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Physics33. Geometric OpticsRay Diagrams For Lenses

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