# Really simple and quick question about optics

1. Oct 19, 2014

1. The problem statement, all variables and given/known data
ZnSe is a II-VI semiconductor and a very useful optical material used in various applications such as optical windows (especially high power laser windows), lenses, and prisms. It transmits over 0.50-19 μm. Refractive index, n in the 1-11 μm range is described by a Cauchy expression of the form
in which λ is in μm. What are the n-2, n0, n2, and n4 coefficients?

2. Relevant equations

3. The attempt at a solution
Given the index of refraction by the following Cauchy equation relationship
with λ given in micrometers, I am to find the Cauchy coefficients
n-2, n0, n2, and n4 such that
Using
I managed to get the first expression to look like
Since n is dimensionless λ can't have a dimension either... but the fact that it is given in micrometers is important, clearly, because a 1 meter value λ would require me to plug in
λ = 1000000​
into the very first expression for n.

Typically hυ is given in eV (electron volts). For n to be dimensionless, wouldn't hc have to be equal to eV (a unit of energy) as well? But obviously energy has units of hc/λ. So something is wrong. How do I take into account the fact that λ is given in micrometers in finding the coefficients?

I know I am close, but I can't finish it up. :(

2. Oct 20, 2014

### BvU

Hello log, and welcome to PF :)

Don't know what I'm talking about, but as a physicist I expect your Cauchy coefficients to have dimensions: 0.0485 $\mu m^2$, etc..

Last edited: Oct 20, 2014
3. Oct 20, 2014

Why would you expect them to have dimensions of length squared?

I'm confused what to do with, for example, what to do with (hc)^2?

Which units should I use? How do I make this equal to eV^2 as the Cauchy coefficient n-2 is normally quoted as?

4. Oct 20, 2014

### BvU

As I said, don't know what I'm talking about. But your coefficients all have different dimensions. The correct expression for n is$$n= 2.4365 + {0.0485\over (\lambda/1\mu m)^2 }+ {0.0061\over (\lambda/1\mu m)^4 } - 0.0003 (\lambda/1\mu m)^2$$

My interpretation of the problem is that the coefficients have to be found for $h\nu$ expressed in eV. You kind of indicate that clarification of the OP in your post # 3.

So start with: how many eV corresponds to 1 $\mu m$.

5. Oct 20, 2014

Thanks for the response, BvU.

I see what you mean about the differing dimensions. So, 1 eV corresponds to 1.23984193 electron volts.

How did I do this? hc/λ for λ = 1 μm.

6. Oct 21, 2014

### BvU

You correctly pulled in e too, but in "1 eV corresponds to 1.23984193 electron volts." the 1 eV is probaby 1 $\mu m$ :) .

So now think about what happens to the various coefficients. You can easily check by filling in e.g. $\lambda = 5\; \mu m$ in the one and $h\nu=6.2$ eV in the other

7. Oct 23, 2014

Thanks. I got how it works because of you! :)