## From the nature of light unit.

Hi and thank you for viewing this post.
I'll begin with the question from the text book;

Q: An electromagnetic radiation has a frequency of 5.00 x 10^14 Hz.
a) Calculate its wavelength in a vacuum, in meters and nanometers.
b) Calculate its wavelength in water.
c) Is this radiation visible? If so, what is its colour?
d) What is the index of refraction of a medium in which the speed of this radiation
is 2.54 x 10^8 m/s?

and here are my attempts to the above questions;

A: a) given that the speed of light in "vacuum" is 3.00 x 10^8 m/s,
(not sure how to type-in the wavelength sign)
wavelength = 3.00 x 10^8 m/s / 5.00 x 10^14 Hz
Then it comes out to be, 6.00 x 10^-17 m which also equals to 600nm.
b) given that the speed of light in "water" is 2.2 x 10^8 m/s,
wavelength = 2.2 x 10^8 m/s / 5.00 x 10^14 Hz
which turns out to be 4.4 x 10^-7 m, and 440 nm.
c) I had a confusion because I wasn't really sure whether the question was asking
about the radiation in vacuum or in water (Sorry this is an independent study course
so there aren't any teachers whom I can contact). So, I gave the answer for both
situations. Orange(590-610nm) for a), and Violet(400-450nm) for b).
d) Again, confusion from the question c.. not really sure whether to use
orange(neon gas) or violet(mercury vapour) as the source. According to the textbook
the equation looks like this, n=c/v (n=absolute index of refraction, c=speed of light,
v=speed of light in substance) so am I looking to solve for the variable n?
e) Our ordinary main way to gather information about the world.

So it would be greatly appreciated if anyone could read this over
and correct me any wrongs. Any suggestions or advice are also appreciated,
Thanks again!
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 Quote by nblu A: a) given that the speed of light in "vacuum" is 3.00 x 10^8 m/s, (not sure how to type-in the wavelength sign) wavelength = 3.00 x 10^8 m/s / 5.00 x 10^14 Hz Then it comes out to be, 6.00 x 10^-17 m which also equals to 600nm. b) given that the speed of light in "water" is 2.2 x 10^8 m/s, wavelength = 2.2 x 10^8 m/s / 5.00 x 10^14 Hz which turns out to be 4.4 x 10^-7 m, and 440 nm.
Good. (Small typo in part a: -7, not -17.) You could also just have used the index of refraction for water.
 c) I had a confusion because I wasn't really sure whether the question was asking about the radiation in vacuum or in water (Sorry this is an independent study course so there aren't any teachers whom I can contact). So, I gave the answer for both situations. Orange(590-610nm) for a), and Violet(400-450nm) for b).
Color depends on frequency, not wavelength. (You don't see color until it hits your eye, so it doesn't matter what happened to it along the way as long as the frequency hasn't changed.)
 d) Again, confusion from the question c.. not really sure whether to use orange(neon gas) or violet(mercury vapour) as the source. According to the textbook the equation looks like this, n=c/v (n=absolute index of refraction, c=speed of light, v=speed of light in substance) so am I looking to solve for the variable n?
The source doesn't matter as you are given the speed in the medium.
 e) Our ordinary main way to gather information about the world.
Sure.

 Quote by Doc Al Good. (Small typo in part a: -7, not -17.) You could also just have used the index of refraction for water. Color depends on frequency, not wavelength. (You don't see color until it hits your eye, so it doesn't matter what happened to it along the way as long as the frequency hasn't changed.) The source doesn't matter as you are given the speed in the medium. Sure.
Thank you Doc Al,
I have read the textbook over again and I've realized that my prediction was incorrect.
I'm aware of the fact that the frequency range of visible light is 10^14 Hz to 10^15 and therefore, saying "Yes" to question c) is correct, however I'm not sure which colour I
should put down as my answer.
I have this chart called "Visible Spectrum" and it has all the colours listed with respect
to their "Range of wavelength in a vacuum".
I was thinking "Orange(590nm - 610nm)" is the only answer to write down because
the chart is discussing only about the wavelength in a vacuum.

Am I correct?

Thanks again Doc!

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## From the nature of light unit.

You are correct!

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