Physics - EM Waves Question with Solution | TestHub

The intensity $I$ of the laser beam is given by $I=\frac{P}{A}$, where $P$ is the power and $A$ is the cross-sectional area. Area $A=\pi r^2=\pi (1\times 10^{-3}{)}^2=\pi \times 10^{-6}$ m${}^2$. So, $I=\frac{100}{\pi \times 10^{-6}}=\frac{10^8}{\pi }$ W/m${}^2$. The intensity $I$ is also related to the peak electric field $E_0$ by $I=\frac{1}{2}c{ϵ}_0{E}_0^2$. Rearranging for $E_0^2$: $E_0^2=\frac{2I}{c{ϵ}_0}=\frac{2\times (10^8/\pi )}{3\times 10^8\times 8.85\times 10^{-12}}=\frac{2\times 10^8}{3\pi \times 8.85\times 10^{-4}}=\frac{2\times 10^{12}}{3\pi \times 8.85}$. Substituting $\pi \approx 3.14$: $E_0^2=\frac{2\times 10^{12}}{3\times 3.14\times 8.85}\approx \frac{2\times 10^{12}}{83.3}\approx 0.0240\times 10^{12}=2.40\times 10^{10}$.$E_0=\sqrt{2.40\times 10^{10}}\approx 1.549\times 10^5$ V/m. In units of $10^3$ V/m, $E_0=154.9\times 10^3$ V/m. Rounding to the nearest integer, we get 155.