Really simple and quick question about optics

[loginorsinup]

Homework Statement

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?

Homework Equations

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. :(

 

[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..

 

[loginorsinup]

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?

 

[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 $\mu m$.

 

[loginorsinup]

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.

 

[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

 

[loginorsinup]

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