Optics

Ajoy Ghatak

Chapter 7

Simple Harmonic Motion, Forced Vibrations and Origin of Refractive Index - all with Video Answers

Educators

Chapter Questions

Problem 1

The displacement in a string is given by the following equation:
$$y(x, t)=a \cos \left(\frac{2 \pi}{\lambda} x-2 \pi v t\right)$$
where $a, \lambda$ and $v$ represent the amplitude, wavelength and the frequency of the wave. Assume $a=0.1 \mathrm{~cm}, \lambda=4 \mathrm{~cm}$,
$V=1 \mathrm{sec}^{-1}$. Plot the time dependence of the displacement at $x=0,0.5 \mathrm{~cm}, 1.0 \mathrm{~cm}, 1.5 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}$ and $4 \mathrm{~cm} .$
Interpret the plots physically.

Ameer Said

Ameer Said

Numerade Educator

Problem 2

The displacement associated with a standing wave on a sonometer is given by the following equation:
$$y(x, t)=2 a \sin \left(\frac{2 \pi}{\lambda} x\right) \cos 2 \pi v$$
If the length of the string is $L$ then the allowed values of $\lambda$ are $2 L, 2 L / 2,2 L / 3, \ldots$ (see Sec. 13.2). Consider the case when $\lambda=2 L / 5$; study the time variation of displacement in each loop and show that alternate loops vibrate in phase (with different points in a loop having different amplitudes) and adjacent loops vibrate out of phase.

Ameer Said

Numerade Educator

Problem 3

A tunnel is dug through the earth as shown in Fig. 7.15. A mass is dropped at the point $A$ along the tunnel. Show that it will execute simple harmonic motion. What will the time period be?

Khoobchandra Agrawal

Khoobchandra Agrawal

Numerade Educator

Problem 4

A 1 g mass is suspended from a vertical spring. It executes simple harmonic motion with period $0.1$ sec. By how much distance had the spring stretched when the mass was attached?

Khoobchandra Agrawal

Khoobchandra Agrawal

Numerade Educator

Problem 5

A stretched string is given simultaneous displacement in the $x-$ and $y-$ directions such that
$$x(z, t)=a \cos \left(\frac{2 \pi}{\lambda} z-2 \pi v t\right)$$
and
$$y(z, t)=a \cos \left(\frac{2 \pi}{\lambda} z-2 \pi v t\right)$$
Show that the string will vibrate along a direction making an angle $\pi / 4$ with the $x$ and $y$ axes.

Ameer Said

Numerade Educator

Problem 6

In Problem $7.5$, if
$$x(z, t)=a \cos \left(\frac{2 \pi}{\lambda} z-2 \pi v t\right)$$
and
$$y(z, t)=a \sin \left(\frac{2 \pi}{\lambda} z-2 \pi v t\right)$$
what will be the resultant displacement?

Ameer Said

Numerade Educator

Problem 7

As mentioned in Sec. $7.5$, alkali metals are transparent to ultraviolet light. Assuming that the refractive index is primarily due to the free electrons and that there is one free electron per atom, calculate $\lambda_{p}\left(=\frac{2 \pi c}{\omega_{p}}\right)$ for $\mathrm{Li}, \mathrm{K}$ and $\mathrm{Rb}$. You may assume that the atomic weights of $\mathrm{Li}, \mathrm{K}$ and $\mathrm{Rb}$ are 6.94, $39.10$ and $85.48$, respectively and that the corresponding densities are $0.534,0.870$ and $1.532 \mathrm{~g} / \mathrm{cm}^{3}$. Also, the values of various physical constants are:
$m=9.109 \times 10^{-31} \mathrm{~kg}, q=1.602 \times 10^{-19} \mathrm{C}$ and $\varepsilon_{0}=8.854$
$\times 10^{-12} \mathrm{C} / \mathrm{N}-\mathrm{m}^{2}$

Ameer Said

Numerade Educator

Problem 8

(a) In a metal, the electrons can be assumed to be essentially free. The drift velocity of the electron satisfies the following equation
$$m \frac{d \mathbf{v}}{d t}+m \mathbf{v} v=\mathbf{F}=-q \mathbf{E}_{\mathbf{0}} e^{-i, x}$$
where $v$ represents the collision frequency. Calculate the steady state current density (J = $-N q \mathbf{v}$ ) and show that the conductivity is given by
$$\sigma(\omega)=\frac{N q^{2}}{m} \frac{1}{v-i \omega}$$
(b) If $\mathbf{r}$ represents the displacement of the electron, show that
$$\mathbf{P}=-N q \mathbf{r}=-\frac{N q^{2}}{m\left(\omega^{2}+i \omega v\right)} \mathbf{E}$$
which represents the polarization. Using the above equation show that
$$\kappa(\omega)=1-\frac{N q^{2}}{m \varepsilon_{0}\left(\omega^{2}+i \omega v\right)}$$
which represents the dielectric constant variation for a free electron gas.

Ameer Said

Numerade Educator

Problem 9

Assuming that each atom of copper contributes one free electron and that the low frequency conductivity $\sigma$ is about $6 \times 10^{7}$ mhos/metre, show that $\mathrm{Y}^{\prime}=4 \times 10^{13} \mathrm{~s}^{-1}$. Using this
$<10^{11} \mathrm{~s}^{-1}$. For $\omega=10^{8} \mathrm{~s}^{-1}$ calculate the complex dielectric constant and compare its value with the one obtained for infrared frequencies.
It may be noted that for small frequencies, only one of the electrons of a copper atom can be considered to be free. On the other hand, for X-ray frequencies all the electrons may be assumed to be free (see Problems 7.10, $7.11$ and 7.12). Discuss the validity of the above argument.

Ameer Said

Numerade Educator

Problem 10

Show that for high frequencies $(\omega \gg v)$ the dielectric constant (as derived in Problem $7.8$ ) is essentially real with frequency dependence of the form
$$\kappa^{*}=1-\frac{\omega_{p}^{2}}{\omega^{2}}$$
where $\omega_{p}=\left(\frac{N q^{2}}{m \varepsilon_{0}}\right)^{1 / 2}$ is known as the plasma frequency. The above dielectric constant variation is indeed valid for X-ray wavelengths in many metals. Assuming that at such frequencies all the electrons can be assumed to be free, calculate $\omega_{p}$ for copper for which the atomic number is 29, mass number is 63 , and density is $9 \mathrm{~g} / \mathrm{cm}^{3}$.

Ameer Said

Numerade Educator

Problem 11

For sodium, at $\lambda=1 A$, all the electrons can be assumed to be free; under this assumption show that $\omega_{p} \approx 3 \times 10^{16} \mathrm{~s}^{-1}$ and $n^{2}=1$ and the metal will be completely transparent.

Ameer Said

Numerade Educator

Problem 12

In an ionic crystal (like $\mathrm{NaCl}, \mathrm{CaF}_{2}$, etc.), one has to take into account infrared resonance oscillations of the ions and Eq. (7.68) modifies to
$$n^{2}=1+\frac{N q^{2}}{m \varepsilon_{0}\left(\omega_{1}^{2}-\omega^{2}\right)}+\frac{p N q^{2}}{M \varepsilon_{0}\left(\omega_{2}^{2}-\omega^{2}\right)}$$
where $M$ represents the reduced mass of the two ions and $p$ represents the valency of the ion $\left(p=1\right.$ for $\mathrm{Na}^{+}, \mathrm{Cl}^{-} ;$ $p=2$ for $\left.\mathrm{Ca}^{++}, \mathrm{F}_{2}^{-}\right)$. Show that the above equation can be written in the form $^{*}$
$$n^{2}=n^{2}+\frac{A_{1}}{\lambda^{2}-\lambda_{1}^{2}}+\frac{A_{2}}{\lambda^{2}-\lambda_{2}^{2}}$$
where $n^{2}=1+\frac{A_{1}}{i_{-1}^{2}}+\frac{A_{2}}{i_{-2}^{2}}$
$$\lambda_{1}=\frac{2 \pi c}{\omega_{1}}, \quad \lambda_{2}=\frac{2 \pi c}{\omega_{2}}$$
$A_{1}=\frac{N q^{2}}{4 \pi^{2} c^{2} \varepsilon_{0} m} \lambda_{1}^{4}, A_{2}=\frac{p N q^{2}}{4 \pi^{2} c^{2} \varepsilon_{0} M} \lambda_{2}^{4}$

Ameer Said

Numerade Educator

Problem 13

The refractive index variation for $\mathrm{CaF}_{2}$ (in the visible region of the spectrum) can be written in the form $^{*}$
$$n^{2}=6.09+\frac{6.12 \times 10^{-15}}{\lambda^{2}-8.88 \times 10^{-15}}+\frac{5.10 \times 10^{-9}}{\lambda^{2}-1.26 \times 10^{-9}}$$
where $\lambda$ is in meters
(a) Plot the variation of $n^{2}$ with $\lambda$ in the visible region.
(b) From the values of $A_{1}$ and $A_{2}$ show that $m / M \approx 2.07 \times$ $10^{-5}$ and compare this with the exact value.
(c) Show that using the constants $A_{1}, A_{2}, \lambda_{1}$ and $\lambda_{2}$ we obtain $n^{2} \approx 5.73$ which agrees reasonably well with the experimental value given above.

Ameer Said

Numerade Educator

Problem 14

(a) The refractive index of a plasma (neglecting collisions) is approximately given by (see Sec. 7.6)
$$n^{2}=1-\frac{\omega_{p}^{2}}{\omega^{2}}$$
where
$$\omega_{p}=\left(\frac{N q^{2}}{m \varepsilon_{0}}\right)^{1 / 2}=56.414 \mathrm{~N}^{1 / 2} \mathrm{~s}^{-1}$$
is known as the plasma frequency. In the ionosphere, the maximum value of $N_{0}$ is $\approx 10^{10}-10^{12}$ electrons $/ \mathrm{m}^{3}$ Calculate the plasma frequency. Notice that at high frequencies $n^{2} \approx 1 ;$ thus high frequency waves (like the one used in TV) are not reflected by the ionosphere. On the other hand, for low frequencies, the refractive index is imaginary (like in a conductor-see $\operatorname{Sec} .24 .3$ ) and the beam gets reflected. This fact is used in long distance radio communications (see Fig. 3.20).
(b) Assume that for $x \approx 200 \mathrm{~km}, N=10^{12}$ electrons $/ \mathrm{m}^{3}$ and that the electron density increases to $2 \times 10^{12}$ electrons/m $^{3}$ at $x=300 \mathrm{~km}$. For $x<300 \mathrm{~km}$, the electron density decreases. Assuming a parabolic variation of $N$, plot the corresponding refractive index variation.

Ameer Said

Numerade Educator