1. Radiation Pressure of a Laser Used to Balance a Particle.
A laser light of power \(3\text{ W}\) is spread evenly across the cross-section of the beam of diameter \(2\text{ mm}\text{.}\) When the laser light is incident on a perfectly reflecting surface of a spherical particle of diameter \(1.5\text{ mm}\) in vacuum, the particle is observed to be suspended in space. Find the density of the particle.
Hint.
Pressure times area of cross-section will give the force that balaces weight.
Answer.
\(0.2\text{ kg/m}^3\)
Solution.
Here, the force due to the radiation pressure is able to balance the force of gravity. Let \(\rho\) be the desired density of the spherical particle and \(R\) its radius. Let \(P\) be the radiation pressure at the site of the particle. The balancing of forces yields the following relation.
\begin{equation*}
P\, A_{\text{cross-section}} = (\rho V_{\text{particle}})\, g.
\end{equation*}
Here, the radiation pressure acts on the cross-section area of the particle, which is simply a circle of radius \(R\text{,}\) and the volume of the particle is just the volume of a sphere.
\begin{equation*}
P\, \pi R^2 =\left( \rho \frac{4}{3}\pi R^3 \right)\, g.
\end{equation*}
Hence the required density is
\begin{equation*}
\rho = \frac{3 P}{4 g R}.
\end{equation*}
Now, we put in the numerical values given here. The values of \(R\text{,}\) \(g\text{,}\) and \(P\) are
\begin{align*}
\amp R = 0.75\ \text{mm} = 7.5\times 10^{-4}\ \text{m}\\
\amp g = 9.81\ \text{m/s}^2\\
\amp P = 2\frac{I}{c} = 2\times \frac{3\ \text{W}}{\pi(1\times 10^{-3}\ \text{m})^{2}} \times \frac{1}{3\times 10^8\ \text{m/s}} = 2\times 10^{-3}\ \text{Pa}.
\end{align*}
Hence, \(\rho = 0.2\text{ kg/m}^3\text{.}\) It is interesting to compare the density found to the density of air at standard temperature and pressure, which is approximately \(1.2\text{ kg/m}^3\text{.}\) Clearly, one will need a very powerful laser to suspend an ordinary material.
