Chapter 40, Problem 38P

(a)

To determine

The amplitude of the electric field of the helium-neon laser.

Answer to Problem 38P

The amplitude of the electric field is $\text{14}\text{.0}\text{kV/m}$.

Explanation of Solution

The diameter of the beam of laser is $1.75\text{mm}$. The laser delivers $2.00\times {10}^{18}\text{photons/s}$. The wavelength of the photon is $633\text{nm}$.

Write the formula for the energy of a single photon.

    $E=\frac{hc}{\lambda }$                                                                                                             (I)

Here, $E$ is the energy of single photon, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum, and $\lambda$ is the wavelength.

Write the formula for the power carried by the laser beam.

    $P=En$                                                                                                           (II)

Here, $P$ is the power carried by the beam, $n$ is the number of photons emitted per second.

Write the formula for the average poynting vector.

    $S_{\text{avg}}=\frac{P}{\pi r^2}$                                                                                                     (III)

Here, $S_{\text{avg}}$ is the average poynting vector, $P$ is the power, and $r$ is the radius of the beam.

Write the formula for the average poynting vector in terms of electric field.

    $S_{\text{avg}}=\frac{E_{\max }^2}{2{\mu }_{0}c}$

Here, $E_{\max }$ is the amplitude of electric field, and ${\mu }_0$ is the permeability of free space.

Re-write the above equation to get an expression for $E_{\max }$.

    $E_{\max }=\sqrt{2{\mu }_{0}c{S}_{\text{avg}}}$                                                                                               (IV)

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $\text{3}\text{.00}\times {\text{10}}^{\text{8}}\text{m/s}$ for $c$, $633\text{nm}$ for $\lambda$ in equation (I) to get $E$.

    $\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(\text{3}\text{.00}\times {\text{10}}^{\text{8}}\text{m/s}\right)}{633\text{nm}}\\ =\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(\text{3}\text{.00}\times {\text{10}}^{\text{8}}\text{m/s}\right)}{633\times {10}^{-9}\text{m}}\\ =3.14\times {10}^{-19}\text{J}\end{array}$

Substitute $3.14\times {10}^{-19}\text{J}$ for $E$, $2.00\times {10}^{18}\text{photons/s}$ for $n$ in equation (II) to get $P$.

    $\begin{array}{c}P=\left(2.00\times {10}^{18}\text{photons/s}\right)\left(3.14\times {10}^{-19}\text{J/photon}\right)\\ =0.628\text{W}\end{array}$

Substitute $0.628\text{W}$ for $P$, $1.75\text{mm}$ for $r$ in equation (III) to get $S_{\text{avg}}$.

    $\begin{array}{c}{S}_{\text{avg}}=\frac{0.628\text{W}}{\pi {\left(\frac{1.75\text{mm}}{2}\right)}^2}\\ =\frac{0.628\text{W}}{\pi {\left(\frac{1.75\times {10}^{-3}\text{m}}{2}\right)}^2}\\ =2.61\times {10}^5\text{W}/{\text{m}}^{\text{2}}\end{array}$

Substitute $2.61\times {10}^5\text{W}/{\text{m}}^{\text{2}}$ for $S_{\text{avg}}$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$, $3.00\times {10}^8\text{m/s}$ for $c$ in equation (IV) to determine $E_{\max }$.

    $\begin{array}{c}{E}_{\text{max}}=\sqrt{2\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(3.00\times {10}^8\text{m/s}\right)\left(2.61\times {10}^5{\text{W/m}}^{\text{2}}\right)}\\ =1.40\times {10}^4\text{N/C}\\ \text{=}1.40\times {10}^4\text{V/m}\\ \text{=14}\text{.0}\text{kV/m}\end{array}$

The amplitude of the electric field is $\text{14}\text{.0}\text{kV/m}$.

(b)

To determine

The amplitude of the magnetic field of the helium-neon laser.

Answer to Problem 38P

The amplitude of the magnetic field is $\text{46}\text{.8}\text{μT}$.

Explanation of Solution

The diameter of the beam of laser is $1.75\text{mm}$. The laser delivers $2.00\times {10}^{18}\text{photons/s}$. The wavelength of the photon is $633\text{nm}$.

Write the formula for the magnitude of the magnetic field.

    $B_{\max }=\frac{E_{\max }}{c}$

Here, $E_{\max }$ is the amplitude of electric field, $B_{\max }$ is the amplitude of the magnetic field, and $c$ is the speed of light in free space.

Conclusion:

Substitute $\text{14}\text{.0}\text{kV/m}$ for $E_{\max }$, $\text{3}\text{.00}\times {\text{10}}^{\text{8}}\text{m/s}$ for $c$ to get $B_{\max }$.

    $\begin{array}{c}{B}_{\max }=\frac{\text{14}\text{.0}\text{kV/m}}{3.00\times {10}^8\text{m/s}}\\ =\frac{1.40\times {10}^4\text{N/C}}{3.00\times {10}^8\text{m/s}}\\ =4.68\times {10}^{-5}\text{T}\\ \text{=46}\text{.8}\text{μT}\end{array}$

The amplitude of the magnetic field is $\text{46}\text{.8}\text{μT}$.

(c)

To determine

The force exerted by the beam on a perfectly reflecting surface when it is incident perpendicularly.

Answer to Problem 38P

The force exerted on the perfectly reflecting surface is $\text{4}\text{.19}\text{nN}$.

Explanation of Solution

The diameter of the beam of laser is $1.75\text{mm}$. The laser delivers $2.00\times {10}^{18}\text{photons/s}$. The wavelength of the photon is $633\text{nm}$.

Write the formula for the force exerted on the perfectly reflecting surface.

    $F=\frac{2P}{c}$

Here, $F$ is the force, $P$ is the power, and $c$ is the speed of light in vacuum.

Conclusion:

Substitute $0.628\text{W}$ for $P$, $\text{3}\text{.00}\times {\text{10}}^{\text{8}}\text{m/s}$ for $c$ to get $F$.

    $\begin{array}{c}F=\frac{2\left(0.628\text{W}\right)}{3.00\times {10}^8\text{ m/s}}\\ =4.19\times {10}^{-9}\text{N}\\ \text{=4}\text{.19}\text{nN}\end{array}$

The force exerted on the perfectly reflecting surface is $\text{4}\text{.19}\text{nN}$.

(d)

To determine

The mass of the ice melted if the beam of laser is absorbed by an ice cube.

Answer to Problem 38P

The mass of the ice melted if the beam of laser is absorbed by an ice cube is $\text{10}\text{.2}\text{g}$.

Explanation of Solution

The diameter of the beam of laser is $1.75\text{mm}$. The laser delivers $2.00\times {10}^{18}\text{photons/s}$. The wavelength of the photon is $633\text{nm}$.

Write the formula for the mass of ice melted.

    $m=\frac{Pt}{L}$

Here, $m$ is the mass of the ice melted, $P$ is the power, $L$ is the latent heat of fusion and $t$ is the time.

Conclusion:

Substitute $0.628\text{W}$ for $P$, $1.50\text{h}$ for $t$, $3.33\times {10}^5\text{J/kg}$ for $L$ to get $m$.

    $\begin{array}{c}m=\frac{\left(0.628\text{W}\right)\left(1.50\text{h}\right)}{3.33\times {10}^5\text{J/kg}}\\ =\frac{\left(0.628\text{W}\right)\left[1.50\times \text{3600}\text{s}\right]}{3.33\times {10}^5\text{J/kg}}\\ =1.02\times {10}^{-2}\text{kg}\\ \text{=10}\text{.2}\text{g}\end{array}$

The mass of the ice melted if the beam of laser is absorbed by an ice cube is $\text{10}\text{.2}\text{g}$.