A laser pointer delivers 0.10-mW average power in a beam 0.90 mm in diameter. Find (a) the average intensity, (b) the peak electric field, and (c) the peak magnetic field.

Sure, let's solve the problem step by step.

(a) To find the average intensity, we use the formula for intensity $I$:

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

where $P$ is the power and $A$ is the area of the beam. The power $P$ is given as 0.10 mW, which is 0.10 \times 10^{-3} W. The diameter of the beam is 0.90 mm, which is 0.90 \times 10^{-3} m. The area $A$ of the beam (assuming it is circular) is:

$A = \pi \left(\frac{d}{2}\right)^2$

Substituting the diameter $d$:

$A = \pi \left(\frac{0.90 \times 10^{-3}}{2}\right)^2$ $A = \pi \left(0.45 \times 10^{-3}\right)^2$ $A = \pi \times (0.45 \times 10^{-3})^2$ $A = \pi \times 0.2025 \times 10^{-6}$ $A = 0.2025 \pi \times 10^{-6}$ $A \approx 0.636 \times 10^{-6} \, \text{m}^2$

Now, we can find the intensity:

$I = \frac{0.10 \times 10^{-3}}{0.636 \times 10^{-6}}$ $I \approx 157 \, \text{W/m}^2$

(b) To find the peak electric field $E_0$, we use the relationship between intensity and the electric field:

$I = \frac{1}{2} \epsilon_0 c E_0^2$

where $\epsilon_0$ is the permittivity of free space ($\epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m}$) and $c$ is the speed of light in a vacuum ($c \approx 3 \times 10^8 \, \text{m/s}$). Rearranging for $E_0$:

$E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$

Substituting the values:

$E_0 = \sqrt{\frac{2 \times 157}{8.85 \times 10^{-12} \times 3 \times 10^8}}$ $E_0 = \sqrt{\frac{314}{2.655 \times 10^{-3}}}$ $E_0 = \sqrt{118.3 \times 10^3}$ $E_0 \approx 10.9 \times 10^3 \, \text{V/m}$ $E_0 \approx 10.9 \, \text{kV/m}$

(c) To find the peak magnetic field $B_0$, we use the relationship between the electric and magnetic fields in an electromagnetic wave:

$B_0 = \frac{E_0}{c}$

Substituting the values:

$B_0 = \frac{10.9 \times 10^3}{3 \times 10^8}$ $B_0 \approx 3.63 \times 10^{-5} \, \text{T}$ $B_0 \approx 36.3 \, \text{μT}$

So, the answers are: (a) The average intensity is approximately 157 W/m². (b) The peak electric field is approximately 10.9 kV/m. (c) The peak magnetic field is approximately 36.3 μT.