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COULD SOMEONE PLEASE HELP A DISABLED WAR VET WITH THE FOLLOWING I NEED HELP QUICKLY?

what is the frequency "in Mhz" of a beam of blue light whos wave length is 4000 angstroms

I know there is a simple formula for this question please help a veteran home back from Iraq after a 27 month tour get this figured out I need this help in getting into a vocationable rehap program and this question will be on a test I have to take please show me the formula of how you figured it and the answeer and it's up to me to remember it for monday at 0400 hrs when I take the test. I appreciate any assistance someone can give me so that I can see how it was calculated and what the final answer was and i'll drill it into my brain but I need the listed above information. Thank you in advance from a soldier who really appreciates the assistance of whomever took the time to help me help myself. It has been a hard transition and at this point I have no other source but to hope someone has the SOLUTION

2007-02-17 14:12:38 · 5 answers · asked by Wayne D 1 in Science & Mathematics Physics

5 answers

the wavelength of blue light is 4 X 10^-7 m
the speed of light is 3 X 10^8 m / s

3 X 10^8 m / s divided by 4 X 10^-7 m / s is

0.75 X 10^15 c/s or 7.5 X 10^14 c/s or Hz

or 7.5 X 10^8 MHz

2007-02-17 14:28:05 · answer #1 · answered by ivorytowerboy 5 · 0 0

Well if you have the wavelength you can use this equation to convert back and forth between the two.

F=c/L where c is the speed of light and L is the wavelength in meters. With would be 4000*10^-10.

Edit: By the way, the answer for this would be in hertz which you can then convert to MHz by dividing by 1X10^6. Good luck on your test!

2007-02-17 14:20:03 · answer #2 · answered by Anonymous · 0 0

f * lambda = c

f = frequency in hertz
lambda = wavelength in meters
c = speed of light = 3 10^8 m/s

4000 angstroms = 4 10^-7 meters

f = 3 10^8 / 4 10^-10
f= 7.5 10^14 Hertz = 7.5 10^8Mhz

2007-02-17 14:35:35 · answer #3 · answered by ZincFinger 1 · 0 0

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2016-12-17 12:39:34 · answer #4 · answered by Anonymous · 0 0

If you take a look at

http://en.wikipedia.org/wiki/Angstrom

and

http://en.wikipedia.org/wiki/Wavelength

I think you will find everything you need

2007-02-17 14:26:53 · answer #5 · answered by oldhippypaul 6 · 0 0

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