NCERT Exemplar Q & A

Hint

(c) Here it is given, the energy required to dissociate a carbon monoxide molecule into carbon and oxygen atoms is $E=11 \mathrm{eV}$
We know that, $E=h f$, where $h=6.62 \times 10^{-34} \mathrm{~J}$-s
$f=\text { frequency }$
$\begin{aligned} \Rightarrow \quad 11 \mathrm{eV} & =h f \quad \\ f & =\frac{11 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}}=2.65 \times 10^{15} \mathrm{~Hz} \end{aligned}$
This frequency radiation belongs to ultraviolet region.

Hint

(b) When a wave is reflected from denser medium, then the type of wave doesn’t change but only its phase changes by $180^{\circ}$ or $\pi$ radian.
Thus, for the reflected wave $\hat{\mathbf{z}}=-\hat{\mathbf{z}}, \hat{\mathbf{i}}=-\hat{\mathbf{i}}$ and additional phase of $\pi$ in the incident wave.
Given, here the incident electromagnetic wave is,
$\mathbf{E}=E_0 \hat{\mathbf{i}} \cos (k z-\omega t)$
The reflected electromagnetic wave is given by
$\begin{aligned} \mathbf{E}_r & =E_0(-\hat{\mathbf{i}}) \cos [k(-z)-\omega t+\pi] \\ & =-E_0 \hat{\mathbf{i}} \cos [-(k z+\omega t)+\pi] \\ & =E_0 \hat{\mathbf{i}} \cos \left[-\left(k_z+\omega t\right)=E_0 \hat{\mathbf{i}} \cos (k z+\omega t)\right] \end{aligned}$

Hint

(b) Given, energy flux $\phi=20 \mathrm{~W} / \mathrm{cm}^2$
Area, $\quad A=30 \mathrm{~cm}^2$
Time, $t=30 \mathrm{~min}=30 \times 60 \mathrm{~s}$
Now, total energy falling on the surface in time $t$ is,
$U=\phi A t=20 \times 30 \times(30 \times 60) \mathrm{J}$
Momentum of the incident light $=\frac{U}{c}$
$=\frac{20 \times 30 \times(30 \times 60)}{3 \times 10^8}=36 \times 10^{-4} \mathrm{~kg}-\mathrm{ms}^{-1}$
Momentum of the reflected light $=0$
$\therefore$ Momentum delivered to the surface
$=36 \times 10^{-4}-0=36 \times 10^{-4} \mathrm{~kg}-\mathrm{ms}^{-1}$

As no reflection from the surface and for complete absorption so momentum of reflected radiation is zero.
Momentum delivered to surface $=$ Change in momentum
$=p_f-p_i=0-36 \times 10^{-4}=-36 \times 10^{-4} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
(-) sign shows the direction of momentum.

Hint

(c) Electric field intensity on a surface due to incident radiation is,
$I_{a v} \propto E_o^2$
$\begin{aligned} & \frac{P_{a V .}}{A} \propto E_o^2 \\ & P_{a v} \propto E_o^2 [ [latex]\because A[latex] is same in both cases] \end{aligned}$
$\begin{aligned} &\text { We know that, } E_0 \propto \sqrt{P_{a v}}\\ &\begin{aligned} & \therefore \frac{\left(E_0\right)_1}{\left(E_0\right)_2}=\sqrt{\frac{\left(P_{a v}\right)_1}{\left(P_{a v}\right)_2}} \\ & \Rightarrow \frac{E}{\left(E_0\right)_2}=\sqrt{\frac{100}{50}} \\ & \left(E_0\right)_2=E / \sqrt{2} \end{aligned} \end{aligned}$

Hint

(d) Key concept: A changing electric field produces a changing magnetic field and vice versa which gives rise to a transverse wave known as electromagnetic wave. The time varying electric and magnetic field are mutually perpendicular to each other and also perpendicular to the direction of propagation of this wave. The electric vector is responsible for the optical effects of an EM wave and is called the light vector.

The direction of propagation of electromagnetic wave is perpendicular to both electric field vector $(\vec{E})$ and $\vec{B}$ magnetic field vector $B$, i.e., in the direction of $\vec{E} \times \vec{B}$.
Here, electromagnetic wave is along the $z$-direction which is given by the cross product of $E$ and $B$.

Hint

Intensity in terms of electric field $\quad U_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 E_o^2$
Intensity in terms of magnetic field $U_{\mathrm{av}}=\frac{1}{2} \frac{B_0^2}{\mu_0}$
Now taking the intensity in terms of electric field.
$\left(U_{\mathrm{av}}\right) \text { electric field }=\frac{1}{2} \varepsilon_0 E_o^2$
$=\frac{1}{2} \varepsilon_0\left(c B_0\right)^2 \quad\left(\because E_0=c B_0\right)$
$=\frac{1}{2} \varepsilon_0 \times C^2 B^2$
$\begin{aligned} &\text { But, }\\ &c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \end{aligned}$
$\therefore \quad\left(U_{\mathrm{av}}\right)_{\text {Electric field }}=\frac{1}{2} \varepsilon_0 \times \frac{1}{\mu_0 \varepsilon_0} B_o^2=\frac{1}{2} \frac{B_0^2}{\mu_0}$
$=\left(U_{\mathrm{av}}\right)_{\text {magnetic field }}$
Thus, the energy in electromagnetic wave is divided equally between electric field vector and magnetic field vector.
Therefore, the ratio of contributions by the electric field and magnetic field components to the intensity of an electromagnetic wave is $1: 1$.

Note:
When the incident EM wave is completely absorbed by a surface, it delivers energy U and momentum $\mathrm{U} / \mathrm{c}$ to the surface.
When a wave of energy $U$ is totally reflected from the surface, the momentum delivered to surface is $2 \mathrm{U} / \mathrm{c}$.

Hint

(c) From a diode antenna, the electromagnetic waves are radiated outwards.
The amplitude of electric field vector $\left(E_0\right)$ which transports significant energy from the source falls off intensity inversely as the distance $(r)$ from the antenna, i.e., $E_0 \propto \frac{1}{r}$.

Hint

(a, d) We are given that the electric field vector of an electromagnetic wave travels in a vacuum along z-direction as,
$\vec{E}=\left(E_2 \hat{i}+E_2 \hat{j}\right) \cos (k z-\omega t)$
The magnitude of the electric and the magnetic fields in an electromagnetic wave are related as
$\begin{aligned} & B_0=\frac{E_0}{c} \\ & \vec{B}=\frac{\vec{E}}{c}=\frac{E_1 \hat{i}+E_2 \hat{j}}{c} \cos (k z-\omega t) \end{aligned}$
Also, $\vec{E}$ and $\vec{B}$ are perpendicular to each other and the propagation of electromagnetic wave is perpendicular to $\vec{E}$ as well as $\vec{B}$, so the given electromagnetic wave is plane polarized.

Hint

(a, b, c) The direction of propagation of an electromagnetic wave is always along the direction of vector product $\vec{E} \times \vec{B}$. Refer to Figure.

$\begin{aligned} \vec{B} & =B \hat{j}=B(\hat{k} \times \hat{i})=\frac{E}{c}(\hat{k} \times \hat{i}) \\ & =\frac{1}{c}[k \times E \hat{i}]=\frac{1}{c}[\hat{k} \times \vec{E}] \quad\left(\text { as } \frac{E}{B}=c\right) \end{aligned}$
(b) $\vec{E}=E \hat{i}=c B(\hat{j} \times \hat{k})=c(B \hat{j} \times \hat{k})=c(\vec{B} \times \hat{k})$
(c) $\hat{k} \cdot \vec{E}=\hat{k} \cdot(E \hat{i})=0, \vec{k} \cdot \vec{B}=\hat{k} \cdot(B \hat{j})=0$
(d)
$\begin{aligned} & \hat{k} \times \vec{E}=\hat{k} \times(E \hat{i})=E(\hat{k} \times \hat{i})=E \hat{j} \\ & \text { and } \hat{k} \times \vec{B}=\hat{k} \times(B \hat{j})=B(\hat{k} \times \hat{j})=-B \hat{i} \end{aligned}$

Hint

(b, d) As electric and magnetic field vectors $\mathbf{E}$ and $\mathbf{B}$ are perpendicular to each other as well as perpendicular to the direction of propagation of electromagnetic wave.
Here in the question electromagnetic wave is propagating along $x$-direction. So, electric and magnetic field vectors should have either $y$-direction or $z$-direction.

Hint

(a, c, d) Here we are given the frequency by which the charged particles oscillates about its mean equilibrium position, it is equal to $10^9 \mathrm{~Hz}$. The frequency of electromagnetic waves produced by a charged particle is equal to the frequency by which it oscillates about its mean equilibrium position.
So, frequency of electromagnetic waves produced by the charged particle is $\nu=10^9 \mathrm{~Hz}$.
Wavelength $\lambda=\frac{c}{\nu}=\frac{3 \times 10^8}{10^9}=0.3 \mathrm{~m}$
The frequency of $10^9 \mathrm{~Hz}$ falls in the region of radiowaves.

Hint

(b, d) Key concept:
An electromagnetic wave can be produced by accelerated or oscillating charge.
An oscillating charge is accelerating continuously, it will radiate electromagnetic waves continuously.
Electromagnetic waves are also produced when fast moving electrons are suddenly stopped by a metal target of high atomic number.

Here, in option (b) charge is moving in a circular orbit.
In circular motion, the direction of the motion of charge is changing continuously, thus it is an accelerated motion and this option is correct.
In option (d), the charge is falling in electric field. If a charged particle is moving in electric field it experiences a force or we can say it accelerates. We know an accelerating charge particle radiates electromagnetic waves. Hence option (d) is also correct.

Note: Also, we know that a charge starts accelerating when it falls in an electric field. Important points:
In an atom an electron is circulating around the nucleus in a stable orbit, although accelerating does not emit electromagnetic waves; it does so only when it jumps from a higher energy orbit to a lower energy orbit.
A simple LC oscillator and energy source can produce waves of desired frequency.

Hint

(a, c, d) Radiation pressure $(p)$ is the force exerted by electromagnetic wave on unit area of the surface, i.e., rate of change of momentum per unit area of the surface.
Momentum per unit time per unit area
$=\frac{\text { Intensity }}{\text { Speed of wave }}=\frac{I}{c}$
Change in momentum per unit time per unit area $=\frac{\Delta I}{c}=$ radiation pressure $(p)$
i.e., $p=\frac{\Delta I}{c}$
Momentum of incident wave per unit time per unit area $=\frac{I}{\mathrm{C}}$
When wave is fully absorbed by the surface, the momentum of the reflected wave per unit time per unit area $=0$.
Radiation pressure $(p)=$ change in momentum per unit time per unit area $=\frac{\Delta I}{\mathrm{c}}=\frac{I}{\mathrm{c}}-0=\frac{I}{\mathrm{c}}$.
When wave is totally reflected, then momentum of the reflected wave per unit time per unit area $=-\frac{I}{\mathrm{C}}$, Radiation pressure $p=\frac{I}{\mathrm{c}}-\left(-\frac{I}{\mathrm{c}}\right)=\frac{2 I}{\mathrm{c}}$.
Here, $p$ lies between $\frac{I}{c}$ and $\frac{2 I}{c}$.