Quantcast

Index of refraction

Get Shadowing and a Virtual Clinical Education
This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

swimmingchemist

New Member
Joined
Dec 6, 2009
Messages
6
Reaction score
0
Hey all- just trying to understand a concept that seems to condradict itself. If anybody has any clear explanations for this I would hugely appreciate it.

Ok, so wavelength and frequency are inversely proportional to each other, so that the speed of light is constant:

c=f*l

We also know that the index of refraction "n" compares the speed "c" of light in a vacuum to the speed "v" of light in a particular medium:

n=c/v

When light crosses into a new medium, the frequency remains the same right? and only the wavelength changes. If the medium's index of refraction is higher, the wavelengths become shorter; if the index is lower, then the wavelengths become longer.

It's this last part that's bugging me- if the wavelength decreases, but frequency stays constant, then based on c=f/l, the speed of light increases above 3E8 m/s. Does this sound crazy to you too? I did not think light could actually exceed the speed of...light.

Any thoughts?
 
Last edited:

dxr

Full Member
7+ Year Member
Joined
Mar 5, 2010
Messages
17
Reaction score
0
i'm not a physicist by any means, but I think the simple explanation is that you simply can't have an n<1. Doesn't exist because a vacuum is 1 (and air is approximated to be effectively 1). So while you can have light going from a substance with higher n to lower n, that "higher n" substance has to have an n>1, because light can never go from vacuum to a substance of lower index of refraction.
 

joshto

Nervous&Neurotic
10+ Year Member
Joined
Jan 30, 2006
Messages
267
Reaction score
0
[/CENTER]

When light crosses into a new medium, the frequency remains the same right? and only the wavelength changes. If the medium's index of refraction is higher, the wavelengths become shorter; if the index is lower, then the wavelengths become longer.


Any thoughts?

the velocity of a wave, when entering a new medium, changes
 

swimmingchemist

New Member
Joined
Dec 6, 2009
Messages
6
Reaction score
0
Ok, yes that makes sense- I think I was mistakenly reading into it to mean that when light passes from a vacuum into a medium with n<1- Thank you dxr for reminding me that n can't be below 1. I guess these rules are intended to be used when looking at a system when light goes from a medium with n>1 (water, glass etc.) to a medium with a lower n (vacuum).

Thanks for giving me a better sense of perspective- it was driving me crazy!
 

puravida85

Full Member
10+ Year Member
7+ Year Member
Joined
Dec 15, 2007
Messages
121
Reaction score
0
I think you started it all wrong by saying that:

c = f / l. Mathematically you are saying these components (frequency and wavelength) are directly proportional.

It should be c(velocity) = f(frequencu)* l(wavelength). <-- an equation you should know over and over because frequency and wavelength are inversely proportional.

Therefore when light hits a higher index of refraction, we know if frequency has to stay the same, wavelength and speed of light has to decrease together. A side note, sound increases speed in the case of a longitudinal. wave traveling from air to say liquid or solid.

Also if we know n = c / v. Well, nothing can go faster than the speed of light that is why n is always greater than one... v can only be a fraction of c. GL with your studies.:thumbup:
 

swimmingchemist

New Member
Joined
Dec 6, 2009
Messages
6
Reaction score
0
Wow- Thank you PuraVida85- Not sure how I mistyped that equation (c=f*l). This was the way I had it organized in my head while I was struggling with this problem, infact it's what had me confused originally, because when an index of refraction decreases, the wavelengths also increase.

What I got wrong originally was that n can not be smaller than 1, because this takes place in a vacuum. Thank you for pointing out my error, which I easily overlooked- I think the important part is that I understand the concepts much better now. Taking the test tomorrow!
 
Top