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 frequency that you observe. Well, it turns out that same Doppler effect also applies to electromagnetic waves, otherwise known as light. That's what we're gonna talk about in this video. Alright? 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, the source, and the observer. The classic example that we saw for sound waves was you, for example, running towards an ambulance siren, or maybe an ambulance siren moving towards you, or something like that. So basically, we had the velocity of the source, and we had the velocity of the listener. 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 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 \(1 \pm\) the relative velocity divided by \(c\). Alright?
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 \(\text{4 m/s}\), and the source is also moving towards you with \(\text{6 m/s}\), then that means that the relative speed, \(v_{\text{relative}}\), is actually the addition of those two velocities. It's \( \text{4 m/s} + \text{6 m/s} = \text{10 m/s} \). 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 if, for example, 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 going to 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 \( \text{630 nm} \). So, that's what theoretically should radiate. Right? So, basically, what this says here is that the wavelength of the source is equal to \( \text{630 nm} \). 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 \( \lambda \) that we actually observe. Alright?
Now you might notice that in this problem, we're using lambdas instead of frequencies, but remember, we can always change them by using the equation for the speed of light. So let's just go ahead and set up our equation here. Alright. So we have that \( f_{\text{observed}} \) is equal to \( f_{\text{source}} \times (1 \pm \frac{v_{\text{relative}}}{c}) \). Alright? By the way, we actually know what our \( v_{\text{relative}} \) is. The \( v_{\text{relative}} \)is just the speed that we're told that this star is receding from the Earth, \( 3 \times 10^{6} \text{ m/s} \). Alright. So its \( 3 \times 10^{6} \). Okay. So first, we have to figure out well, the \( f_{\text{observed}}\) is what we're looking for right here. So what about the \( f_{\text{source}} \)? That really just comes from the \( \lambda_{\text{source}} \). Remember, we can always just get to \( f \). So \( f_s \) is just going to be equal to \( \frac{c}{\lambda_s} \). So this is going to be \( \frac{3 \times 10^{8}}{ 630 \times 10^{-9}} \) and so that \( f_s \) is just going to be \( 4.76 \times 10^{14} \text{ Hz} \). So that's what you plug into this, over here.
So now we have that \( f_{\text{observed}} \) is going to equal \( 4.76 \times 10^{14} \). Now we have \( 1 \), and then this is going to be a plus or a minus sign. But what happens is, remember, this thing is receding, so that means that because this is receding over here, we're going to use a minus sign. Alright. So this is receding. We use a minus sign here, and this is going to be \( v_{\text{relative}} \). So this is just going to be \( \frac{3 \times 10^{6}}{ 3 \times 10^{8}} \). 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 \( 4.715 \times 10^{14} \) Hz. Now remember, this is just the frequency that you observe. Now all we have to do is we just have to change it back into \( \lambda_{\text{observed}} \) by using the same exact equation over here. So in other words, \( \lambda_{\text{observed}} \) is just going to be \( \frac{c}{f_{\text{observed}}} \), and this is just going to be \( \frac{3 \times 10^{8}}{4.715 \times 10^{14}} \). What you should get, by the way, is you should get \( 630 \text{ nm}\). So this is actually going to be very close to the wavelength of the source. So you can see here that even when objects have really, really insane velocities, like tremendous velocidads, the wavelength actually doesn't change a whole lot. It really requires significant fractions of the speed of light for stuff to start Doppler shifting in a way that's perceivable. Anyways, that's it for this one. Let me know if you have any questions.
