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

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SUMMARY

The discussion focuses on the calculations related to a light wave with a frequency of 6 x 1014 Hz. The period is calculated using the formula T = 1/f, resulting in T = 1.67 x 10-15 seconds. The wavelength in a vacuum is determined using λ = c/f, yielding λ = 5 x 10-7 meters. When the light wave enters water, its velocity decreases to 0.75 times its velocity in a vacuum, affecting the wavelength while the frequency remains constant, resulting in a new wavelength of 3.75 x 10-7 meters.

PREREQUISITES
  • Understanding of wave properties: frequency, wavelength, and period
  • Familiarity with the speed of light in a vacuum (c = 3.0 x 108 m/s)
  • Knowledge of the relationship between velocity, frequency, and wavelength (c = λf)
  • Basic understanding of refraction and how velocity changes in different media
NEXT STEPS
  • Explore the concept of wave refraction and its effects on wavelength and velocity
  • Learn about the implications of the index of refraction on light waves in various media
  • Investigate the relationship between frequency and energy of light waves
  • Study advanced wave equations and their applications in optics
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Students studying physics, particularly those focusing on wave mechanics, optics, and anyone involved in understanding light wave behavior in different media.

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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?
 
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Yes, the frequency of the wave will remain constant.
 
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?
 
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.
 
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!
 
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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