What is the relationship between critical angle and negative refractive index?

[Tanishq Nandan]

Homework Statement

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Homework Equations

Snell's Law:
n1sin(i)=n2sin(r)
Critical angle=sininverse(n1/n2) [**only works when n2>n1]
If a light beam is incident at an interface of 2 mediums(with r.i n1 and n2) at an angle greater than the critical angle,the interface acts as a mirror,reflecting the entire beam.

The Attempt at a Solution

I thought of it in terms of critical angle,and saw that according to the formula,critical angle's coming out to be some negative value.Doesn't this mean that any angle of incidence greater than that(which is practically every possible light ray) will have i>critical angle and undergo reflection??
So,I thought option C to be correct.
However,answer's A.
Help appreciated.

 

[berkeman]

I haven't heard of this before, but it would just seem to be a result of substituting -n for n in Snell's Law...

https://skullsinthestars.files.wordpress.com/2008/08/negativerefraction.jpg

 

[TSny]

I don't know much about this either. I found this to helpful
http://faculty.washington.edu/goussiou/486_W15/Harvey_NIM.pdf

 

[Tanishq Nandan]

I haven't heard of this before, but it would just seem to be a result of substituting -n for n in Snell's Law...

https://skullsinthestars.files.wordpress.com/2008/08/negativerefraction.jpg
View attachment 211601

I know,I accounted for the given answer in the same way,but only after I had consulted the answer key.Before that,I was thinking about critical angles..and I still can't find the wrong in that.The only criterion for the formula,as I stated earlier,is n1>n2,which is obviously true since n2 is negative

 

[TSny]

I know,I accounted for the given answer in the same way,but only after I had consulted the answer key.Before that,I was thinking about critical angles..and I still can't find the wrong in that.The only criterion for the formula,as I stated earlier,is n1>n2,which is obviously true since n2 is negative

OK, good point! But the condition for the existence of a critical angle would not be n₁ > n₂. I think it would be n₁ > |n₂|.

In this question, medium 1 is air. So n₁ ≈ 1. Then, in order to have total reflection, you need |n₂| < 1.

Some of the articles I found on the net mention materials with n₂ = -0.3. So, this does satisfy |n₂| < 1.

What would be the value of the critical angle in this case where n₁ = 1.0 and n₂ = -0.3?

So, you can see that for this case, answer C looks correct.

For n₂ < -0.1, |θ₂| is always less than θ₁, and answer A would be correct.

 

[Tanishq Nandan]

n1 > |n2|.

Ok...
But the question just says negative..and nothing about the value.I found your link interesting. .but a few(a lot,actually) of things looked kinda out of my league

 

[TSny]

Ok...
But the question just says negative..and nothing about the value.

Yes. So, it seems A or C could be correct. The only thing that I see that might dismiss C is that the angle of reflection does not look quite equal to the angle of incidence.

I found your link interesting. .but a few(a lot,actually) of things looked kinda out of my league

Same here. Some of the slides do use advanced physics concepts. But overall I was able to get a better understanding of the topic.