Finding Wavelength of Light in Water (n=1.33)

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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?
 

Thread 'Modeling a graph that shows age in relation to depth of an ice sample'

To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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