• anonymous
as the wavelength increases, the frequency
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • jamiebookeater
I got my questions answered at in under 10 minutes. Go to now for free help!
  • 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?
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.