Derivation of wavelength in a waveguide

Question

I'm trying to derive this expression for the wave length in a wave guide. I'm following this derivation from the Feynman lectures https://www.feynmanlectures.caltech.edu/II_24.html#Ch24-F4. I do not understand how he went from eqn 24.17 to eqn 24.19. Where did $\lambda_{g}$ and $\lambda_{0}$ come from?

Answer 1

As stated in the text $$k_z=\frac{2\pi }{\lambda_g} \ \ \text{and} \ \ \ \lambda_0=\frac{2\pi c}{\omega}$$ Further $$k_z=\sqrt{\left(\frac{\omega}{c}\right)^2-\left(\frac{\pi }{a}\right)^2}$$ From above (first line) $$\frac{2\pi }{\lambda_g}=\sqrt{\left(\frac{2\pi}{\lambda_0}\right)^2-\left(\frac{\pi }{a}\right)^2}=\frac{2\pi}{\lambda_0}\sqrt{1-\left(\frac{\lambda_0}{2a}\right)^2}$$ Invert both side $$\lambda_g=\frac{\lambda_0}{\sqrt{1-\left(\frac{\lambda_0}{2a}\right)^2}}$$ As required!