An EM wave from air enters a medium. The electric fields are E1 = E01 x^ cos[2πν(z/c - t)] in air an

Q: Why did we expand the terms inside the cosine?

Expanding the terms allows us to directly compare the equation with the standard form $E = E_0 \cos(kz - \omega t)$, making it easy to identify the wave number $k$ (coefficient of $z$) and angular frequency $\omega$ (coefficient of $t$).

Q: What does it mean for a medium to be non-magnetic?

A non-magnetic medium does not affect magnetic fields significantly differently than a vacuum does. Mathematically, this means its relative permeability $\mu_r$ is approximately $1$, so its absolute permeability $\mu$ is just $\mu_0$.

Q: Why is the speed of light $c$ in the first equation but $c/2$ in the second?

The first equation describes the wave in air (which is very close to a vacuum), where it travels at speed $c$. The second equation describes the wave in a denser medium, where its speed is reduced to $c/2$ due to the medium's higher permittivity.

Q: Can the frequency of a wave change when it enters a new medium?

No, the frequency $\nu$ (and thus angular frequency $\omega$) of a wave is determined by its source and remains constant when crossing boundaries. However, in this specific problem, the equations are given with different apparent $\omega$ values because the variable $k$ in the second equation is just a constant parameter, not the actual wave number of the second medium.

Q: Why do we square the ratio at the end?

The ratio of the speeds gives us the square root of the inverse ratio of the permittivities. To find the actual ratio of the permittivities, we must square both sides of the equation to eliminate the square root.

Question

An EM wave from air enters a medium. The electric fields are $\mathbf{E}_1 = E_{01} \hat{\mathbf{x}} \cos\left[2\pi
u\left(\frac{z}{c} - tight)ight]$ in air and $\mathbf{E}_2 = E_{02} \hat{\mathbf{x}} \cos[k(2z - ct)]$ in medium, where the wave number $k$ and frequency $
u$ refer to their values in air. The medium is non-magnetic. If $\varepsilon_{r_1}$ and $\varepsilon_{r_2}$ refer to relative permittivities of air and medium respectively, which of the following options is correct?

Options:

(A) $\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = 4$

(B) $\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = 2$

(C) $\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \frac{1}{4}$

(D) $\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \frac{1}{2}$

Sigma Percentile · Accepted Answer

Have you ever wondered what happens to light when it plunges from the thin air into a dense medium like glass or water? It slows down! But how do we describe this mathematically? In this problem, we are going to decode the secret language of electromagnetic waves and uncover the relationship between a wave's speed and the electrical properties of the medium it travels through.

### Analyzing the Wave in Air

We are given the electric field of an electromagnetic wave traveling in air:

$\mathbf{E}_1 = E_{01} \hat{\mathbf{x}} \cos\left[2\pi
u\left(\frac{z}{c} - tight)ight]$

To make sense of this, we need to compare it to the standard wave equation, which looks like $E = E_0 \cos(kz - \omega t)$. Let's expand the terms inside our cosine function:

$\mathbf{E}_1 = E_{01} \hat{\mathbf{x}} \cos\left(\frac{2\pi
u}{c}z - 2\pi
u tight)$

By comparing the coefficients, we can immediately spot the wave number $k_1$ and the angular frequency $\omega_1$. The coefficient of $t$ gives us $\omega_1 = 2\pi
u$, and the coefficient of $z$ gives us $k_1 = \frac{2\pi
u}{c}$.

Why do we care about these? Because the speed of any wave is simply the ratio of its angular frequency to its wave number!

$v_1 = \frac{\omega_1}{k_1} = \frac{2\pi
u}{\frac{2\pi
u}{c}} = c$

As expected, the wave travels at the speed of light, $c$, in air.

### The Master Equation of Electromagnetism

Now, let's connect this kinematic speed to the fundamental properties of electromagnetism. James Clerk Maxwell taught us that the speed of an electromagnetic wave is governed by the permittivity ($\varepsilon$) and permeability ($\mu$) of the medium:

$v = \frac{1}{\sqrt{\mu \varepsilon}}$

For air, we can write this as:

$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0 \varepsilon_{r_1}}}$

Here, $\mu_0$ and $\varepsilon_0$ are the constants for a vacuum, and $\varepsilon_{r_1}$ is the relative permittivity of air.

### Entering the Denser Medium

Next, the wave crashes into a new, non-magnetic medium. The electric field transforms into:

$\mathbf{E}_2 = E_{02} \hat{\mathbf{x}} \cos[k(2z - ct)]$

Let's expand this just like we did before:

$\mathbf{E}_2 = E_{02} \hat{\mathbf{x}} \cos(2kz - kct)$

Comparing this to our standard form, the new angular frequency is $\omega_2 = kc$, and the new wave number is $k_2 = 2k$. Let's find the new speed!

$v_2 = \frac{\omega_2}{k_2} = \frac{kc}{2k} = \frac{c}{2}$

The wave has hit the brakes! It is now traveling at half the speed of light.

### The Final Calculation

Since the medium is non-magnetic, its permeability remains $\mu_0$. We can write the speed equation for this new medium as:

$\frac{c}{2} = \frac{1}{\sqrt{\mu_0 \varepsilon_0 \varepsilon_{r_2}}}$

We now have two beautiful equations for the speeds in both media. To find the relationship between the relative permittivities, we simply divide the first equation by the second:

$\frac{c}{c/2} = \frac{\frac{1}{\sqrt{\mu_0 \varepsilon_0 \varepsilon_{r_1}}}}{\frac{1}{\sqrt{\mu_0 \varepsilon_0 \varepsilon_{r_2}}}}$

The $c$, $\mu_0$, and $\varepsilon_0$ terms cancel out perfectly, leaving us with:

$2 = \sqrt{\frac{\varepsilon_{r_2}}{\varepsilon_{r_1}}}$

To get rid of the square root, we square both sides:

$4 = \frac{\varepsilon_{r_2}}{\varepsilon_{r_1}}$

Finally, rearranging for the ratio we need:

$\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \frac{1}{4}$

**And there we have it!** The relative permittivity of the first medium is exactly one-fourth that of the second medium. The denser the medium electrically, the slower the wave travels.

Select Answer:

Visualized Solution

$E_{1}$

$v_{1} = \frac{ω _{1}}{k _{1}}$

$v = \frac{1}{μ ε}$

$E_{2}$

$v_{2} = \frac{ω _{2}}{k _{2}}$

$v_{2} = \frac{1}{μ ε}$

$\frac{ε _{r_{1}}}{ε _{r_{2}}}$

The Sigma Insight: Characteristics of Electromagnetic Waves

Analyzing the Wave in Air

The Master Equation of Electromagnetism

Entering the Denser Medium

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v1​=k1​ω1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=με​1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v2​=k2​ω2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v2​=με​1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } εr2​​εr1​​​