Answer to Question #102518 in Optics for VV

Question

Light waves are propagating in vacuum. Derive the wave equation for the associated magnetic field vector. On the basis of this equation, calculate the speed of light.

Answer 1

$\overrightarrow{E}(x,t)=E(x,t)\overrightarrow{j}$

$\overrightarrow{B}(x,t)=B(x,t)\overrightarrow{k}$

$\nabla \cdot \overrightarrow{E}=0$

$\nabla \cdot \overrightarrow{B}=0$

$\nabla \times \overrightarrow{E}=-\frac{d\overrightarrow{B}}{dt}$

$\nabla \times \overrightarrow{B}=\mu_0 \epsilon_0\frac{d\overrightarrow{E}}{dt}$

$\nabla \times E(x,t)\overrightarrow{j}=\frac{dE}{dx}\overrightarrow{k}$

$\frac{dE}{dx}=-\frac{dB}{dt}$

$\frac{d^2B}{dx^2}=-\frac{d}{dx}(\mu_0\epsilon_0\frac{dE}{dt})=-\mu_0\epsilon_0\frac{d}{dt}\frac{dE}{dx}=\mu_0\epsilon_0\frac{d}{dt}\frac{dB}{dt}=\mu_0\epsilon_0\frac{d^2B}{dt^2}$

So, we have

$\frac{d^2B}{dx^2}=\mu_0\epsilon_0\frac{d^2B}{dt^2}$

The general wave equation for a wave $\psi(x,t)$ traveling in the $x$ direction with speed $v$

$\frac{d^2\psi}{dx^2}=\frac{1}{v^2}\frac{d^2\psi}{dt^2}$

So,

$\mu_0\epsilon_0=\frac{1}{v^2}\to c=v=\frac{1}{\sqrt{\mu_0\epsilon_0}}=2.997\cdot10^8 m/s$