A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets refle

Q: Why can't I just take 96% of the incident electric field amplitude?

That's a very common trap! The 96% refers to the *energy* or *intensity* of the wave, not its amplitude. Intensity is proportional to the *square* of the amplitude, and it also depends on the properties of the medium (permittivity and wave speed). You must equate the intensities, not the amplitudes directly.

Q: Why did we use $v$ instead of $c$ for the transmitted wave's intensity?

When an electromagnetic wave enters a denser medium like glass, it slows down. The speed of light in vacuum is $c$, but in a medium with refractive index $n$, the speed becomes $v = c/n$. Since intensity depends on the wave's speed, we must use $v$ for the glass medium.

Q: How did we know that $\varepsilon_g = \varepsilon_0 n^2$?

The refractive index $n$ is defined as $n = \sqrt{\varepsilon_r \mu_r}$. For non-magnetic materials like glass, the relative permeability $\mu_r$ is approximately 1. Therefore, $n = \sqrt{\varepsilon_r}$, which means $\varepsilon_r = n^2$. Since the absolute permittivity is $\varepsilon_g = \varepsilon_0 \varepsilon_r$, we get $\varepsilon_g = \varepsilon_0 n^2$.

Q: Does the frequency of the light wave change when it enters the glass?

No, the frequency of a wave is a fundamental property determined by its source and remains absolutely constant across different media. Only the wave's speed and wavelength change when it crosses a boundary.

Q: What happens to the magnetic field amplitude in the glass?

In any medium, the electric and magnetic field amplitudes are related by $E_0 = v B_0$, where $v$ is the speed of the wave in that medium. So, in the glass, the magnetic field amplitude would be $B_t = A_t / v_g$. Since $v_g = c/n$, the magnetic field amplitude actually increases by a factor of $n$ relative to the electric field!

Question

A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is $30	ext{ V/m}$, then the amplitude of the electric field for the wave propogating in the glass medium will be

Options:

(A) $30	ext{ V/m}$

(B) $6	ext{ V/m}$

(C) $10	ext{ V/m}$

(D) $24	ext{ V/m}$

Sigma Percentile · Accepted Answer

### The Dance of Light: Unraveling Electric Field Amplitudes Across Boundaries

Imagine a light wave traveling freely through the vast emptiness of space, or simply through the air in your room. Suddenly, it encounters a solid boundary—a glass slab. What happens next is a beautiful physical phenomenon governed by the strict laws of energy conservation and electromagnetism. Some of the light bounces back, reflecting off the surface, while the rest plunges into the denser medium, transmitting through the glass.

In this problem, we are tasked with finding exactly how the amplitude of the electric field changes as the wave crosses this boundary. It might seem intuitive to just take a percentage of the initial amplitude, but as we will see, the physics of waves demands a much deeper look into the concept of **intensity**.

### The Power of Intensity

To understand what happens to the wave, we must first understand how it carries energy. The energy transported by an electromagnetic wave per unit area per unit time is called its **intensity** ($I$). The intensity is directly linked to the amplitude of the oscillating electric field ($E_0$), but it also depends heavily on the properties of the medium it travels through.

The fundamental formula for the intensity of an electromagnetic wave is:

$$I = \frac{1}{2} \varepsilon v E_0^2$$

Here, $\varepsilon$ represents the permittivity of the medium, $v$ is the speed of the wave in that medium, and $E_0$ is the peak amplitude of the electric field. Notice that intensity is proportional to the *square* of the amplitude. This is a critical detail that often trips up students!

### The Boundary Conditions and Energy Conservation

When our light wave hits the glass slab, the problem states that $4\%$ of the light gets reflected. By the unbreakable law of conservation of energy, the remaining energy must pass into the glass. Therefore, $96\%$ of the light is transmitted.

We can express this mathematically by relating the transmitted intensity ($I_t$) to the incident intensity ($I_i$):

$$I_t = 0.96 I_i$$

This is our master equation. It is the bridge that connects the wave in the air to the wave in the glass.

### Translating Medium Properties

Before we substitute our intensity formula into the master equation, we need to account for how the medium changes. The glass slab has a refractive index of $n = 1.5$.

How does this refractive index alter the wave's environment? First, it slows the wave down. The speed of light in the glass ($v_g$) is reduced compared to the speed of light in a vacuum ($c$):

$$v_g = \frac{c}{n}$$

Second, the permittivity of the medium changes. For a non-magnetic material like glass (where the relative permeability $\mu_r \approx 1$), the relative permittivity $\varepsilon_r$ is simply the square of the refractive index ($n^2$). Thus, the absolute permittivity of the glass ($\varepsilon_g$) becomes:

$$\varepsilon_g = \varepsilon_0 n^2$$

### The Algebraic Symphony

Now, let's substitute the full intensity expressions into our energy conservation equation. For the transmitted wave in the glass, we use $\varepsilon_g$, $v_g$, and the unknown transmitted amplitude $A_t$. For the incident wave in the air, we use the vacuum permittivity $\varepsilon_0$, the speed of light $c$, and the given incident amplitude $A_i$:

$$\frac{1}{2} \varepsilon_g v_g A_t^2 = 0.96 \left( \frac{1}{2} \varepsilon_0 c A_i^2 ight)$$

Next, we substitute our medium property translations ($\varepsilon_g = \varepsilon_0 n^2$ and $v_g = c/n$) into the left side of the equation:

$$\frac{1}{2} (\varepsilon_0 n^2) \left(\frac{c}{n}ight) A_t^2 = 0.96 \left( \frac{1}{2} \varepsilon_0 c A_i^2 ight)$$

Notice the beautiful algebraic cancellation that is about to happen. The $n^2$ in the numerator partially cancels with the $n$ in the denominator, leaving just a single $n$. Furthermore, the $\frac{1}{2}$, $\varepsilon_0$, and $c$ appear on both sides of the equation and vanish completely!

$$n A_t^2 = 0.96 A_i^2$$

### The Final Calculation

We have arrived at a remarkably elegant and simple relationship. To find the transmitted amplitude, we just need to isolate $A_t^2$:

$$A_t^2 = \frac{0.96}{n} A_i^2$$

Now, we simply plug in the given values. The refractive index $n$ is $1.5$, and the incident amplitude $A_i$ is $30	ext{ V/m}$:

$$A_t^2 = \frac{0.96}{1.5} (30)^2$$

$$A_t^2 = 0.64 	imes 900$$

$$A_t^2 = 576$$

Taking the square root of both sides reveals our final answer:

**$$A_t = 24 	ext{ V/m}$$**

The amplitude of the electric field propagating through the glass medium is $24	ext{ V/m}$. This problem is a fantastic reminder that when dealing with waves crossing boundaries, we must always anchor our calculations in the conservation of energy and the fundamental definition of intensity.

Select Answer:

Visualized Solution

\text{Visualizing the Interface}

\text{Intensity of EM Wave}

\text{Energy Conservation}

\text{Substituting Intensity Formula}

\text{Medium Properties}

\text{Simplifying the Equation}

\text{Isolating } A_t

\text{Plugging in Values}

\text{Final Amplitude}

\text{What about the Magnetic Field?}

The Sigma Insight: Characteristics of Electromagnetic Waves

The Dance of Light

The Power of Intensity

The Boundary Conditions and Energy Conservation

Translating Medium Properties

The Algebraic Symphony

The Final Calculation

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