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

  • Thread starter Violagirl
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  • #1
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Homework Statement


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?


Homework Equations


c = λf

v = c/n

T = 1/f


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?
 

Answers and Replies

  • #2
798
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Yes, the frequency of the wave will remain constant.
 
  • #3
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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
798
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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
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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
798
1
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.
 

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