What is the wavelength of light in water when it enters a swimming pool?

AI Thread Summary
The discussion revolves around calculating the wavelength of light in water when it enters a swimming pool, starting with a wavelength of 0.527 µm in air. The speed of light in water is 0.700 times that in air, and the index of refraction for water is given as 1.33. Participants highlight that the frequency of light remains unchanged when transitioning between mediums, allowing for the calculation of the new wavelength in water. The relationship between speed, wavelength, and frequency is emphasized, noting that as the speed decreases, the wavelength also decreases proportionally. The conversation concludes with the suggestion that understanding these relationships simplifies the calculation process.
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Homework Statement



"Light of wavelength 0.527 µm (in air) enters the water in a swimming pool. The speed of light in water is 0.700 times the speed in air. What is the wavelength of the light in water?"


Homework Equations



v = c /n, where n = 1.33 ,the index of refraction

The Attempt at a Solution



The wavelength of light in air is given. \lambda=0.527
I looked up the speed of light in air -- 299704644.54 m/s
I also know that the speed of light in water is 0.700(299704644.54)

I am not sure how to relate these numbers to v = c/n though, or the equation to use to find the \lambda of light in water. Can anyone offer any insight? Thanks!
 
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What's the relationship between speed, wavelength, and frequency for any wave? Which of those parameters remains unchanged when light travels from one medium to another?
 
Doc Al said:
What's the relationship between speed, wavelength, and frequency for any wave? Which of those parameters remains unchanged when light travels from one medium to another?

frequency = v / \lambda, so I would think the frequency would be left unchanged.

So I'm thinking I could do:

f = (speed of light in air)/0.527

to solve for the frequency, then use that value to solve for the wavelength of light in water?
 
Soojin said:
frequency = v / \lambda, so I would think the frequency would be left unchanged.
Good.

So I'm thinking I could do:

f = (speed of light in air)/0.527

to solve for the frequency, then use that value to solve for the wavelength of light in water?
OK, but even easier is to realize that the ratio v/λ must remain fixed, so λ decreases by the same factor that the speed decreases.
 
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