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

Click For Summary

Homework Help Overview

The discussion revolves around the properties of a light wave, specifically its period, frequency, wavelength, and velocity as it transitions from a vacuum to water. The original poster presents a problem involving calculations related to these properties and seeks clarification on the effects of changing velocity on wavelength.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the period and wavelength of a light wave using given frequency and speed in a vacuum. They express confusion regarding the implications of reduced velocity in water on frequency and wavelength. Participants discuss the constancy of frequency and how to relate the new velocity to wavelength.

Discussion Status

Some participants confirm that the frequency remains constant when the light wave enters water, while others provide insights on how to compute the new wavelength using the reduced velocity. There appears to be a productive exchange of ideas, with participants verifying their understanding and calculations.

Contextual Notes

The original poster is working within the constraints of a homework assignment, which may limit the information available for solving the problem. The discussion includes assumptions about the behavior of light as it transitions between mediums.

Violagirl
Messages
112
Reaction score
0

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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
Replies
8
Views
2K
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
941
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
Replies
4
Views
2K