Chapter 42, Problem 64P

(a)

To determine

The number of waves constituting the laser light and the wavelength of each waves precise to eight digits.

Answer to Problem 64P

The laser light is made up of $\underset{\_}{\text{three}}$ waves. The wavelength of first wave is $\underset{\_}{632.80914\text{nm}}$, wavelength of second wave is $\underset{\_}{632.80857\text{nm}}$, and wavelength of third wave is $\underset{\_}{632.80971\text{nm}}$.

Explanation of Solution

The reflecting surface of the mirror is metallic in nature. So the waves formed has nodes at both the ends. The distance between two nodes is $\frac{\lambda }{2}$.

Write the expression for the distance between $N$ nodes.

    $d=N\frac{\lambda }{2}$                                                                                                               (I)

Here, $d$ is the distance between the nodes, and $N$ is the number of nodes.

The active medium between the reflecting mirrors amplify lights with the wavelengths in between $632.80840\text{nm}$ and $632.80980\text{nm}$.

Write the expression for the midpoint of the wavelength range.

    $\lambda =\frac{{\lambda }_1+{\lambda }_2}{2}$                                                                                                            (II)

Here, $\lambda$ is the midpoint of range of wavelength, ${\lambda }_1$ is the minimum wavelength, and ${\lambda }_2$ is the maximum wavelength.

Using these values gathered, a trial value of number of nodes can be calculated.

Use expression (I) to find a trial value of $N$.

    $N_{\text{trial}}=\frac{2d}{\lambda }$                                                                                                             (III)

Here, $N_{\text{trial}}$ is a trial value of number of nodes.

Now assume some values for $N$ and find out the wavelength of the laser lights which lies within the given range of wavelength.

Use expression (I) to find the wavelength of the first wave of laser light.

    ${\lambda }_1=\frac{2d}{N_1}$                                                                                                               (IV)

Here, ${\lambda }_1$ is the wavelength of the first wave, and $N_1$ is the number of nodes.

Write the expression for wavelength of second wave.

    ${\lambda }_2=\frac{2d}{N_2}$                                                                                                              (V)

Here, ${\lambda }_2$ is the wavelength of the second wave, and $N_2$ is the number of nodes.

Write the expression for wavelength of third wave.

    ${\lambda }_3=\frac{2d}{N_3}$                                                                                                               (VI)

Here, ${\lambda }_3$ is the wavelength of the third wave, and $N_3$ is the number of nodes.

Write the expression for wavelength of fourth wave.

    ${\lambda }_4=\frac{2d}{N_4}$                                                                                                              (VII)

Here, ${\lambda }_4$ is the wavelength of the fourth wave, and $N_4$ is the number of nodes.

Conclusion:

Substitute $632.80840\text{nm}$ for ${\lambda }_1$, and $632.80980\text{nm}$ for ${\lambda }_2$ in equation (II) to find $\lambda$.

    $\begin{array}{c}\lambda =\frac{632.80840\text{nm}+632.80980\text{nm}}{2}\\ =632.80910\text{nm}\end{array}$

Substitute $632.80910\text{nm}$ for $\lambda$, and $35.124103\text{cm}$ for $d$ in (III) to find $N_{\text{trial}}$.

    $\begin{array}{c}{N}_{\text{trial}}=\frac{2\left(35.124103\text{cm}\times \frac{1\text{m}}{100\text{m}}\right)}{\left(632.80910\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)}\\ =1110101.07\end{array}$

The trial value of $N$ is $1110101.07$. So assume values which are smaller than and greater than $1110101.07$.

Therefore the possible $N$ values are $1110100,1110101,1110102,1110103,\mathrm{...}$

Substitute $1110101$ for $N_1$, and $35.124103\text{cm}$ for $d$ in equation (IV) to find ${\lambda }_1$.

    $\begin{array}{c}{\lambda }_1=\frac{2\left(35.124103\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{1110101}\\ =632.80914\text{nm}\end{array}$

Substitute $1110102$ for $N_2$, and $35.124103\text{cm}$ for $d$ in equation (V) to find ${\lambda }_2$.

    $\begin{array}{c}{\lambda }_2=\frac{2\left(35.124103\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{1110102}\\ =632.80857\text{nm}\end{array}$

Substitute $1110100$ for $N_3$, and $35.124103\text{cm}$ for $d$ in equation (VI) to find ${\lambda }_3$.

    $\begin{array}{c}{\lambda }_3=\frac{2\left(35.124103\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{1110100}\\ =632.80971\text{nm}\end{array}$

Substitute $1110103$ for $N_4$, and $35.124103\text{cm}$ for $d$ in equation (VII) to find ${\lambda }_4$.

    $\begin{array}{c}{\lambda }_4=\frac{2\left(35.124103\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{1110103}\\ =632.80800\text{nm}\end{array}$

The active medium between the reflecting mirrors amplify lights with the wavelengths in between $632.80840\text{nm}$ and $632.80980\text{nm}$ only. Thus all waves except ${\lambda }_4$ can be amplified. Thus three waves are present in the laser light.

Therefore, the laser light is made up of $\underset{\_}{\text{three}}$ waves. The wavelength of first wave is $\underset{\_}{632.80914\text{nm}}$, wavelength of second wave is $\underset{\_}{632.80857\text{nm}}$, and wavelength of third wave is $\underset{\_}{632.80971\text{nm}}$.

(b)

To determine

The root mean square speed for neon atom at $120°\text{C}$.

Answer to Problem 64P

The root mean square value of speed for the neon atom at $120°\text{C}$ is $\underset{\_}{697\text{m/s}}$.

Explanation of Solution

Write the expression for the rms speed of the atom.

    $v=\sqrt{\frac{3kT\left(°\text{K}\right)}{m_0}}$                                                                                                     (VIII)

Here, $v$ is the root mean square value of speed, $k$ is Boltzmann constant, $T\left(°\text{K}\right)$ is the temperature in Kelvin scale, and $m_0$ is the mass of the atom.

Write the expression to convert temperature from Celsius scale to Kelvin scale.

    $T\left(\text{K}\right)=T\left(°\text{C}\right)+273$                                                                                               (IX)

Here, $T\left(\text{K}\right)$ is the temperature in Kelvin scale, and $T\left(°\text{C}\right)$ is the temperature in Celsius scale.

Conclusion:

Substitute $120°\text{C}$ for $T\left(°\text{C}\right)$ in equation (IX) to find $T°\text{K}$.

    $\begin{array}{c}T°\text{K}=120°\text{C}+273\\ =393\text{K}\end{array}$

Substitute $393\text{K}$ for $T\left(\text{K}\right)$, $1.38\times {10}^{-23}\text{J/K}$ for $k$, and $20.18\text{u}$ for $m_0$ in equation (VIII) to find $v$.

    $\begin{array}{c}v=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(393\text{K}\right)}{20.18\text{u}}\left(\frac{1\text{u}}{1.66\times {10}^{-27}\text{kg}}\right)}\\ =697\text{m/s}\end{array}$

Therefore, the root mean square value of speed for the neon atom at $120°\text{C}$ is $\underset{\_}{697\text{m/s}}$.

(c)

To determine

Show that the Doppler effect for light emitted by the moving Neon atoms makes the bandwidth of the amplifier larger than $0.00140\text{nm}$.

Answer to Problem 64P

It is shown that the Doppler effect for light emitted by the moving Neon atoms makes the bandwidth of the amplifier larger than $0.00140\text{nm}$.

Explanation of Solution

Write the expression for the Doppler shift in the frequency of light emitted by the moving Neon atom.

    $f^{\prime }=f\sqrt{\frac{c+v}{c-v}}$                                                                                                       (X)

Here, $f^{\prime }$ is the shifted frequency, $f$ is the frequency before motion, $c$ is the speed of light, and $v$ is the speed of atom.

Write the expression for frequency after Doppler shift.

    $f^{\prime }=\frac{c}{{\lambda }^{\prime }}$                                                                                                            (XII)

Here, ${\lambda }^{\prime }$ is the wavelength after the shift.

Write the expression for frequency before Doppler shift.

    $f=\frac{c}{\lambda }$                                                                                                             (XIII)

Here, $\lambda$ is the wavelength before the shift.

Use expressions (XII) and (XIII) in (X) and solve for ${\lambda }^{\prime }$.

    $\begin{array}{c}\frac{c}{{\lambda }^{\prime }}=\frac{c}{\lambda }\sqrt{\frac{c+v}{c-v}}\\ {\lambda }^{\prime }=\lambda \sqrt{\frac{c-v}{c+v}}\end{array}$                                                                                                    (XIV)

Conclusion:

Substitute $632.8091\text{nm}$ for $\lambda$, $3.00\times {10}^8\text{m/s}$ for $c$, and $697\text{m/s}$ for $v$ in equation (XIV) to find ${\lambda }^{\prime }$.

  $\begin{array}{c}{\lambda }^{\prime }=\left(632.8091\text{nm}\right)\sqrt{\frac{\left(3.00\times {10}^8\text{m/s}\right)-\left(697\text{m/s}\right)}{\left(3.00\times {10}^8\text{m/s}\right)+\left(697\text{m/s}\right)}}\\ =632.80763\text{nm}\end{array}$

Therefore, the new wavelength is greater than $0.00140\text{nm}$. So it is shown that the Doppler effect for light emitted by the moving Neon atoms makes the bandwidth of the amplifier larger than $0.00140\text{nm}$.