lens</font> <font color="#ffffff">and now it's going to make a pattern that is shifted slightly.</font> <font color="#ffffff">Okay, it doesn't really matter which one shifted up or down, they're going to be spread out.</font> <font color="#ffffff">And if they are spread out, it's tough to distinguish the two stars.</font> <font color="#ffffff">Okay? The patterns start to overlap.</font> <font color="#ffffff">So, how do we figure out when we can distinguish the two and when we can't?</font> <font color="#ffffff">To do that, we need to understand the resolving power of a telescope.</font> <font color="#ffffff">And if you look in your chapter it tells us the minimum resolvable angle,</font> <font color="#ffffff">and we talked about this in class,</font> <font color="#ffffff">it is given by theta, which is 1.22 lambda over D.</font> <font color="#ffffff">What is theta in this picture? Theta is this angle right here.</font> <font color="#ffffff">What is the angle between the rays from those two different objects?</font> <font color="#ffffff">D is the diameter of the lens, lambda is the wavelength that is given to us.</font> <font color="#ffffff">Okay, and they gave us the lambda for mine, lambda is 650 nanometers.</font> <font color="#ffffff">Alright, we're looking for D in this problem, so we need to figure out what theta is.</font> <font color="#ffffff">Now if I think about this triangle here, there is some </font> <font color="#ffffff">distance, H, between the two objects that are emitting light.</font> <font color="#ffffff">There is also some distance L between those objects and the telescope that we're using.</font> <font color="#ffffff">Now, H -- they give us to us. H is the following, it is 149.6 times ten to the six kilometers.</font> <font color="#ffffff">Which is ten to the nine meters.</font> <font color="#ffffff">And they also give us the distance L.</font> <font color="#ffffff">L is equal to four lightyears. Four lightyears.</font> <font color="#ffffff">And each lightyear is a distance of 9.461 times ten to the fifteen meters.</font> <font color="#ffffff">All right, so it looks like we have enough here to calculate theta</font> <font color="#ffffff">if we remember the small angle approximation.</font> <font color="#ffffff">And the small angle approximation says the following:</font> <font color="#ffffff">theta is essentially the same as tangent of theta.</font> <font color="#ffffff">And tangent of theta here, is just H over L.</font> <font color="#ffffff">Okay, this number H is clearly much smaller than this distance L</font> <font color="#ffffff">and so we are satisfying the small angle approximation,</font> <font color="#ffffff">Okay, and now we can put everything together.</font> <font color="#ffffff">Let's solve for D.</font> <font color="#ffffff">D is equal to 1.22 lambda divided by theta.</font> <font color="#ffffff">Just solving this equation for D.</font> <font color="#ffffff">And now we have all these numbers, 1.22 times lambda, which is 650 times ten to the minus nine meters.</font> <font color="#ffffff">That's nano.</font> <font color="#ffffff">And then we have theta, but theta is just H over L.</font> <font color="#ffffff">H is 149.6 times ten to the nine meters.</font> <font color="#ffffff">L was four times 9.461 times ten to the fifteen meters</font> <font color="#ffffff">and all of this is in SI units.</font> <font color="#ffffff">And so we're going to end up with a calculation in meters and you can check it yourself,</font> <font color="#ffffff">But I ended up with 0.2 meters.</font> <font color="#ffffff">Okay, 20 centimeters for the diameter of this lens, right?</font> <font color="#ffffff">You'll have some different numbers particularly on your wavelength.</font> <font color="#ffffff">Everything else should be the same so you should get a diameter that is roughly on that order.</font> <font color="#ffffff">Which is pretty remarkable when you think about it, right?</font> <font color="#ffffff">These things are 4 light years away and somehow we can resolve them</font> <font color="#ffffff">with a diameter telescope that's 20 centimeters, right?</font> <font color="#ffffff">20 centimeters is like that, that's not that big, okay?</font> <font color="#ffffff">So it's pretty impressive what you can see with a normal-sized telescope.</font>
<font color="#ffffff">The nearest neighboring star to the Sun is about four light years away.</font> <font color="#ffffff">If a planet happened to be orbiting this star at an orbital radius </font> <font color="#ffffff">equal to that of the Earth-Sun distance,</font> <font color="#ffffff">what minimum diameter would an earth-based telescopes aperture have to be</font> <font color="#ffffff">in order to obtain an image that is resolved by this star planet system?</font> <font color="#ffffff">Assume the light emitted by the star and planet has a wavelength of 650 nanometers</font> <font color="#ffffff">and then it tells us about the Earth-Sun distance and one lightyear.</font> <font color="#ffffff">So, let's see if we can attack this problem.</font> <font color="#ffffff">And what we need is to understand the resolving power of a telescope.</font> <font color="#ffffff">These are new pens they don't squeak. Yeah, brand new pens don't squeak.</font> <font color="#ffffff">Okay, and so if we understand the resolving power of a telescope</font> <font color="#ffffff">then maybe we can understand how to attack this problem.</font> <font color="#ffffff">So let's see if we can draw a simple picture of what we're trying to do.</font> <font color="#ffffff">Here is a star, here is a planet.</font> <font color="#ffffff">Both of them are gonna admit light, so just pretend they're two light sources.</font> <font color="#ffffff">This one is emitting light, this one is emitting light.</font> <font color="#ffffff">You are way over here with your telescope</font> <font color="#ffffff">and you have a plane of film right there or CCD</font> <font color="#ffffff">that you're gonna record the image of the star on.</font> <font color="#ffffff">So light rays from each star come to our telescope</font> <font color="#ffffff">and when those light rays come into our lens,</font> <font color="#ffffff">they don't just make a single point on the film.</font> <font color="#ffffff">Okay? A point source here doesn't just become a point source here</font> <font color="#ffffff">because there is some diameter D of this system </font> <font color="#ffffff">that causes diffraction.</font> <font color="#ffffff">Any time light goes through a slit, no matter how big it is,</font> <font color="#ffffff">it could be a lens, it could be some aperture, it's going to diffract.</font> <font color="#ffffff">And so the light from this star in fact makes a pattern that looks like this on the film.</font> <font color="#ffffff">on the film, okay?</font> <font color="#ffffff">Light from the other star makes a similar pattern but it's shifted in space.</font> <font color="#ffffff">Let's change the color of that light just so we can distinguish them on the film.</font> <font color="#ffffff">So here's the other star.</font> <font color="#ffffff">Okay, and it is emitting light which goes to our lens</font> <font color="#ffffff">and now it's going to make a pattern that is shifted slightly.</font> <font color="#ffffff">Okay, it doesn't really matter which one shifted up or down, they're going to be spread out.</font> <font color="#ffffff">And if they are spread out, it's tough to distinguish the two stars.</font> <font color="#ffffff">Okay? The patterns start to overlap.</font> <font color="#ffffff">So, how do we figure out when we can distinguish the two and when we can't?</font> <font color="#ffffff">To do that, we need to understand the resolving power of a telescope.</font> <font color="#ffffff">And if you look in your chapter it tells us the minimum resolvable angle,</font> <font color="#ffffff">and we talked about this in class,</font> <font color="#ffffff">it is given by theta, which is 1.22 lambda over D.</font> <font color="#ffffff">What is theta in this picture? Theta is this angle right here.</font> <font color="#ffffff">What is the angle between the rays from those two different objects?</font> <font color="#ffffff">D is the diameter of the lens, lambda is the wavelength that is given to us.</font> <font color="#ffffff">Okay, and they gave us the lambda for mine, lambda is 650 nanometers.</font> <font color="#ffffff">Alright, we're looking for D in this problem, so we need to figure out what theta is.</font> <font color="#ffffff">Now if I think about this triangle here, there is some </font> <font color="#ffffff">distance, H, between the two objects that are emitting light.</font> <font color="#ffffff">There is also some distance L between those objects and the telescope that we're using.</font> <font color="#ffffff">Now, H -- they give us to us. H is the following, it is 149.6 times ten to the six kilometers.</font> <font color="#ffffff">Which is ten to the nine meters.</font> <font color="#ffffff">And they also give us the distance L.</font> <font color="#ffffff">L is equal to four lightyears. Four lightyears.</font> <font color="#ffffff">And each lightyear is a distance of 9.461 times ten to the fifteen meters.</font> <font color="#ffffff">All right, so it looks like we have enough here to calculate theta</font> <font color="#ffffff">if we remember the small angle approximation.</font> <font color="#ffffff">And the small angle approximation says the following:</font> <font color="#ffffff">theta is essentially the same as tangent of theta.</font> <font color="#ffffff">And tangent of theta here, is just H over L.</font> <font color="#ffffff">Okay, this number H is clearly much smaller than this distance L</font> <font color="#ffffff">and so we are satisfying the small angle approximation,</font> <font color="#ffffff">Okay, and now we can put everything together.</font> <font color="#ffffff">Let's solve for D.</font> <font color="#ffffff">D is equal to 1.22 lambda divided by theta.</font> <font color="#ffffff">Just solving this equation for D.</font> <font color="#ffffff">And now we have all these numbers, 1.22 times lambda, which is 650 times ten to the minus nine meters.</font> <font color="#ffffff">That's nano.</font> <font color="#ffffff">And then we have theta, but theta is just H over L.</font> <font color="#ffffff">H is 149.6 times ten to the nine meters.</font> <font color="#ffffff">L was four times 9.461 times ten to the fifteen meters</font> <font color="#ffffff">and all of this is in SI units.</font> <font color="#ffffff">And so we're going to end up with a calculation in meters and you can check it yourself,</font> <font color="#ffffff">But I ended up with 0.2 meters.</font> <font color="#ffffff">Okay, 20 centimeters for the diameter of this lens, right?</font> <font color="#ffffff">You'll have some different numbers particularly on your wavelength.</font> <font color="#ffffff">Everything else should be the same so you should get a diameter that is roughly on that order.</font> <font color="#ffffff">Which is pretty remarkable when you think about it, right?</font> <font color="#ffffff">These things are 4 light years away and somehow we can resolve them</font> <font color="#ffffff">with a diameter telescope that's 20 centimeters, right?</font> <font color="#ffffff">20 centimeters is like that, that's not that big, okay?</font> <font color="#ffffff">So it's pretty impressive what you can see with a normal-sized telescope.</font>