Question

An electromagnetic wave travels from vacuum into a non-magnetic dielectric medium with permittivity $\epsilon = 4\epsilon_0$. If the ratio ($E_0 / B_0$) of the electric field amplitude $E_0$ to the magnetic field amplitude $B_0$ in vacuum is equal to the speed of light $c$, then the corresponding ratio in the given medium is

• A. $c / 2$
• B. $c$
• C. $c / \sqrt{2}$
• D. $c / 4$

Hint:

The refractive index of a non-magnetic medium is $n = \sqrt{\epsilon_r} = \sqrt{\frac{\epsilon}{\epsilon_0}}$.
The speed of light in the medium is $v = \frac{c}{n}$.
Since $\epsilon_r = 4$, we get $n = 2$, which immediately yields $v = \frac{c}{2}$.

Answer 1

The Correct Option is A

Step 1: Understand the ratio being asked.
For a light wave the electric and magnetic field amplitudes are tied together. Their ratio $E_0 / B_0$ always equals the speed at which that wave actually travels through the material.

Step 2: Recall the rule in vacuum.
In empty space the wave moves at $c$, so $E_0 / B_0 = c$, which matches what the question states.

Step 3: See what changes inside the medium.
Inside a material the wave slows down to a speed $v$, and the same relation holds, so now $E_0 / B_0 = v$. Our whole job is to find $v$.

Step 4: Use the speed formula for a medium.
The speed in a non-magnetic medium is \[ v = \frac{c}{\sqrt{\epsilon_r}}, \] where $\epsilon_r$ is the relative permittivity.

Step 5: Plug in the given permittivity.
Here $\epsilon = 4\epsilon_0$, so $\epsilon_r = 4$ and $\sqrt{\epsilon_r} = 2$.

Step 6: Get the final ratio.
Therefore $v = c / 2$, and the field ratio in the medium is the same value. \[ \boxed{\dfrac{E_0}{B_0} = \dfrac{c}{2}} \]