Refractive index and critical angle

[Cheman]

The equation relating these is :

sine(critical angle) = refractive index between 2 meterials.

Could someone please supply a proof for this?


Also, as a side question , is there a proof for the sin i / sin r = v1/ v2 , not including the one in which you draw triangles and substitute d1 for v1t, etc?

Thanks.

 

[ZapperZ]

The equation relating these is :

sine(critical angle) = refractive index between 2 meterials.

Could someone please supply a proof for this?

It's hard to know why, for example, you can't derive this yourself. May I suggest that you first find the definition of what is meant by the "critical angle"? This is because, once you understand what it means, then you'll realize that all that is involved in finding the "proof" of this is simply setting one of the appropriate angle in Snell's Law to be 90 degrees (upon which, the sine of that angle "magically" becomes 1). This doesn't involve a rigorous proof. It does, however, involve the understanding of the meaning of the variables that are being used.

Zz.

 

[Cheman]

I am not stupid so please try to avoid patronising me. I do know what all of these terms mean, but I believe that if you actually carry out the proof you are saying it will not work out.

Let me show you:

sin i / sin r = mu ( refractive index)

sin i = sin c
sin r = sin 90 = 1

Therefore, sin c = mu; which cannot be true - most refractive indices are above one and you cannot inverse sine something larger than one - there is no solution.

Now I have checked up what the resulting formula should be and apparently it should be sin c = 1/ mu. Now do you understand why I am asking the question - I can't get the proof to work.
I would really appreciate if someone could please give me a hand - cause I'm stumped.

Thanks.

 

[ZapperZ]

I am not stupid so please try to avoid patronising me. I do know what all of these terms mean, but I believe that if you actually carry out the proof you are saying it will not work out.

Let me show you:

sin i / sin r = mu ( refractive index)

sin i = sin c
sin r = sin 90 = 1

Therefore, sin c = mu; which cannot be true - most refractive indices are above one and you cannot inverse sine something larger than one - there is no solution.

Now I have checked up what the resulting formula should be and apparently it should be sin c = 1/ mu. Now do you understand why I am asking the question - I can't get the proof to work.
I would really appreciate if someone could please give me a hand - cause I'm stumped.

Thanks.

This is why I said it is HARD to understand why you are having a problem doing this, because you NEGLECTED to describe in detail WHERE you are stuck! I'm sorry, but contrary to popular belief, I am not a mind-reader, so I have no idea WHERE exactly you are stuck and what you actually know! (Oy vey!)

Secondly, "mu" is the index of refraction? Are you sure? Or is "mu" simply the RATIO of the index of refractions of the two medium, and that it is in particular a ratio when compared to vacuum where n=1? Start with the Snell's Law. There are TWO different index of refractions involved! It has to, the light is going from one medium to another, so it makes no sense to only have it as a function of one index. However, the RATIO of the two indexes would be relevant. ONLY when one of the medium is a vacuum (or even air as the first approximation) would this ratio would be simply the index of refraction of one of the medium.

I assume that I do not need to work out the detail of this since I hate to be accused of "patronizing" you with baby algebra that you already know.

Zz.

 

[Doc Al]

Let me show you:

sin i / sin r = mu ( refractive index)

Start with Snell's law: $n_1 sin\theta_1 = n_2sin\theta_2$
If the second medium is air ($n_2 = 1$) and the angle of refraction is 90 degrees ($sin\theta_2 = 1$), then you can solve for the critical angle.

This is what ZapperZ was saying. But you've got to start with the correct Snell's law.

 

[Cheman]

But 1mu2 is defined as the refractive index. ie - it equals v1/ v2. Therefore, can we not start by using this?

 

[Doc Al]

But 1mu2 is defined as the refractive index. ie - it equals v1/ v2. Therefore, can we not start by using this?

The definition of index of refraction is $n_1 = c/v_1$, where c is the speed of light in vacuum and v is the speed of light in the medium of interest. Thus, $v_1/v_2 = n_2/n_1$.

 

[Cheman]

Could someone please prove the equation using sin i / sin r = v1 / v2 = 1mu2? After all, if the other equation is true, and this one is then we should be able to derive one from the other.

Thanks.

 

[Doc Al]

Could someone please prove the equation using sin i / sin r = v1 / v2 = 1mu2? After all, if the other equation is true, and this one is then we should be able to derive one from the other.

What equation are you trying to prove? What is "1mu2"?

$sin\theta_i/sin\theta_r = v_1/v_2$ is equivalent to Snell's law given the definition of index of refraction ($n_1 = c/v_1$).

 

[Cheman]

Trying to proove sin c = 1mu2, from sin i / sin r = v1 / v2 = 1mu2. 1mu2 is the refractive index between the two materials.

Thanks.

 

[Doc Al]

Trying to proove sin c = 1mu2, from sin i / sin r = v1 / v2 = 1mu2. 1mu2 is the refractive index between the two materials.

I don't know what refractive index "between the two materials" means; v1/v2 equals the inverse ratio of the indices of refraction. (Of course, if the second material is vacuum or air, then $n_2 = 1$.)

Combine: $sin\theta_i/sin\theta_r = v_1/v_2$
With: $n_1 = c/v_1$
Gives: $sin\theta_i/sin\theta_r = n_2/n_1$

For the critical angle $sin\theta_r = 1$ and let $n_2 = 1$,
then: $sin\theta_c = 1/n_1$

 

[tumor]

When diamond is immersed in the water ray of light is not reflected back but passes thru it, how come?