Question about a line from Rosenfeld's paper on optical rotation

Question

There are a lot of papers on optical rotation which cite Rosenfeld's (German) 1928 paper "Quantum mechanical theory of natural optical rotation..." [Quantenmechanische Theorie der naturlichen optischen Aktivitat von Flussigkeiten und Gasen, Zeitschrift für Physik, Volume 52, Issue 3-4, pp. 161-174] as background. Here is a paid link--

Apologies that I cannot link to the paper. I have a paper copy and can't find a paywall-free version.

The symbols below are: $\lambda$--wavelength; $c$ is the speed of light; $\nu$ is frequency; $n$ is the index of refraction; $\rho$ is a quantity that does not change between equations (42) and (43).

The last two equations in that paper (42), (43), are a conclusion about an angle of rotation $\theta:$

$$\Delta n = \frac {2\pi ~\rho}{\lambda~ n} \tag{42}$$

and

$$\theta = \frac {2\pi \nu~\cdot \Delta n}{c~\cdot ~ 2} = \frac{2 \pi^2\rho}{\lambda^2} \tag{43}$$

My question is pretty dumb. If we insert the expression for $\Delta n$ into the equality in (43) we get:

$$\frac{2\pi\nu}{c}\cdot \frac{2\pi \rho}{2\lambda~ n} = \frac{2\pi^2\rho}{\lambda^2}\to \frac{\nu}{c~n}=\frac{1}{\lambda}\to\lambda =\frac{cn}{\nu} $$

We know already--and Rosenfeld mentions this a few lines up and at eq. (21)--that

$$\lambda = \frac{c}{\nu~n}$$

which forces $n$ to be $1$. In general $n \neq 1$.

Can someone tell me what, if anything, has gone awry here?

Edit: After looking at several versions of Rosenfeld's equation I see $\theta$ is proportional to the trace of the rotation matrix...the constant of prop. can take various forms and is not the focus of this approach.

Answer 1

This answer was suggested by Ed V. It resolves the apparent problem nicely and readers with some fluency in the subject would probably not find the transition from $\lambda$ (medium) to $\lambda$ (vacuum) too troublesome.

We begin with $c = \nu \lambda_o$ in a vacuum and $\frac{c}{n} =\nu\lambda $ in a medium, in which $n$ is th refractive index.

In equation (42) of Rosenfeld's paper we have (using $\lambda n =\lambda_o$):

$$\Delta n = \frac{2\pi\rho}{\lambda n}= \frac{2\pi\rho}{\lambda_o}.\hspace 74 mm ~~~(42)$$

Then in equation (43) above we get $\lambda_o^2$ in the denominator:

$$\theta = \frac{2\pi\nu}{c}\cdot \frac{\Delta n}{2}=\frac{2\pi\nu}{c}\cdot\frac{\pi\rho}{\lambda_o}=\frac{2\pi^2\nu\rho}{c\lambda_o}=\frac{2\pi^2\rho}{\lambda_o^2}\hspace 30 mm (43) $$

and finally substitution of (42) into (43) gives

$$\theta = \frac{2\pi\nu}{c}\cdot \frac{2\pi\rho}{2\lambda n}= \frac{2\pi^2\rho}{\lambda_o\lambda n}=\frac{2\pi^2\rho}{\lambda_o^2}. $$