Speed of EM wave in medium with μ_r=2 ε_r=4 — find refractive index

Question

Find the speed of an electromagnetic wave in a medium with relative permeability $\mu_r = 2$ and relative permittivity $\varepsilon_r = 4$ . Also find the refractive index of the medium.

Solution — Step by Step

In free space, EM waves travel at:

c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}

In a medium with permittivity $\varepsilon = \varepsilon_r \varepsilon_0$ and permeability $\mu = \mu_r \mu_0$ :

v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{1}{\sqrt{\mu_r \mu_0 \cdot \varepsilon_r \varepsilon_0}}

v = \frac{1}{\sqrt{\mu_r \varepsilon_r} \cdot \sqrt{\mu_0 \varepsilon_0}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}

This is the key relation. The speed in a medium is $c$ divided by $\sqrt{\mu_r \varepsilon_r}$ .

v = \frac{3 \times 10^8}{\sqrt{2 \times 4}} = \frac{3 \times 10^8}{\sqrt{8}} = \frac{3 \times 10^8}{2\sqrt{2}}

v = \frac{3 \times 10^8}{2 \times 1.414} = \frac{3 \times 10^8}{2.828}

\boxed{v \approx 1.06 \times 10^8 \text{ m/s}}

The refractive index $n$ is defined as:

n = \frac{c}{v} = \frac{c}{\dfrac{c}{\sqrt{\mu_r \varepsilon_r}}} = \sqrt{\mu_r \varepsilon_r}

n = \sqrt{2 \times 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83

\boxed{n = 2\sqrt{2}}

Why This Works

Maxwell’s equations predict that EM waves are self-sustaining oscillations of electric and magnetic fields. The wave equation gives speed $v = 1/\sqrt{\mu\varepsilon}$ . When we enter a denser medium (higher $\varepsilon_r$ or $\mu_r$ ), the “restoring force” effectively increases, slowing the wave down.

The refractive index is simply the ratio $c/v$ — it tells us how much slower light travels in the medium compared to vacuum. A higher $n$ means slower propagation.

For most optical materials, $\mu_r \approx 1$ (they are non-magnetic at optical frequencies), so $n \approx \sqrt{\varepsilon_r}$ . This is why the refractive index of glass (~1.5) is close to $\sqrt{\varepsilon_r}$ for glass (~2.25).

Alternative Method

If you directly remember that $n = \sqrt{\mu_r \varepsilon_r}$ , you can skip the intermediate steps and jump straight to:

n = \sqrt{2 \times 4} = \sqrt{8} = 2\sqrt{2}

Then use $v = c/n = 3 \times 10^8 / 2\sqrt{2}$ to get the speed. This is the faster route in an MCQ.

Common Mistake

A common slip is to write $n = \mu_r \varepsilon_r$ instead of $n = \sqrt{\mu_r \varepsilon_r}$ . Remember: it’s the square root because $v = c/\sqrt{\mu_r \varepsilon_r}$ and $n = c/v$ . If you forget the square root, you get $n = 8$ instead of $n = 2\sqrt{2}$ — a factor of 2 error in $n$ and a factor of 4 error in $v$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next