## rachelris1 one year ago as the wavelength increases, the frequency

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1. 4n1m0s1ty

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

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.