Question

Chapter 33, Problem 026

In the figure, a laser beam of power 5.52 W and diameter D= 2.60 mm is directed upward at one circular face (of diameter d < 2.60 mm) of a perfectly reflecting cylinder. The cylinder is levitated because the upward radiation force matches the downward gravitational force. If the cylinder's density is 1.12 g/cm³, what is its height H?

Answer 1

at equilibrium

$F_{net}=0$

$mg-F_{r}=0=>\frac{\pi HD^{2}g\rho }{4}-\left ( \frac{\pi D^{2}}{4} \right )\left ( \frac{2I}{c} \right )=0$

$H=\frac{2I}{gc\rho }=\left ( \frac{2P}{\frac{\pi D^{2}}{4}} \right )\left ( \frac{1}{gc\rho } \right )$

$H=\left ( \frac{2\times 5.52}{\frac{\pi \times0.0026^{2}}{4} } \right )\left ( \frac{1}{9.81\times 3\times 10^{8}\times 1.12\times 10^{3}} \right )$

H=6.31*10^-7 m