Question

A small laser used as a pointer produces a beam of red light 5 mm in diameter, and has a power output of 5 milliwatts. What is the magnitude of the electric field in the laser beam?

Answer 1

Answer:

434.0 V/m

Explanation:

The power output of the laser is:

$P=5 mW = 0.005 W$

while the radius of the beam is

$r=\frac{5 mm}{2}=2.5 mm = 0.0025 m$

so the cross-sectional area is

$A=\pi r^2 = \pi (0.0025 m)^2=2.0\cdot 10^{-5} m^2$

So the intensity of the laser beam is

$I=\frac{P}{A}=\frac{0.005 W}{2.0\cdot 10^{-5} m^2}=250 W/m^2$

The intensity of a laser beam is related to the magnitude of the electric field by

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

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

Solving the formula for E, we find

$E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(250 W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=434.0 V/m$