# Needing help again ~_~

1. May 31, 2005

### chibi_lenne

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.

2. May 31, 2005

### Chi Meson

You are correct. Remember, the wavelength of any frequency of light is always longest in a vacuum, shorter in a medium. Frequecy of the light stays constant.

3. May 31, 2005

### chibi_lenne

Yay, thanks so much that was just driving me nuts