Period, frequency, wavelength, and velocity of a light wave

1. Dec 1, 2013

Violagirl

1. The problem statement, all variables and given/known data
A light wave has a frequency of 6 x 1014 Hz. A) What is its period? B) What is its wavelength in a vacuum? C) When the light wave enters water, its velocity decreases to 0.75 times its velocity in vacuum. What happens to the frequency and wavelength?

2. Relevant equations
c = λf

v = c/n

T = 1/f

3. The attempt at a solution

I understood how to do parts A and B.

For A, since we know the frequency, we can take the equation for period, T to find the answer:

T = 1/f = 1/(6 x 1014 Hz = 1.67 x 10-15 sec.

For B, we know that c = 3.0 x 108 m/s in a vacuum and we're given the frequency so wavelength is found by taking the equation:

λ = c/f = (3.0 x 108 m/s)/(6 x 1014 Hz) = 5 x 10-7 m.

For part C, however, I got confused.

I believe we can use the equation:

v = c/n

We're told that velocity decreases by 0.75 times in a vacuum so I think that v then would be:

v = 0.75c

So from here, we have to relate it to the equation c = λf.

So would we use the equation:

0.75c = λf to find wavelength since we already what f is? Can we assume that f does not change but that wavelength would?

To get wavelength then, we'd take:

0.75c/f = λ?

2. Dec 1, 2013

sandy.bridge

Yes, the frequency of the wave will remain constant.

3. Dec 1, 2013

Violagirl

So for finding the wavelength then, how exactly do you use the information of v being reduced to 0.75 times its original velocity to find the new wavelength in water then?

4. Dec 1, 2013

sandy.bridge

The new velocity will will be 3/4 the original velocity c. You know what c is, so compute the new velocity. Furthermore, you know what the frequency is. That's two out of three variables. You can solve for lambda.

5. Dec 1, 2013

Violagirl

Oooh! Ok got it. I wanted to verify my understanding when we were told that the velocity was 0.75 times its original velocity in water. Got an answer of 3.75 x 10^-7 m, which makes sense as an answer. Thanks!

6. Dec 1, 2013

sandy.bridge

Perfect. Since the velocity decreased by a factor of 3/4, the wavelength does also. This can be verified by multiplying your wavelength result you attained previously.