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### Fundamentals Of Physics

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/cm3, what is its height H?

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

See the step by step solution

## Step 1: Given

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

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

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

## Step 2: Determining the concept

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:

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

Force due to radiation pressure, F=A.Pr

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

Power, $p={I}_{0}A$

where, Pr 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.

## Step 3: (a) Determining the height of the cylinder, H

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

Fg=Fr

Mg=A.Pr

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:

$\rho V=m$

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

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,

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

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

Therefore, the height of the cylinder,

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

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