Question

A laser beam has diameter 1.30 mm . The beam exerts a force of 3.6×10−9 N on a totally reflecting surface.What is the amplitude of the electric field of the electromagnetic radiation in this beam?

Answer 1

Answer:

$1.75\cdot 10^4 V/m$

Explanation:

For a totally reflecting surface, the radiation pressure is related to the intensity of radiation by

$p=2\frac{I}{c}$ (1)

where

I is the intensity of radiation

c is the speed of light

The radiation pressure is given by

$p=\frac{F}{A}$

where in this case:

$F=3.6\cdot 10^{-9}N$ is the force exerted by the beam

A is the area on which the force is exerted

The beam has a diameter of d = 1.30 mm, so its area is

$A=\pi (\frac{d}{2})^2=\pi (\frac{0.0013 m}{2})^2=1.33\cdot 10^{-6} m^2$

Now we can write eq(1) as

$\frac{F}{A}=\frac{2I}{c}$

From which we find the intensity of radiation, I:

$I=\frac{Fc}{2A}=\frac{(3.6\cdot 10^{-9})(3.0\cdot 10^8)}{2(1.33\cdot 10^{-6})}=4.06\cdot 10^5 W/m^2$

Now we can find the amplitude of the electric field using the equation

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

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

E is the amplitude of the electric field

And solving for E,

$E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(4.06\cdot 10^5)}{(3\cdot 10^8)(8.85\cdot 10^{-12})}}=1.75\cdot 10^4 V/m$