NCERT Exemplar Solutions Class 12 Physics Electromagnetic Waves — Multiple Choice Questions Q1–13

NCERT Exemplar Problems, Class 12 Physics — Electromagnetic Waves, MCQ I and MCQ II. All 13 questions solved.

The book numbers these questions 8.1 to 8.13; here they are Questions 1 to 13. Questions 1 to 7 are MCQ I (exactly one option is correct) and Questions 8 to 13 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Transverse and perpendicular. $\mathbf{E}$, $\mathbf{B}$ and the direction of travel are mutually perpendicular, and the wave travels along $\mathbf{E}\times\mathbf{B}$, with $\mathbf{B}=\frac{1}{c}\hat{\mathbf{k}}\times\mathbf{E}$.
• Energy and momentum. The intensity is proportional to $E_0^2$, the electric and magnetic energy densities are equal, and a wave carrying energy $U$ carries momentum $U/c$.
• Radiation needs acceleration. Only accelerated charges radiate, at the frequency of the motion, and the radiation field falls as $1/r$.
• Reflection. A perfect conductor forces the tangential electric field to zero, so the reflected electric field is inverted.

Key insight. Most of these are checks of direction and proportionality, not heavy calculation. Questions 2, 4 and 8 are the ones where the printed key is wrong.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

One requires $11\ \text{eV}$ of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

Solution. One photon must carry at least $11\ \text{eV}$, so $h\nu = 11\times1.6\times10^{-19}\ \text{J}$ and

$$\nu = \frac{11\times1.6\times10^{-19}}{6.63\times10^{-34}} \approx 2.66\times10^{15}\ \text{Hz}, \qquad \lambda = \frac{c}{\nu}\approx 113\ \text{nm}.$$

Visible light spans roughly $400$ to $700$ nm and has photon energies of about $1.8$ to $3.1$ eV, far too small. A wavelength of $113$ nm lies below $400$ nm, in the ultraviolet. Infrared and microwave photons carry even less energy.

Question 2

A linearly polarized electromagnetic wave given as $\mathbf{E}=E_0\hat{\mathbf{i}}\cos(kz-\omega t)$ is incident normally on a perfectly reflecting infinite wall at $z=a$. Assuming that the material of the wall is optically inactive, the reflected wave will be given as

Solution. Two things must change on reflection.

• Direction. The reflected wave travels along $-z$, so $kz-\omega t$ becomes $kz+\omega t$. This rules out (a) and (d), which still have $kz-\omega t$.
• Sign of $\mathbf{E}$. A perfect conductor cannot sustain a tangential electric field at its surface, so the incident and reflected fields must cancel at the wall. The reflected electric field is therefore inverted, a phase change of $\pi$. That is the minus sign in (c). Option (b) has no phase reversal, so at the wall the two fields would add and double rather than cancel.

The options take the wall as the phase origin. Strictly, with the wall at $z=a$ the reflected wave is $-E_0\cos(kz+\omega t-2ka)$, but among the given choices only (c) has both the reversed direction and the inverted field.

Printed key error. The key prints (b). That reflected wave has the right direction but the wrong sign: it would give a doubled electric field at a perfect conductor, where the field must vanish. The correct answer is (c).

Question 3

Light with an energy flux of $20\ \text{W/cm}^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $30\ \text{cm}^2$, the total momentum delivered (for complete absorption) during $30$ minutes is

Solution. The energy absorbed is flux times area times time:

$$U = 20\times30\times(30\times60) = 20\times30\times1800 = 1.08\times10^{6}\ \text{J}.$$

Radiation carrying energy $U$ carries momentum $U/c$, and on complete absorption all of it is delivered to the surface:

$$p=\frac{U}{c}=\frac{1.08\times10^{6}}{3\times10^{8}}=3.6\times10^{-3}=36\times10^{-4}\ \text{kg m/s}.$$

Options (c) and (d) are the energy itself, not the momentum, and (a) is off by a factor of ten.

Question 4

The electric field intensity produced by the radiations coming from a $100\ \text{W}$ bulb at a $3\ \text{m}$ distance is $E$. The electric field intensity produced by the radiations coming from a $50\ \text{W}$ bulb at the same distance is

Solution. At distance $r$ the intensity is $I=\dfrac{P}{4\pi r^2}$, and the intensity of a wave is proportional to the square of its electric field amplitude, $I\propto E_0^2$. At fixed $r$ this gives $E_0^2\propto P$, so $E_0\propto\sqrt{P}$.

Halving the power from $100$ W to $50$ W therefore divides the field by $\sqrt2$, giving $E/\sqrt2$. It is the intensity that halves; the field falls by a smaller factor because it enters squared.

Printed key error. The key prints (d), $\sqrt2E$, which would be the field of a bulb of twice the power. A weaker bulb cannot give a stronger field. The correct answer is (c).

Question 5

If $\mathbf{E}$ and $\mathbf{B}$ represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along

Solution. An electromagnetic wave is transverse: $\mathbf{E}$ and $\mathbf{B}$ are both perpendicular to the direction of travel, so (a) and (b) are out. The energy flows along the Poynting vector $\mathbf{S}=\frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$, and the wave travels in the same direction. Since $\mathbf{B}\times\mathbf{E}=-\mathbf{E}\times\mathbf{B}$, option (c) points backwards.

Question 6

The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is

Solution. The energy densities of the two fields are $u_E=\frac12\varepsilon_0E^2$ and $u_B=\frac{B^2}{2\mu_0}$. In vacuum $E=cB$ and $c^2=\dfrac{1}{\mu_0\varepsilon_0}$, so

$$u_E=\tfrac12\varepsilon_0c^2B^2=\frac{B^2}{2\mu_0}=u_B.$$

The two fields carry equal energy at every point, so each contributes half of the intensity, a ratio of $1:1$. The factors of $c$ in the other options come from comparing $E$ with $B$ directly, which have different units and are not energies.

Question 7

An EM wave radiates outwards from a dipole antenna, with $E_0$ as the amplitude of its electric field vector. The electric field $E_0$ which transports significant energy from the source falls off as

Solution. The energy passing outward through a sphere of radius $r$ is the intensity times $4\pi r^2$. For this to stay the same at every radius, the intensity must fall as $1/r^2$. Since $I\propto E_0^2$, the amplitude of the radiated field must fall as $1/r$. The $1/r^2$ and $1/r^3$ terms in the field of an oscillating dipole are the induction and static parts. They die away too quickly to carry any net energy to large distances. A field that stays constant (d) would carry infinite energy.

Question 8

An electromagnetic wave travels in vacuum along the $z$ direction: $\mathbf{E}=(E_1\hat{\mathbf{i}}+E_2\hat{\mathbf{j}})\cos(kz-\omega t)$. Choose the correct options from the following:

Solution. For a wave along $+z$, $\mathbf{B}=\frac1c\hat{\mathbf{k}}\times\mathbf{E}$. With $\hat{\mathbf{k}}\times\hat{\mathbf{i}}=\hat{\mathbf{j}}$ and $\hat{\mathbf{k}}\times\hat{\mathbf{j}}=-\hat{\mathbf{i}}$,

$$\mathbf{B}=\frac1c\,(E_1\hat{\mathbf{j}}-E_2\hat{\mathbf{i}})\cos(kz-\omega t).$$

Options (a) and (b) are printed identically in the book, $\frac1c(E_1\hat{\mathbf{i}}-E_2\hat{\mathbf{j}})$, and neither matches this. Their dot product with $\mathbf{E}$ is proportional to $(E_1^2-E_2^2)/c$, which is not zero in general, so that field is not perpendicular to $\mathbf{E}$ as it must be. Both are false as printed.

The two components of $\mathbf{E}$ share the same $\cos(kz-\omega t)$, so they are in phase. The tip of $\mathbf{E}$ then oscillates along a fixed line at angle $\tan^{-1}(E_2/E_1)$ to the $x$-axis: the wave is linearly (plane) polarised, so (d) is true and (c) is false. Circular polarisation would need the components a quarter cycle apart.

Printed key error. The key prints (a), (d). Option (a) presumably meant the correct expression for $\mathbf{B}$, but as printed it is not perpendicular to $\mathbf{E}$, so it cannot be accepted. Only (d) is correct.

Question 9

An electromagnetic wave travelling along the $z$-axis is given as $\mathbf{E}=\mathbf{E}_0\cos(kz-\omega t)$. Choose the correct options from the following:

Solution.

• (a) $\mathbf{B}=\frac1c\hat{\mathbf{k}}\times\mathbf{E}$ is the standard relation. For the second form to be dimensionally right, $\mathbf{k}$ must be the wave vector of magnitude $k$ (the book’s printed symbol is easily read as a unit vector): $\frac1\omega\mathbf{k}\times\mathbf{E}=\frac k\omega\hat{\mathbf{k}}\times\mathbf{E}=\frac1c\hat{\mathbf{k}}\times\mathbf{E}$, since $\omega/k=c$. True.
• (b) $\mathbf{B}\times\hat{\mathbf{k}}=\frac1c(\hat{\mathbf{k}}\times\mathbf{E})\times\hat{\mathbf{k}}=\frac{\mathbf{E}}{c}$, using that $\mathbf{E}$ is perpendicular to $\hat{\mathbf{k}}$. So $\mathbf{E}=c\,(\mathbf{B}\times\hat{\mathbf{k}})$. True.
• (c) The wave is transverse, so both fields are perpendicular to $\hat{\mathbf{k}}$ and the dot products vanish. True.
• (d) Cross products with $\hat{\mathbf{k}}$ would vanish only if the fields were parallel to the direction of travel, the opposite of transverse. False.

Question 10

A plane electromagnetic wave propagating along the $x$ direction can have the following pairs of $\mathbf{E}$ and $\mathbf{B}$

Solution. Both fields must be perpendicular to the direction of travel, so neither can have an $x$-component. That removes (a) and (c). The pair must also be perpendicular to each other and give $\mathbf{E}\times\mathbf{B}$ along $+x$. For (b): $\hat{\mathbf{j}}\times\hat{\mathbf{k}}=\hat{\mathbf{i}}$, which works. For (d): $\hat{\mathbf{k}}\times\hat{\mathbf{j}}=-\hat{\mathbf{i}}$, so $E_z$ with $B_y$ is along $+x$ when $E_z$ and $B_y$ have opposite signs, which a wave can have. Both fields are in the $yz$ plane and mutually perpendicular, so (b) and (d) are allowed.

Question 11

A charged particle oscillates about its mean equilibrium position with a frequency of $10^9\ \text{Hz}$. The electromagnetic waves produced:

Solution. An oscillating charge radiates at the frequency at which it oscillates, so the waves have $\nu=10^9$ Hz: (a) true, (b) false. The wavelength is

$$\lambda=\frac c\nu=\frac{3\times10^8}{10^9}=0.3\ \text{m},$$

so (c) is true. A frequency of $1$ GHz sits at the upper edge of the radio band, at the border of the microwave range; NCERT counts it as radio, so (d) is true.

Question 12

The source of electromagnetic waves can be a charge

Solution. Radiation is produced by accelerated charges. A charge at rest or in uniform motion has no changing field pattern that detaches and travels away, so (a) and (c) are false. A charge in a circular orbit has centripetal acceleration, and a charge falling in an electric field accelerates uniformly. Both radiate.

Question 13

An EM wave of intensity $I$ falls on a surface kept in vacuum and exerts radiation pressure $p$ on it. Which of the following are true?

The book prints the fourth option as “(b)” a second time; it is labelled (d) here.

Solution. In time $\Delta t$ an area $A$ receives energy $IA\Delta t$ and momentum $IA\Delta t/c$. If the wave is totally absorbed, the surface receives this momentum, giving a pressure of $I/c$. If it is totally reflected, the momentum reverses, so the change is twice as large and the pressure is $2I/c$. A real surface absorbs some light and reflects the rest, so its pressure lies between the two limits. Hence (a), (c) and (d) are true and (b) is false.

Common mistakes

• Keeping the same sign on reflection in Question 2. A perfect conductor inverts the electric field; otherwise the field would double at the wall instead of vanishing.
• Forgetting the square in Question 4. Intensity is proportional to $E_0^2$, so halving the power divides the field by $\sqrt2$, not by $2$ and not multiplying by $\sqrt2$.
• Accepting the printed $\mathbf{B}$ in Question 8. Check that $\mathbf{E}\cdot\mathbf{B}=0$; the printed options fail it.
• Treating the electric field as the only carrier of energy in Question 6. The magnetic field carries exactly as much.
• Calling any moving charge a source in Question 12. Only an accelerated charge radiates.
• Using $I/c$ for total reflection in Question 13. Reversing the momentum doubles the pressure to $2I/c$.

Practise next

• Exemplar Electromagnetic Waves: Very Short, Short and Long Answer Questions — where the same ideas are applied to displacement current, the Poynting vector and radiation pressure.
• NCERT textbook Chapter 8 — Electromagnetic Waves for the theory of displacement current, the spectrum and energy in EM waves.
• Exemplar Alternating Current — the previous unit, whose oscillating currents and reactance carry over.
• Exemplar Ray Optics and Optical Instruments — the next unit, which treats light as rays through mirrors, lenses and instruments.