Question

A helium-neon laser emits a $1.0$-mm-diameter laser beam with a power of $1.0mW$. What are the amplitudes of the electric and magnetic fields of the light wave?

Short answer

The amplitudes of the light wave's electric and magnetic fields is,

$\begin{array}{l}{E}_o=976\mathrm{V}/\mathrm{m},\\ B_o=3.3\mu \mathrm{T}.\end{array}$

Step-by-step solution

Step: 1 Finding the values of Iavg:

Finding the world first

$$\begin{gathered} A=\pi \times r^2 \\ A=\pi {\left({10}^{-3}\right)}^2 \\ A=\pi \times {10}^{-6}. \end{gathered}$$

Then finding the values

$\begin{array}{l}{I}_{avg}=\frac{P}{A}\\ I_{avg}=\frac{{10}^{-3}}{\pi \times {10}^{-6}}\\ I_{avg}=1.27\times {10}^3.\end{array}$

Step: 2 Finding the vales of E0:

By using the formula,

$$\begin{gathered} E_o={\left(\frac{2I_{avg}}{c\cdot {\epsilon }_o}\right)}^{1/2} \\ \begin{array}{l}{E}_o={\left(\frac{2I_{avg}}{c\cdot {\epsilon }_o}\right)}^{1/2}\\ E_o={\left(\frac{2\times 1.27\times {10}^3}{3\times {10}^8\times 8.85\times {10}^{-12}}\right)}^{1/2}\\ E_o=976\mathrm{V}/\mathrm{m}.\end{array} \end{gathered}$$

Step: 3 Finding the values of B0:

By using,

$\begin{array}{l}{B}_o=\frac{E_o}{c}\\ B_o=\frac{976}{3\times {10}^8}\\ B_o=\frac{9.7\times {10}^2}{3\times {10}^8}\\ B_o=3.3\mu T.\end{array}$