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rachelris1

  • one year ago

as the wavelength increases, the frequency

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  1. 4n1m0s1ty
    • one year ago
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    Here's the relationship: \[c = f*\lambda\] if c is the speed of light (which is constant in all frames of reference) and you increase \[\lambda\] which is the wavelength, what must happen to f?

  2. theEric
    • one year ago
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    I agree completely with @4n1m0s1ty , noting that \(c\) is appropriate only in a vacuum. Otherwise, you generally use "\(v\)" for velocity. In case you want to take 4n1m0s1ty 's approach another way, you use \(c=f\ \lambda\), rearrange it to be \(f=\Large\frac{c}{\lambda}\), and take out all variables that don't depend on \(f\) or \(\lambda\), but using the \(\alpha\) symbol to show proportionality rather than equality.\[f\ \ \ \alpha\ \ \ \frac{1}{\lambda}\]...It looks better on paper, but people often use that to discuss general proportionality. So, increase \(\lambda\), and see what happens to the other side. This is pretty much the same approach as 4n1m0s1ty, but more formal. That is, it's unnecessarily complex :P But it's also common. If you want to say "frequency is inversely proportional to wavelength," then you can just write \(f\ \ \ \alpha\ \ \ \Large\frac{1}{\lambda}\). Solving for \(f\) might make it easier, no matter how you do it. It just depends on how your mind works.

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