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

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