A 27 mW laser beam has a cross-sectional area of 10 mm^2. The magnitude of the maximum electric fiel

Q: Why do we use the formula with $1/2$ for intensity?

The formula $I = \frac{1}{2} c \varepsilon_0 E_0^2$ gives the *average* intensity of the wave over a full cycle. Since the electric field oscillates sinusoidally, the average of $\sin^2(\omega t)$ over a cycle is $1/2$.

Q: Can I use the RMS electric field instead?

Yes, if you use the RMS electric field, the formula for intensity is $I = c \varepsilon_0 E_{rms}^2$. However, the question specifically asks for the *maximum* electric field ($E_0$), so using the formula with $E_0$ is more direct.

Q: Why did we convert $mm^2$ to $m^2$ by multiplying by $10^{-6}$?

Since $1 \text{ mm} = 10^{-3} \text{ m}$, squaring both sides gives $1 \text{ mm}^2 = (10^{-3} \text{ m})^2 = 10^{-6} \text{ m}^2$. It's a common trap to just multiply by $10^{-3}$!

Q: What is the significance of the refractive index $n$ here?

The general formula for intensity in a medium is $I = \frac{1}{2} n c \varepsilon_0 E_0^2$. Since the laser is traveling in free space (or air), the refractive index $n \approx 1$, which simplifies our formula.

Q: How would I find the magnetic field from this?

Once you have the maximum electric field $E_0$, you can find the maximum magnetic field $B_0$ using the simple relation $E_0 = c B_0$.

Question

A $27 	ext{ mW}$ laser beam has a cross-sectional area of $10 	ext{ mm}^2$. The magnitude of the maximum electric field in this electromagnetic wave is given by
[Take, permittivity of space, $\varepsilon_0 = 9 	imes 10^{-12}$ SI units and speed of light, $c = 3 	imes 10^8 	ext{ m/s}$]

Options:

(A) $1 	ext{ kV/m}$

(B) $0.7 	ext{ kV/m}$

(C) $2 	ext{ kV/m}$

(D) $1.4 	ext{ kV/m}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a laser beam cutting through space. We are given its power as $27 	ext{ mW}$, and it passes through a cross-sectional area of $10 	ext{ mm}^2$. Before we dive into the formulas, it is crucial to convert these into standard SI units to avoid any silly mistakes later on.

The power $P$ is $27 	imes 10^{-3} 	ext{ W}$, and the area $A$ is $10 	imes 10^{-6} 	ext{ m}^2$.

### The Master Equation
Now, what is the intensity of this beam? Intensity is simply the power delivered per unit area. So, we can write:

$$I = \frac{P}{A}$$

But we also know from electromagnetic theory that the average intensity of an EM wave is related to its maximum electric field ($E_0$). The formula is:

$$I = \frac{1}{2} c \varepsilon_0 E_0^2$$

Since both expressions represent the same physical quantity—the intensity of the wave—we can equate them. This gives us our master equation:

$$\frac{P}{A} = \frac{1}{2} c \varepsilon_0 E_0^2$$

### Final Calculation
Our goal is to find the maximum electric field, $E_0$. So, let's rearrange the terms to isolate $E_0$:

$$E_0 = \sqrt{\frac{2P}{A c \varepsilon_0}}$$

Now, let's carefully substitute all the values we converted earlier into this expression:

$$E_0 = \sqrt{\frac{2 	imes 27 	imes 10^{-3}}{10 	imes 10^{-6} 	imes 3 	imes 10^8 	imes 9 	imes 10^{-12}}}$$

Let's simplify the numbers. The denominator becomes $27 	imes 10^{-9}$. The numerator is $54 	imes 10^{-3}$. This beautifully simplifies to:

$$E_0 = \sqrt{\frac{54 	imes 10^{-3}}{27 	imes 10^{-9}}} = \sqrt{2 	imes 10^6}$$

The square root of $2$ is approximately $1.414$, and the square root of $10^6$ is $10^3$. So, our final answer is:

$$E_0 \approx 1.414 	imes 10^3 	ext{ V/m} = \mathbf{1.4 	ext{ kV/m}}$$

Always keep these interconnections in mind! If the question had asked for the maximum magnetic field, you could easily find it using the relation $B_0 = \frac{E_0}{c}$.

Select Answer:

Visualized Solution

\text{Given Parameters}

\text{Intensity of the Beam}

\text{Intensity and Electric Field}

\text{Equating the Intensities}

\text{Rearranging for } E_0

\text{Substituting Values}

\text{Simplifying the Expression}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Characteristics of Electromagnetic Waves

Analyzing the Setup Imagine a laser beam cutting through space

The Master Equation Now, what is the intensity of this beam? Intensity is simply the power delivered per unit area

Final Calculation Our goal is to find the maximum electric field, $E_{0}$

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