Question

The helium-neon lasers most commonly used in student physics laboratories have average power outputs of 0.250 mW.

Answer 1

(a) $72.3 W/m^2$

First of all, we need to find the area of the circular spot, which is given by:

$A=\pi r^2$

where r is the radius of the spot, which is half the diameter, therefore

$r=\frac{d}{2}=\frac{2.10 mm}{2}=1.05 mm=1.05\cdot 10^{-3} m$

So, the area of the spot is

$A=\pi (1.05\cdot 10^{-3}m)^2=3.46\cdot 10^{-6} m^2$

We know that the power output of the laser is

$P=0.250 mW=2.5\cdot 10^{-4} W$

So the intensity of the laser beam is

$I=\frac{P}{A}=\frac{2.5\cdot 10^{-4} W}{3.46\cdot 10^{-6} m^2}=72.3 W/m^2$

(b) $7.8\cdot 10^{-7}T$

The average intensity of the laser is related to the peak magnetic field strength by

$I=\frac{cB_0^2}{2\mu_0}$

where

c is the speed of light

$B_0$ is the peak magnetic field strength

$\mu_0=1.257\cdot 10^{-6} H/m$ is the vacuum magnetic permeability

Solving the formula for $B_0$, we find

$B_0 = \sqrt{\frac{2I\mu_0}{c}}=\sqrt{\frac{2(72.3 W/m^2)(1.257\cdot 10^{-6} H/m)}{3\cdot 10^8 m/s}}=7.8\cdot 10^{-7}T$

(c) 234 V/m

The relationship between magnetic field and electric field in an electromagnetic wave is

$E_0=cB_0$

where

$E_0$ is the peak electric field strength

c is the speed of light

$B_0$ is the peak magnetic field strength

Substituting numbers into the formula, we find

$E_0=(3\cdot 10^8 m/s)(7.8\cdot 10^{-7} T)=234 V/m$