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

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