Finding Wavelength of Light in Water (n=1.33)

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SUMMARY

The wavelength of light in water, where the refractive index (n) is 1.33, can be calculated using the formula λ(water) = λ(vacuum) / n. Given the wavelength in a vacuum as 540 nm, the wavelength in water is determined to be approximately 4.06 x 10-7 m. The discussion highlights the importance of ensuring the correct units are used, as the initial value of 540 mm was incorrect for visible light, which should be in nanometers (nm).

PREREQUISITES
  • Understanding of the refractive index and its implications on light propagation.
  • Familiarity with the wave equation λ = c / ν.
  • Knowledge of unit conversions, particularly between millimeters and nanometers.
  • Basic principles of optics and light behavior in different media.
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  • Research the properties of light in various media, focusing on refractive indices.
  • Learn about the implications of wavelength changes in optical systems.
  • Explore the relationship between frequency, wavelength, and speed of light in different materials.
  • Study advanced optics concepts such as Snell's Law and total internal reflection.
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Given the wavelength of lightwave in a vacuum is 540mm, what will it be in water, where n = 1.33?

To solve this problem, it seems like I would need to find a relationship of the frequency of light to the wavelength and that is easily obtained from the equation lambda = c/nu and the frequency can be solved.

Since n=speed of light in vacuum/speed of light in material.

We can find the speed of light in the material (water) and thus the wavelength.

But it seems like there is a problem somewhere as I cannot seem to obtain a reasonable answer.
 
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You quoted the wavelength as 540mm, are you sure it's not nm? That would make sense if it's visible light. Then I'm pretty sure lambda(water)=lambda(vac)/n is right.
 
You know what. It is nm. I got an answer where the wavelength is 4.06*10^-7m. Anyone wants to double-check my answer?
 

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