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Chapter 4 Electromagnetic Waves

Electromagnetic Waves

Theory, worked examples and 51 previous-year questions with full solutions — IIT JAM, GATE, CSIR NET-JRF and JEST, 2020 onwards.

Theory
Physical motivation

Maxwell's equations in free space predict waves that travel at \(c=1/\sqrt{\muz\eps}\) — the speed of light. This chapter studies these waves in detail: their structure, energy, momentum, polarization, and behavior at interfaces. The results apply to light, microwaves, X-rays, and radio waves alike.

Wave Equation from Maxwell's Equations

In free space (\(\rho=0\), \(\vec{J}=\mathbf{0}\)), Maxwell's equations reduce to: \begin{align} \nabla\cdot\vec{E} &= 0, & \nabla\cdot\vec{B} &= 0, \\ \nabla\times\vec{E} &= -\frac{\partial\vec{B}}{\partial t}, & \nabla\times\vec{B} &= \muz\eps\frac{\partial\vec{E}}{\partial t}. \end{align} Taking \(\nabla\times\) of the third equation and using the fourth: \begin{equation} \boxed{\nabla^2\vec{E} = \muz\eps\frac{\partial^2\vec{E}}{\partial t^2}, \qquad \nabla^2\vec{B} = \muz\eps\frac{\partial^2\vec{B}}{\partial t^2}} \end{equation}

Speed of light

\begin{equation} c = \frac{1}{\sqrt{\muz\eps}} = 2.998\times10^8\,\text{m/s} \end{equation} EM waves are transverse (\(\vec{E}\perp\vec{B}\perp\hat{k}\)), do not require a medium, and propagate at \(c\) in vacuum.

Properties of Maxwell's equations

  • [noitemsep]

  • Linear — obey superposition

  • Relativistically invariant

  • Solutions: plane waves (Cartesian), spherical waves, cylindrical waves

Monochromatic Plane Waves

The general plane-wave solution propagating along \(\hat{k}\): \begin{equation} \vec{E}(\vec{r},t) = \vec{E}_0\,e^{i(\vec{k}\cdot\vec{r}-\omega t)}, \qquad \vec{B}(\vec{r},t) = \vec{B}_0\,e^{i(\vec{k}\cdot\vec{r}-\omega t)} \end{equation} where \(\omega = ck\) and \(\vec{k}=k\hat{k}\).

For a wave propagating along \(+\zhat\): \begin{align} \vec{E}(z,t) &= E_0\,e^{i(kz-\omega t)}\,\xhat \\ \vec{B}(z,t) &= \frac{E_0}{c}\,e^{i(kz-\omega t)}\,\yhat = \frac{1}{c}\,\hat{k}\times\vec{E} \end{align}

Relations between \(\vec{E}\), \(\vec{B}\), \(\hat{k}\)

\begin{equation} \vec{B} = \frac{1}{c}\hat{k}\times\vec{E}, \qquad |\vec{B}| = \frac{|\vec{E}|}{c}, \qquad \vec{E}\perp\vec{B}\perp\hat{k} \end{equation}

Example 4.1   Identifying wave solutions

Which of the following satisfy the wave equation \(\partial^2 f/\partial z^2 = (1/v^2)\,\partial^2 f/\partial t^2\)? (a) \(A\sin(kz)\cos(\omega t)\); (b) \(A e^{-k^2(z^2+v^2t^2)}\); (c) \(A\sin k(z-vt) + B\cos 2k(z+vt)\).

(a): \(\partial^2/\partial z^2 = -Ak^2\sin kz\cos\omega t\); \((1/v^2)\partial^2/\partial t^2 = -A(\omega/v)^2\sin kz\cos\omega t = -Ak^2\sin kz\cos\omega t\). ✓

(b): \(\partial^2/\partial z^2 = A(-2k^2+4k^4z^2)e^{-k^2(z^2+v^2t^2)}\); \((1/v^2)\partial^2/\partial t^2 = A(-2k^2+4k^4v^2t^2)e^{-k^2(z^2+v^2t^2)}\). Not equal.

(c): Both terms are functions of \((z\pm vt)\), so both satisfy the wave equation. ✓

Example 4.2   Writing plane wave fields

Write \(\vec{E}\) and \(\vec{B}\) for a monochromatic plane wave of amplitude \(E_0\), frequency \(\omega\), phase \(\delta=0\): (a) traveling in \(-\zhat\), polarized in \(\xhat\); (b) traveling along \(\hat{n}=(\xhat+\yhat+\zhat)/\sqrt{3}\), polarized \(\parallel\yhat\)-\(\zhat\) plane.

(a): \(\vec{E} = E_0\cos(-kz-\omega t)\,\xhat\);   \(\vec{B} = \frac{E_0}{c}\cos(-kz-\omega t)\,(-\yhat)\) (since \(\hat{k}=-\zhat\), \(\hat{k}\times\xhat=-\yhat\)).

(b): \(\hat{k}=(\xhat+\yhat+\zhat)/\sqrt{3}\); \(k=\omega/c\); \(\vec{k}\cdot\vec{r}=k(x+y+z)/\sqrt{3}\); polarization \(\hat{e}=(\yhat-\zhat)/\sqrt{2}\). \[ \vec{E} = E_0\cos\!\left[\frac{\omega}{\sqrt{3}c}(x+y+z)-\omega t\right] \frac{\yhat-\zhat}{\sqrt{2}}. \] \(\vec{B} = \hat{k}\times\vec{E}/c\).

Concept Checks — Plane Waves, Intensity & Polarization

Concept check.

\(\vec{E}=E_0(\hat{x}+\hat{y})e^{i(kz-\omega t)}/\sqrt{2}\). Polarization: (a) Linear at 45^°   (b) Right circular   (c) Left circular   (d) Elliptical

Show answer

(a). Both components have the same phase: linear polarization at 45^° to \(x\)-axis.

Concept check.

\(\vec{E}=E_0(\hat{x}+i\hat{y})e^{i(kz-\omega t)}\). Represents: (a) Linear   (b) Right circular   (c) Left circular   (d) Elliptical

Show answer

(c). \(\mathrm{Re}\,\vec{E}=E_0[\cos(kz-\omega t)\hat{x}+\sin(\omega t-kz)\hat{y}]\). At fixed \(z\) the vector turns from \(\hat{x}\) towards \(\hat{y}\), i.e. anticlockwise when viewed facing the oncoming wave: left-circular in the optics convention (positive helicity; `right-handed' in the IEEE convention).

Energy and Momentum in EM Waves

Energy Density and Intensity

For a monochromatic plane wave: \begin{equation} u = \eps E^2 = \frac{B^2}{\muz} \quad\text{(electric and magnetic parts are equal)} \end{equation}

Time-averaged energy density: \begin{equation} \langle u\rangle = \frac{\eps}{2}E_0^2 \end{equation}

Intensity (time-averaged Poynting vector)

\begin{equation} I = \langle S\rangle = \frac{1}{2}\eps c E_0^2 = \frac{c\eps}{2}E_0^2 = \frac{E_0^2}{2\muz c} \end{equation}

EM Momentum and Radiation Pressure

Momentum density: \begin{equation} \vec{g} = \frac{\vec{S}}{c^2} = \eps(\vec{E}\times\vec{B}) \end{equation}

Radiation pressure

\begin{align} P_\text{abs} &= \frac{I}{c} \quad\text{(perfect absorber)} \\ P_\text{ref} &= \frac{2I}{c} \quad\text{(perfect reflector)} \end{align}

Wave impedance of free space

\begin{equation} Z_0 = \sqrt{\frac{\muz}{\eps}} = \muz c \approx 377\,\Omega \end{equation}

Example 4.3   Radiation pressure on a reflector

A laser beam of intensity \(I\) hits a perfect mirror at normal incidence. Find the radiation pressure and force on an area \(A\).

Momentum delivered per unit time per unit area (reflection): \(2I/c\). \[\boxed{P = \frac{2I}{c},\quad F = \frac{2IA}{c}}\]

Example 4.4   Electric field from intensity

A radio station radiates \(P=50\,\text{kW}\) isotropically. Find \(E_0\) at \(R=100\,\text{km}\).

\(I = P/(4\pi R^2) = 50\times10^3/(4\pi\times10^{10}) = 3.98\times10^{-7}\) W/m\(^2\).

\(E_0 = \sqrt{2I/(\eps c)} = \sqrt{2\times3.98\times10^{-7}/(8.854\times10^{-12}\times3\times10^8)} \approx 0.0173\) V/m.

\(B_0 = E_0/c \approx 5.77\times10^{-11}\) T.

Polarization

Types of polarization

Type

Condition

Description

Linear

\(E_y/E_x = \) const (real)

\(\vec{E}\) oscillates in a fixed plane

Circular

\(|E_x|=|E_y|\), phase diff \(=\pm\pi/2\)

\(\vec{E}\) traces a circle

Elliptical

General \(E_x\), \(E_y\), phase diff

\(\vec{E}\) traces an ellipse

Right-circular: \(\vec{E}=E_0(\xhat\cos\theta-\yhat\sin\theta)\) (rotates CW viewed toward source).

Identifying polarization from field expressions

Given \(\vec{E}=E_1\xhat\cos(kz-\omega t)+E_2\yhat\cos(kz-\omega t+\delta)\):

  • [noitemsep]

  • \(\delta=0\) or \(\pi\): linear

  • \(E_1=E_2\), \(\delta=\pm\pi/2\): circular

  • otherwise: elliptical

Example 4.5   Identifying polarization state

\(\vec{E} = E_0[\xhat\cos(kz-\omega t) + \yhat\cos(kz-\omega t+\pi/2)]\).

\(E_x = E_0\cos\phi\), \(E_y = E_0\cos(\phi+\pi/2) = -E_0\sin\phi\).

\(E_x^2+E_y^2 = E_0^2\): circle. Sense: as \(\phi\) increases, \((E_x,E_y)\) rotates counterclockwise \(\Rightarrow\) left-circular polarization.

EM Waves in Linear Dielectrics

In a linear, isotropic, homogeneous (LIH) non-conducting medium (\(\varepsilon_r\), \(\mu_r\), \(\sigma=0\)):

Wave speed and index of refraction in a medium

\begin{equation} v = \frac{c}{n},\qquad n = \sqrt{\varepsilon_r\mu_r} \approx \sqrt{\varepsilon_r} \quad(\mu_r\approx1\text{ for most media}) \end{equation} Energy density, Poynting vector, intensity: \begin{equation} u = \frac{1}{2}\varepsilon_r\eps E^2 + \frac{B^2}{2\muz\mu_r},\quad I = \frac{1}{2}\varepsilon_r\eps v E_0^2 = \frac{n}{2\muz c}E_0^2 \end{equation}

Example 4.6   Plane wave in a dielectric

A plane wave propagates in a lossless dielectric: \(\vec{E} = 10\cos(10^8 t - 0.5z)\,\xhat\) V/m.

From the form \(\cos(\omega t - kz)\): \(\omega=10^8\) rad/s, \(k=0.5\) rad/m.

Phase velocity: \(v=\omega/k = 2\times10^8\) m/s.

Refractive index: \(n=c/v=1.5\).

Relative permittivity: \(\varepsilon_r=n^2=2.25\).

Magnetic field: \(\vec{B} = (10/v)\cos(10^8t-0.5z)\,\yhat = 50\cos(10^8t-0.5z)\,\text{nT}\,\yhat\).

Poynting vector: \(\vec{S} = (E_0^2/2\muz v)\,\zhat \approx 132.6\,\text{mW/m}^2\,\zhat\).

Concept Checks — Reflection, Brewster & Critical Angle

Concept check.

Unpolarized light hits glass (\(n_2\)) from air at angle \(\arctan(n_2/n_1)\). Reflected light: (a) Unpolarized   (b) \(s\)-polarized   (c) \(p\)-polarized   (d) Circularly polarized

Show answer

(b). This is Brewster's angle: \(r_p=0\); reflected is purely \(s\)-polarized (perpendicular to plane of incidence).

EM Waves in Conductors

In a conductor with conductivity \(\sigma\), the modified wave equation gives a complex wave number \(\tilde{k} = k + i\kappa\):

Complex wave number in a conductor

\begin{align} k &= \omega\sqrt{\frac{\varepsilon_r\eps\muz}{2}} \left[\sqrt{1+\left(\frac{\sigma}{\varepsilon_r\eps\omega}\right)^2}+1\right]^{1/2}\\[6pt] \kappa &= \omega\sqrt{\frac{\varepsilon_r\eps\muz}{2}} \left[\sqrt{1+\left(\frac{\sigma}{\varepsilon_r\eps\omega}\right)^2}-1\right]^{1/2} \end{align} Wave form: \(\vec{E} = E_0\,e^{-\kappa z}\cos(kz-\omega t)\,\xhat\)

Skin depth

\begin{equation} \boxed{\delta_s = \frac{1}{\kappa} \approx \sqrt{\frac{2}{\sigma\muz\omega}}\quad(\text{good conductor: }\sigma\gg\varepsilon\omega)} \end{equation} The amplitude decays to \(1/e\) in one skin depth.

Good conductor limit (\(\sigma\gg\varepsilon_r\eps\omega\))

\begin{equation} k \approx \kappa \approx \sqrt{\frac{\muz\sigma\omega}{2}} = \frac{1}{\delta_s}, \qquad \tilde{n} = (1+i)\sqrt{\frac{\sigma}{2\muz\eps\omega^2}} \end{equation} \(\vec{B}\) lags \(\vec{E}\) by \(45\ensuremath{^\circ}\) inside the conductor.

Relaxation time in a conductor

\begin{equation} \tau = \frac{\varepsilon_r\eps}{\sigma} \end{equation} Free charge density decays as \(\rho(t)=\rho_0 e^{-t/\tau}\). For good conductors \(\tau\ll 1\) ps — charge distributes to surface almost instantly.

Example 4.7   Skin depth calculation

Find the skin depth in copper (\(\sigma=5.96\times10^7\) S/m) at \(f=60\) Hz and \(f=1\) GHz.

\(\delta_s = \sqrt{2/(\sigma\muz\omega)}\).

At 60 Hz: \(\omega=2\pi\times60=377\) rad/s. \(\delta_s = \sqrt{2/(5.96\times10^7\times4\pi\times10^{-7}\times377)} = 8.5\times10^{-3}\) m \(= 8.5\) mm.

At 1 GHz: \(\omega=2\pi\times10^9\). \(\delta_s = \sqrt{2/(5.96\times10^7\times4\pi\times10^{-7}\times2\pi\times10^9)} \approx 2.1\times10^{-6}\) m \(= 2.1\,\mu\)m. \[\boxed{\delta_s\propto1/\sqrt{f}}\]

Example 4.8   Phase lag in a conductor

An EM wave enters a conductor. The electric field inside is \(E=E_0e^{-z/\delta_s}\cos(z/\delta_s-\omega t)\). Find the phase of \(\vec{B}\) relative to \(\vec{E}\).

In a good conductor \(\tilde{k}=(1+i)/\delta_s\), so \(|\tilde{k}|e^{i\phi_k}\) where \(\phi_k=45\ensuremath{^\circ}\). Since \(\vec{B}=(\tilde{k}/\omega)\times\vec{E}\), \(\vec{B}\) lags \(\vec{E}\) by \(45\ensuremath{^\circ}\). \[\boxed{\vec{B}\text{ lags }\vec{E}\text{ by }45\ensuremath{^\circ}\text{ inside a good conductor.}}\]

Concept Checks — Waves in Conductors & Skin Depth

Concept check.

EM wave in conductor, \(\vec{E}=E_0 e^{-z/\delta_s}\cos(z/\delta_s-\omega t)\hat{x}\). Phase difference between \(\vec{B}\) and \(\vec{E}\): (a) \(45\ensuremath{^\circ}\), \(\vec{B}\) lags   (b) \(45\ensuremath{^\circ}\), \(\vec{E}\) lags   (c) \(90\ensuremath{^\circ}\), \(\vec{B}\) lags   (d) \(90\ensuremath{^\circ}\), \(\vec{E}\) lags

Show answer

(a). In a good conductor \(\tilde{k}=(1+i)/\delta_s\); the phase of \(\tilde{k}\) is \(45\ensuremath{^\circ}\), so \(\vec{B}\) lags \(\vec{E}\) by \(45\ensuremath{^\circ}\).

Reflection and Transmission at Normal Incidence

A plane wave hits a planar interface between two media (\(n_1\), \(n_2\)) at normal incidence. Matching boundary conditions (\(E_\parallel\) continuous, \(B_\parallel/\mu_r\) continuous):

Fresnel coefficients — Normal incidence

\begin{align} r &= \frac{E_{0R}}{E_{0I}} = \frac{n_1-n_2}{n_1+n_2} \\[6pt] t &= \frac{E_{0T}}{E_{0I}} = \frac{2n_1}{n_1+n_2} \end{align} Reflectance and transmittance: \begin{align} R &= r^2 = \left(\frac{n_1-n_2}{n_1+n_2}\right)^2 \\[6pt] T &= \frac{n_2}{n_1}\,t^2 = \frac{4n_1 n_2}{(n_1+n_2)^2}, \quad R+T=1 \end{align}

Example 4.9   Reflection from glass

Light travels in air (\(n_1=1\)) and hits glass (\(n_2=1.5\)) at normal incidence.

\(R = (1-1.5)^2/(1+1.5)^2 = 0.25/6.25 = 4\%\).

\(T = 1-R = 96\%\). \[\boxed{R=4\%,\quad T=96\%}\]

Example 4.10   Finding \(n\) from transmittance

Transmitted amplitude is \(2E_0/3\). The incident amplitude is \(E_0\) (normal incidence, air to medium). Find \(n\).

\(t=E_T/E_I = 2/(1+n) = 2/3 \Rightarrow 1+n=3 \Rightarrow n=2\).

Check: \(R=(1-2)^2/(1+2)^2=1/9\); \(T=(4\times1\times2)/(3)^2=8/9\). Power: \(R+T=1/9+8/9=1\). ✓ \[\boxed{n=2}\]

Reflection and Transmission at Oblique Incidence

Snell's Law

At any interface: \begin{equation} \boxed{n_1\sin\theta_i = n_2\sin\theta_t} \quad\text{(Snell's law)} \end{equation} The angle of reflection equals the angle of incidence: \(\theta_r=\theta_i\).

Fresnel Equations

Fresnel equations — Oblique incidence

TE (s-polarization): \(\vec{E}\perp\) plane of incidence \begin{align} r_s &= \frac{n_1\cos\theta_i - n_2\cos\theta_t} {n_1\cos\theta_i + n_2\cos\theta_t},\quad t_s = \frac{2n_1\cos\theta_i} {n_1\cos\theta_i + n_2\cos\theta_t} \end{align} TM (p-polarization): \(\vec{E}\parallel\) plane of incidence \begin{align} r_p &= \frac{n_2\cos\theta_i - n_1\cos\theta_t} {n_2\cos\theta_i + n_1\cos\theta_t},\quad t_p = \frac{2n_1\cos\theta_i} {n_2\cos\theta_i + n_1\cos\theta_t} \end{align}

Brewster's Angle

\(r_p=0\) when numerator vanishes: \(n_2\cos\theta_B = n_1\cos\theta_t\). Combined with Snell's law: \begin{equation} \boxed{\tan\theta_B = \frac{n_2}{n_1}} \end{equation} At Brewster's angle, reflected light is completely s-polarized. Also: \(\theta_B + \theta_t = 90\ensuremath{^\circ}\).

Total Internal Reflection

When \(n_1>n_2\), for \(\theta_i>\theta_c\): \begin{equation} \boxed{\sin\theta_c = \frac{n_2}{n_1}} \end{equation} All energy reflects back; transmitted wave is evanescent.

Example 4.11   Brewster's angle and critical angle

Air (\(n_1=1\)) to glass (\(n_2=1.5\)): \(\theta_B = \arctan(1.5/1) = 56.3\ensuremath{^\circ}\). Reflected light at this angle is purely s-polarized.

Glass (\(n_1=1.5\)) to air (\(n_2=1\)): \(\theta_c = \arcsin(1/1.5) = 41.8\ensuremath{^\circ}\). Beyond this angle, total internal reflection.

Example 4.12   Circularly polarized at Brewster's angle

Circularly polarized light hits a dielectric interface at Brewster's angle.

Circular polarization = equal TE and TM components with \(\pi/2\) phase difference. At Brewster's angle \(r_p=0\) so TM reflected amplitude \(=0\); TE is reflected. \[\boxed{\text{Reflected: purely s-polarized (TE). Transmitted: elliptically polarized.}}\]

EM Waves in Dispersive and Absorbing Media

Phase velocity and group velocity

\begin{equation} v_p = \frac{\omega}{k},\qquad v_g = \frac{d\omega}{dk} \end{equation} In a dispersive medium \(v_g\neq v_p\). Energy travels at \(v_g\).

Complex refractive index

\begin{equation} \tilde{n} = n_r + in_i \end{equation} \(n_r\) determines phase velocity; \(n_i\) determines absorption (skin depth). For a good conductor: \(n_r\approx n_i \approx \sqrt{\sigma/2\muz\eps\omega^2}\).

Example 4.13   Dispersion relation and group velocity

A wave has dispersion relation \(\omega^2 = \omega_0^2 + c^2k^2\) (e.g., plasma). Find \(v_p\) and \(v_g\).

\(v_p = \omega/k = \sqrt{\omega_0^2/k^2+c^2} > c\).

\(v_g = d\omega/dk = c^2k/\omega = c^2/v_p < c\).

\[\boxed{v_p v_g = c^2}\] Phase velocity exceeds \(c\) but group velocity (energy transport) does not.

Worked Examples

Example 4.14   Standing wave ratio

A normal-incidence wave produces a standing wave in front of an interface. The ratio of max to min electric field amplitude is 5. Find \(R\).

\(\text{SWR} = (1+\sqrt{R})/(1-\sqrt{R}) = 5 \Rightarrow \sqrt{R}=4/6=2/3 \Rightarrow R=4/9\). \[\boxed{R = \frac{4}{9} \approx 44\%}\]

Example 4.15   EM wave in a dielectric slab

A plane wave of wavelength \(\lambda_0\) passes through a dielectric slab (thickness \(d\), permittivity \(\varepsilon_r\)). Find the phase change compared to propagation in vacuum.

In slab: wavelength \(\lambda = \lambda_0/n = \lambda_0/\sqrt{\varepsilon_r}\); phase accumulated \(= 2\pi d/\lambda = 2\pi nd/\lambda_0\).

In vacuum: phase \(= 2\pi d/\lambda_0\).

\[\boxed{\Delta\phi = \frac{2\pi d}{\lambda_0}(n-1) = \frac{2\pi d(\sqrt{\varepsilon_r}-1)}{\lambda_0}}\]

Example 4.16   Poynting vector — proton beam

A narrow beam of protons moves in vacuum at velocity \(v\) along \(+\zhat\). Find the Poynting vector outside the beam at radial distance \(r\).

Linear charge density \(\lambda\); surface current \(K=\lambda v\).

\(E_r = \lambda/(2\pi\eps r)\) (Gauss); \(B_\phi = \muz\lambda v/(2\pi r)\) (Ampère).

\(\vec{S} = (E_r B_\phi/\muz)\,\zhat = \frac{\lambda^2 v}{4\pi^2\eps r^2}\,\zhat\).

Direction: \(+\zhat\) (along beam). Magnitude \(\propto 1/r^2\).

Example 4.17   Reflectance at oblique incidence — Brewster check

Unpolarized light in medium \(n_1=1\) hits \(n_2=2\) at \(\theta_i=\theta_B=\arctan 2\). At this angle \(r_p=0\), so only s-component reflects.

\(\cos\theta_B = 1/\sqrt{5}\); Snell: \(\sin\theta_t = \sin\theta_B/2 = 1/\sqrt{5}\), \(\cos\theta_t=2/\sqrt{5}\).

\(r_s = (1\cdot1/\sqrt{5}-2\cdot2/\sqrt{5})/(1\cdot1/\sqrt{5}+2\cdot2/\sqrt{5}) = (1-4)/(1+4) = -3/5\).

\(R_s = 9/25 = 36\%\); \(R_p=0\). For unpolarized light: \(R=18\%\).

Chapter Summary

EM Waves — Key Results

Quantity

Vacuum

Medium (\(n\))

Wave speed

\(c=1/\sqrt{\muz\eps}\)

\(v=c/n\)

\(\vec{E}\)-\(\vec{B}\) relation

\(B=E/c\)

\(B=nE/c\)

Intensity

\(\frac{\eps c}{2}E_0^2\)

\(\frac{n\eps c}{2}E_0^2\)

Impedance

\(Z_0=377\,\Omega\)

\(Z=Z_0/n\)

Normal incidence: \(r=(n_1-n_2)/(n_1+n_2)\); \(R=r^2\); \(T=1-R\)

Brewster's angle: \(\tan\theta_B=n_2/n_1\)

Critical angle: \(\sin\theta_c=n_2/n_1\) (\(n_1>n_2\))

Skin depth: \(\delta_s=\sqrt{2/\sigma\muz\omega}\) (good conductor)

Radiation pressure: \(I/c\) (absorbed); \(2I/c\) (reflected)

Polarization: Linear (\(\delta=0,\pi\)); Circular (\(|E_x|=|E_y|\), \(\delta=\pm\pi/2\)); Elliptical (general)

Previous-Year Questions

IIT-JAM Questions

Q1★★IIT JAM 2022

The electric field associated with an electromagnetic radiation is given by \(E=a\left(1+\cos\omega_1t\right)\cos\omega_2t\). Which of the following frequencies are present in the field?

  1. \(\omega_1\)

  2. \(\omega_1+\omega_2\)

  3. \(\left|\omega_1-\omega_2\right|\)

  4. \(\omega_2\)

Show solution

Answer: (b), (c) and (d) \(\boxed{\omega_2,\ \omega_1+\omega_2,\ \left|\omega_1-\omega_2\right|}\)

1. Expand the product. \[ E=a\cos\omega_2t+a\cos\omega_1t\cos\omega_2t \] The first term already contributes the frequency \(\omega_2\). (d) present.

2. Apply the product-to-sum identity. \[ \cos\omega_1t\cos\omega_2t =\tfrac12\left[\cos\left(\omega_1+\omega_2\right)t+\cos\left(\omega_1-\omega_2\right)t\right] \]

3. Read off the sidebands. This contributes \(\omega_1+\omega_2\) and \(\left|\omega_1-\omega_2\right|\). (b) and (c) present.

4. Why \(\omega_1\) itself is absent. The modulating factor \(\cos\omega_1t\) never appears alone — it is always multiplied by \(\cos\omega_2t\), which shifts it into the two sidebands. (a) is not present. This is standard amplitude modulation: a carrier at \(\omega_2\) with sidebands at \(\omega_2\pm\omega_1\), and no component at the modulation frequency itself.

Q2★★IIT JAM 2022

A current density for a fluid flow is given by \[ \vec{J}(x,y,z,t)=\frac{8e^{t}}{\left(1+x^{2}+y^{2}+z^{2}\right)}\hat{x} \] At time \(t=0\), the mass density \(\rho(x,y,z,0)=1\). Using the equation of continuity, \(\rho(1,1,1,1)\) is found to be (Round off to \(2\) decimal places.)

Show solution

Answer: \(\boxed{\rho(1,1,1,1)=e\approx2.72}\)

1. Write the continuity equation. \[ \frac{\partial\rho}{\partial t}=-\nabla\cdot\vec{J} \]

2. Take the divergence. Only the \(\hat{x}\) component exists, so \[ \nabla\cdot\vec{J}=\frac{\partial}{\partial x}\left[\frac{8e^{t}}{1+x^{2}+y^{2}+z^{2}}\right] =\frac{-16xe^{t}}{\left(1+x^{2}+y^{2}+z^{2}\right)^{2}} \]

3. Evaluate at the point \((1,1,1)\), where \(1+x^{2}+y^{2}+z^{2}=4\): \[ \nabla\cdot\vec{J}=\frac{-16e^{t}}{16}=-e^{t} \;\Longrightarrow\;\frac{\partial\rho}{\partial t}=e^{t} \]

4. Integrate from \(t=0\) to \(t=1\) with \(\rho(0)=1\): \[ \rho=1+\int_0^{1}e^{t}dt=1+\left(e-1\right)=e \]

Official accepted band

The official key accepts 2.70 to 2.74, and the exact value \(\rho=e=2.7183\) sits inside it. The band allows for candidates who round \(e\) at intermediate steps.

Q3★IIT JAM 2020

A collimated beam of laser light of wavelength \(514\) nm is normally incident on a smooth glass slab placed in air. Given the refractive indices of glass and air are \(1.47\) and \(1.0\) respectively, the percentage of light intensity reflected back is

  1. \(2.0\)

  2. \(4.0\)

  3. \(3.6\)

  4. \(4.2\)

Show solution

Answer: (c) \(\boxed{3.6\%}\)

1. Reflectance at normal incidence. \[ R=\left(\frac{n_2-n_1}{n_2+n_1}\right)^{2} \]

2. Substitute \(n_1=1.0\), \(n_2=1.47\). \[ R=\left(\frac{1.47-1.00}{1.47+1.00}\right)^{2}=\left(\frac{0.47}{2.47}\right)^{2} \]

3. Evaluate. \[ R=\left(0.19028\right)^{2}=0.036206 \]

4. Express as a percentage. \[ \boxed{R\approx3.6\%} \] The wavelength is a red herring — at normal incidence \(R\) depends only on the two indices. Note \(4\%\) (option (b)) is the familiar value for \(n=1.5\); the slightly lower index here gives \(3.6\%\).

Q4★IIT JAM 2024

A plane electromagnetic wave is incident on an interface AB separating two media (refractive indices \(n_1=1.5\) and \(n_2=2.0\)) at Brewster angle \(\theta_B\), as schematically shown in the figure. The angle \(\alpha\) (in degrees) between the reflected and the refracted wave is

  1. \(120\)

  2. \(116\)

  3. \(90\)

  4. \(74\)

Show solution

Answer: (c) \(\boxed{\alpha=90^\circ}\)

1. Recall what defines the Brewster angle. At \(\theta_B\) the reflected wave is completely polarized, and the defining geometric condition is that the reflected and refracted rays are

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