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26P

Expert-verifiedFound in: Page 1002

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig.33-38, a laser beam of power 4.60W and diameter d=2.60mm is directed upward at one circular face (of diameter d<2.60mm) 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 1.20 g/cm ^{3}, what is its height H?**

Height of cylinder, $\text{H}=4.91\times {10}^{-7}\text{\hspace{0.17em}m}$

Diameter of the laser beam,$D=\text{2.60 mm}\left(\frac{{\text{10}}^{\text{-3}}\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}}mm}\right)=\text{2.60}\times {\text{10}}^{\text{-3}}\text{\hspace{0.17em}m}$

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

Cylinder density, $\text{\rho}=1.20{\text{gm/cm}}^{\text{3}}\left(\frac{{10}^{-3}\text{\hspace{0.17em}kg}}{1\text{\hspace{0.17em}gm}}\right)\left(\frac{1{\text{\hspace{0.17em}cm}}^{\text{3}}}{1\text{\hspace{0.17em}}{0}^{-6}{\text{\hspace{0.17em}m}}^{3}}\right)\text{=1.2}\times {\text{10}}^{\text{3}}{\text{\hspace{0.17em}kg/m}}^{\text{3}}$

**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 on ****the ****cylinder is zero. From this ****and after balancing forces on the cylinder, ****solve the problem for ****the ****cylinder 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,

${P}_{r}=\frac{2{I}_{0}}{c}$

Force due to radiation pressure, F=A.P_{r}

Density, $\rho =\frac{m}{V}$

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 $A=\frac{\pi {D}^{2}}{4}$ 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:

$\rho V=m$

$m=\rho \times \pi {\left(\frac{D}{2}\right)}^{2}H$

Where, H is the height of the cylinder.

$g\rho \pi {\left(\frac{D}{2}\right)}^{2}H=A{P}_{r}$

Hence,

$\begin{array}{c}H=\frac{4A{P}_{r}}{g\rho \pi {D}^{2}}\\ =\frac{4\left(\frac{\pi {D}^{2}}{4}\right){P}_{r}}{g\rho \pi {D}^{2}}\\ =\frac{\pi {D}^{2}{P}_{r}}{g\rho \pi {D}^{2}}\\ =\frac{\pi {D}^{2}}{g\rho \pi {D}^{2}}\cdot \frac{2{I}_{0}}{c}\\ =\frac{2{I}_{0}}{g\rho c}\end{array}$

Now using the relation between power and intensity,

$\begin{array}{c}H=\frac{1}{\frac{\pi {D}^{2}}{4}}\cdot \frac{2p}{g\rho c}\\ =\frac{4}{\pi {D}^{2}}\left(\frac{2\times 4.60\text{\hspace{0.17em}}W}{9.8\text{\hspace{0.17em}}m/{s}^{2}\times \text{1.2}\times {\text{10}}^{\text{3}}{\text{\hspace{0.17em}kg/m}}^{\text{3}}\times 3.0\times {10}^{8}\text{\hspace{0.17em}}m/s}\right)\end{array}$

$\begin{array}{c}H=\frac{4\times 2\times 4.60\text{\hspace{0.17em}}W}{{(2.6\times {10}^{-3}\text{\hspace{0.17em}}m)}^{2}\times 9.8\text{\hspace{0.17em}}m/{s}^{2}\times \text{1.2}\times {\text{10}}^{\text{3}}{\text{\hspace{0.17em}kg/m}}^{\text{3}}\times 3.0\times {10}^{8}\text{\hspace{0.17em}}m/s}\\ =4.{\text{91\xd710}}^{\text{-7}}\text{\hspace{0.17em}m}\end{array}$

Therefore, the height of the cylinder,

$\text{H}=4.91\times {10}^{-7}\text{\hspace{0.17em}}m$

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

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