The Doppler Effect of Light - Video Tutorials & Practice Problems
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The Doppler Effect (Light)
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Video transcript
Hey folks, welcome back. So remember when we talked about sound waves, there was a phenomenon known as the Doppler effect, which was basically a shift in the frequency that you observe. Well, it turns out that same Doppler effect also applies for electromagnetic waves otherwise known as light. That's what we're gonna talk about in this video. All right, let's just jump right into it. So remember that the Doppler effect really just shifts the frequency that you observe based on the relative speeds of the listener and the source and the observer. The classic, the classic example that we saw for sound waves was you, for example, running towards an ambulance siren or maybe an ambulance siren is moving towards you or something like that. So basically, we had the velocity of the source and we had the velocity of the listener. And so for example, the ambulance siren would sound different as it was going towards you versus as it was going away from you. And we use this equation over here. Now, the main difference between the sound wave version and the light version of the Doppler effect is in the equation. So we saw this equation here. And the main thing was we had to figure out the velocity of the listener and the source and those things could have different values and signs, we had to kind of figure out which one was positive, which one was negative and it was really tricky. So for the uh electromagnetic wave version, it's actually even simpler. The equation turns out to be that the observed frequency is going to be the frequency of the source times one plus or minus the relative velocity divided by C. All right. So basically what happens here is that you only need the relative speeds of the source and the observer. And basically what that means here is that for example, if you have the speed towards the source and this is equal to, let's say four and the source is also moving towards you with let's say six, then that means that the relative speed V relative is actually the addition of those two velocities, it's four plus six and that equals 10 m per second. So this is actually what you need for this equation, the relative velocity that gets plugged in here. And that by the way, always gets plugged in as a positive number. So you will always plug this in as a positive number. The real tricky part only happens when you have to figure out which sign do you use in this equation? Do you use a plus or a minus? And it really just comes down to these two basic rules, you'll use a plus sign whenever the observer and the source are getting closer together. So for example, the this person and this light bulb are getting closer together. So you use a positive, but if they're going away from each other, you're gonna use a negative sign. So you use the negative sign when they're getting farther apart, that's really all there is to it. So let's just go ahead and jump right into a problem here. So we have a distant star and it's radiating light with a wavelength of 630 nanometers. So that's what theoretically should we or should radiate, right? So basically what, what this says here is that the wavelength of the source is equal to 630 nanometers. Now, what we want to do is we want to calculate the wavelength of light that we measure when looking at the star. So that's the source. So really what we're looking at is what is the uh what is the lambda that we actually observe? All right. Now, you might notice that in this problem, we're using Lambdas instead of frequencies. But remember, we can always change them by using uh the equation for the speed of light. So let's just go ahead and set up our equation here. All right. So we have, that's uh let's see. I'm just gonna put this over here. So we have that F observed is equal to uh this is gonna be F source times one plus or minus V relative divided by C. All right. So by the way, we actually know what our V relative is. The V relative is just the speed that we're told that this, this star is receding from the earth three times 10 to the sixth meters per second. All right. So it's three times 10 to the sixth. OK. So first, we have to figure out, well, the F observed is what we're looking for right here. So what about the f of the source that really just comes from the Lambda of the source? Remember we can always just get to F so, FS is just gonna be equal to C divided by lambda S. So this is gonna be three times 10 to the eighth, divide by, this is gonna be 630 times 10 to the negative nine. And so that FS is just going to be 4.76 times 10 to the 14 and that's in Hertz. So that's what you plug into this uh over here. So now we have that F observed is gonna equal, this is gonna be 4.76 times 10 to the 14. Now we have one and there is gonna be a plus or a minus sign. What happens is remember this thing is receding. So that means that because this is receding over here, we're gonna use a minus sign all right. So this is receding. We use a minus sign here and this is gonna be V relative. So this is just gonna be three times 10 to the sixth divided by the speed of light, three times 10 to the eighth. You always want to put this in parentheses when you're doing this, by the way, just so you don't get the wrong answer. And then basically what you should get here is you should get 4.71. This is gonna be 715 times 10 to the 14th. Now, remember this is just the frequency that you observe. Now, what we have to do is we just have to ch change it back into Lambda observe by using the same exact equation over here. So in other words, Lambda observed is just going to be C over F observed and this is just gonna be three times 10 to the eighth divided by 4.715 times 10 to the 14. What you should get by the way is you should get 636 nanometers. So this is actually gonna be very close to the uh the wavelength of the source. So you can, so we can see here that even when objects have really, really insane velocities, like tremendous velocities, the wavelength actually doesn't change a whole lot. It really requires like significant fractions of the speed of light uh for stuff to start Doppler shifting in a way that's perceivable anyway. So that's it for this one. Let me know if you have any questions.
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