Calculating Wavelength in Water for Electromagnetic Radiation

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SUMMARY

The discussion focuses on calculating the wavelength of electromagnetic radiation in water, specifically for a frequency of 5.00 x 1014 Hz. The correct formula used is wavelength(med) = wavelength(vac)/n, where n is the index of refraction for water (1.33). The calculated wavelength in water is 4.51 x 10-7 m or 451 nm, confirming that wavelengths are longer in a vacuum and shorter in a medium. The user successfully applies the principles of wave mechanics to derive the solution.

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chibi_lenne
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Seriously, if I didn't need physics I probably wouldn't take it.I've don the work for this one problem, but then I'm not sure if it's the right formula for what I need. Here's what I have so far(the parts in bold are the work I've done)

10. An electromagnetic radiation has a frequency of 5.00 x 10^14Hz.
b) Calculate its wavelength in water.

Given: wavelength = 0.6 x 10^-6m or 600nm
n = 1.33 for water​
Required: wavelength(med)
Analysis: c=(f)(wavelength{vac}) and c=(n)(wavelength{med})

therefore wavelength(vac) = (n)(wavelength{med}) and​
wavelength(med) = wavelength(vac)/n​
Solution: wavelength(med) = (0.6 x 10^-6m)/1.33
= 4.51 x 10^-7 or 451 nm​
Now I get looking at this and I'm not sure if that's exactly how I go about doing this or if this is what I'm supposed to do for finding the index of refraction (which comes later) Please help, my brain is fried.
 
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You are correct. Remember, the wavelength of any frequency of light is always longest in a vacuum, shorter in a medium. frequency of the light stays constant.
 
Yay, thanks so much that was just driving me nuts
 

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