91Ó°ÊÓ

Diameter of the laser beam,$D=\text{2.60mm}\left(\frac{{\text{10}}^{\text{-3}}\text{″¾}}{1\text{ }mm}\right)=\text{2.60}×{\text{10}}^{\text{-3}}\text{″¾}$

Power or intensity of the beam,$\text{p}=4.\text{60W}$

Cylinder density,$\text{ÒÏ}=1.20{\text{gm/cm}}^{\text{3}}\left(\frac{{10}^{−3}\text{ k²µ}}{1\text{ g³¾}}\right)\left(\frac{1{\text{ c³¾}}^{\text{3}}}{1\text{ }{0}^{−6}{\text{″¾}}^3}\right)\text{=1.2}×{\text{10}}^{\text{3}}{\text{ k²µ/m}}^{\text{3}}$

In this problem, use force due to radiation and definition of radiation intensity in terms of power.In the equilibrium state of the cylinder, the net force acting onthecylinder is zero. From thisand after balancing forces on the cylinder,solve the problem forthecylinder height.The force acting on a perfectly reflective surface of the area A as a result of incident radiation of intensity I is given by F=2IAc, where c is the speed of light. The radiation intensity is defined as the power per unit solid angle, which is the power incident on that portion of the surface of a sphere that subtends an angle of one radian at the center of the sphere in both the horizontal and the vertical planes.

Formulae are as follows:

Radiation pressure for reflection,

$P_r=\frac{2I_0}{c}$

Force due to radiation pressure, F=A.P_r

Density, $\mathrm{ÒÏ}=\frac{m}{V}$

Power, $p=I_{0}A$

where, P_r is the radiation pressure for reflection, c is the speed of light, F is the force, A is the area, p is the density, m is the mass, V is the volume, and p is the power.

As mentioned in the problem, the cylinder is levitating because the forces the on the cylinder are balanced. It is in equilibrium,

F_g=F_r

M_g=A.P_r

where, m is the mass of the cylinder, and $A=\frac{\mathrm{Ï€}{D}^2}{4}$is the cross-sectional area of the laser beam falling on the cylinder.

Express the mass of the cylinder in terms of the density of the cylinder as follows:

$\mathrm{ÒÏ}V=m$

$m=\mathrm{ÒÏ}×\mathrm{Ï€}{\left(\frac{D}{2}\right)}^{2}H$

Where, H is the height of the cylinder.

$g\mathrm{ÒÏ}\mathrm{Ï€}{\left(\frac{D}{2}\right)}^{2}H=A{P}_r$

Hence,

$\begin{array}{c}H=\frac{4A{P}_r}{g\mathrm{ÒÏ}\mathrm{Ï€}{D}^2}\\ =\frac{4\left(\frac{\mathrm{Ï€}{D}^2}{4}\right)P_r}{g\mathrm{ÒÏ}\mathrm{Ï€}{D}^2}\\ =\frac{\mathrm{Ï€}{D}^2{P}_r}{g\mathrm{ÒÏ}\mathrm{Ï€}{D}^2}\\ =\frac{\mathrm{Ï€}{D}^2}{g\mathrm{ÒÏ}\mathrm{Ï€}{D}^2}â‹\dots \frac{2I_0}{c}\\ =\frac{2I_0}{g\mathrm{ÒÏ}c}\end{array}$

Now using the relation between power and intensity,

$\begin{array}{c}H=\frac{1}{\frac{\mathrm{Ï€}{D}^2}{4}}â‹\dots \frac{2p}{g\mathrm{ÒÏ}c}\\ =\frac{4}{\mathrm{Ï€}{D}^2}\left(\frac{2×4.60\text{ }W}{9.8\text{ }m/s^2×\text{1.2}×{\text{10}}^{\text{3}}{\text{ k²µ/m}}^{\text{3}}×3.0×{10}^8\text{ }m/s}\right)\end{array}$

$\begin{array}{c}H=\frac{4×2×4.60\text{ }W}{{(2.6×{10}^{−3}\text{ }m)}^2×9.8\text{ }m/s^2×\text{1.2}×{\text{10}}^{\text{3}}{\text{ k²µ/m}}^{\text{3}}×3.0×{10}^8\text{ }m/s}\\ =4.{\text{91×10}}^{\text{-7}}\text{″¾}\end{array}$

Therefore, the height of the cylinder,

$\text{H}=4.91×{10}^{−7}\text{ }m$

The height of the cylinder using the equilibrium condition of forces can be found.