A charged particle oscillates about its mean equilibrium position with a frequency of $10^9\ \text{Hz}$. The electromagnetic waves produced: (a) will have frequency of $10^9$ Hz. (b) will have frequency of $2 \times 10^9$ Hz. (c) will have a wavelength of 0.3 m. (d) fall in the region of radiowaves.

A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹Hz. The electromagnet

NCERT Exemplar · Q30

10^9\ \text{Hz}

(a) will have frequency of

10^9

Hz.

(b) will have frequency of

2 \times 10^9

Hz.

(d) fall in the region of radiowaves.

CBSEMCQ· 1mImportance★★★★★

91% · 42/46 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Speed Of Light

Speed of Light

What It Is

The speed of light is the speed at which light — and every other electromagnetic wave — travels through empty space (vacuum). It is one of the most important constants in physics, denoted $c$ :

c \approx 3 \times 10^8 \ \text{m/s} = 3\times10^{5}\ \text{km/s}

More precisely $c = 2.998 \times 10^8\ \text{m/s}$ . In one second light travels about 300,000 km — roughly seven and a half times around the Earth.

Where the Value Comes From (Maxwell)

The speed of light is not just measured; it is predicted by Maxwell's equations. When Maxwell combined his laws of electricity and magnetism, he found that electromagnetic waves must travel through vacuum at a speed fixed entirely by two constants of free space:

c = \frac{1}{\sqrt{\mu_0\,\varepsilon_0}}

where

$\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space, and
$\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1}$ is the permeability of free space.

Plugging in these numbers gives $c \approx 3\times10^8$ m/s — matching the measured speed of light. This agreement was the decisive clue that light itself is an electromagnetic wave.

Key Properties

Same for all electromagnetic waves. Radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays all travel at $c$ in vacuum, regardless of their frequency or wavelength.
Universal constant. In vacuum, $c$ is the same for every observer and does not depend on the motion of the source — the starting postulate of Einstein's special relativity.
The cosmic speed limit. No material object or signal carrying information can travel faster than $c$ .
Links wavelength and frequency. For any EM wave in vacuum,

c = f\lambda

so a high-frequency wave has a short wavelength and vice versa.

Speed of Light in a Medium

Inside a transparent material (glass, water, etc.) light slows down. Its speed becomes

v = \frac{c}{n} = \frac{1}{\sqrt{\mu\,\varepsilon}}

where $n = \sqrt{\mu_r\varepsilon_r}$ is the refractive index of the medium and is always greater than 1. For example, in water $n \approx 1.33$ , so light travels at about $2.25\times10^8$ m/s. The frequency stays the same, but the wavelength shortens because $v = f\lambda$ . …

Why this formula?

Speed of Light: Why the Formula Holds

The speed of light ( $c$ ) is not just a number — it emerges from the fundamental laws of electricity and magnetism. Let's understand why its value is fixed and where the formula comes from.

1. The Core Formula

The speed of light in vacuum is given by:

c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}

Where:

$\mu_0$ = permeability of free space (how easily a magnetic field forms)
$\varepsilon_0$ = permittivity of free space (how easily an electric field forms)

2. Why This Formula? — The Derivation

Step 1: Maxwell's Equations

James Clerk Maxwell unified electricity and magnetism into four equations. Two key ones for light:

Faraday's Law: A changing magnetic field creates an electric field

\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}

Ampère's Law (with Maxwell's correction): A changing electric field creates a magnetic field

\nabla \times \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}

Step 2: The Wave Emerges

Take the curl of Faraday's Law:

\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B})

Using the vector identity $\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}$ and noting that in vacuum $\nabla \cdot \mathbf{E} = 0$ , we get:

-\nabla^2 \mathbf{E} = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B})

Now substitute Ampère's Law for $\nabla \times \mathbf{B}$ :

-\nabla^2 \mathbf{E} = -\frac{\partial}{\partial t} \left( \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} \right)

Step 3: The Wave Equation

This simplifies to:

\nabla^2 \mathbf{E} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}

This is the wave equation. For any wave, the general form is:

\nabla^2 \mathbf{E} = \frac{1}{v^2} \frac{\partial^2 \mathbf{E}}{\partial t^2}

Comparing the two, the wave speed $v$ must satisfy:

\frac{1}{v^2} = \mu_0 \varepsilon_0

Hence:

v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}

This $v$ is the speed of light — denoted $c$ .

3. Why Is It Constant?

$\mu_0$ and $\varepsilon_0$ are fundamental constants of nature — they don't depend on the observer or the source.
Therefore, $c$ is also a universal constant. …

Start your 14-day free trial to unlock the full solution →

An oscillating charge radiates EM waves at exactly its own oscillation frequency, so with $f=10^9$ Hz: (a) is true, (b) [doubled frequency] is false, (c) [ $\lambda=0.3$ m] is true, and (d) [radio-wave region] is true. The correct options are (a), (c), and (d).

The governing principle

An oscillating charged particle is an accelerating charge, and Maxwell's theory shows that an accelerating charge radiates electromagnetic waves at exactly the same frequency as its own oscillation -- the emitted wave's electric and magnetic fields vary in step with the source's motion. There is no classical mechanism by which a simply-oscillating dipole charge would radiate at double (or any other multiple of) its own frequency.

Checking each option

(a) Frequency $=10^9$ Hz. Since the emitted wave's frequency equals the source's oscillation frequency, $f=10^9$ Hz. TRUE.

(b) Frequency $=2\times10^9$ Hz. This would require a frequency-doubling mechanism (e.g. a nonlinear process), which doesn't apply to a simple oscillating charge. FALSE.

(c) Wavelength $=0.3$ m. In vacuum, $c=f\lambda$ , so

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

TRUE. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$c=\dfrac{1}{\sqrt{\mu_0\epsilon_0}}$
(i) Both A and R are correct and R is correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) A is correct but R is incorrect.
(iv) Both A and R are incorrect.
›Reveal solutionSolution

Both statements are true, and the formula in R is exactly why speed is wavelength-independent.

Assertion is correct: in vacuum every electromagnetic wave, regardless of wavelength, travels at the same speed $c$ (vacuum is non-dispersive for EM waves). Reason is also correct: $c=\dfrac{1}{\sqrt{\mu_0\epsilon_0}}$ , and since $\mu_0$ and $\epsilon_0$ are constants of free space (not functions of wavelength or frequency), t …

Start your 14-day free trial to see the full solution →
CBSE 2025Set 55/5/11 markMCQ
$(\mu\varepsilon)^{-1}$
›Reveal solutionSolution

The product $\mu\varepsilon$ has dimensions of $[T^2 L^{-2}]$ , so $(\mu\varepsilon)^{-1}$ has dimensions of velocity squared: $[L^2 T^{-2}]$ . The answer is (B).

The key insight here is recognizing what $\mu$ and $\varepsilon$ represent physically and how they combine in electromagnetic theory. The speed of light (or electromagnetic waves) in a medium is given by $c = \frac{1}{\sqrt{\mu\varepsilon}}$ , which immediately tells us that $\mu\varepsilon$ must have dimensions of inverse velocity squared.

Let me work through the dimensional analysis systematically.

Finding the dimensions of permittivity $\varepsilon$ :

Permittivity appears in Coulomb's law and the relation for electric field. From the capacitance of a parallel-plate capacitor:
$C = \frac{\varepsilon A}{d}$
where $C$ is capacitance, $A$ is area, and $d$ is separation. Since capacitance has dimensions $[Q^2 T^2 M^{-1} L^{-2}]$ (from $C = Q/V$ and energy $U = \frac{1}{2}CV^2$ ), we get:
$[\varepsilon] = \frac{[C][L]}{[L^2]} = [M^{-1} L^{-3} T^4 A^2]$
In SI base units with charge $Q = AT$ , this becomes $[\varepsilon] = [M^{-1} L^{-3} T^4 A^2]$ .

Finding the dimensions of permeability $\mu$ :

Permeability appears in Ampère's law. The magnetic field around a wire is $B = \frac{\mu I}{2\pi r}$ , giving:

$[\mu] = \frac{[B][L]}{[A]}$ …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The speed of electromagnetic waves in vacuum is given by the equation$
(a) c = sqrt(mu0 * epsilon0)
(b) c = 1 / sqrt(mu0 * epsilon0)
(c) c = sqrt(mu0 / epsilon0)
(d) c = sqrt(epsilon0 / mu0)
›Reveal solutionSolution

Maxwell showed that combining the wave equations for E and B derived from his equations in free space gives a wave speed c = 1/sqrt(mu0 epsilon0), which matched the known speed of light and led him to conclude light itself is an electromagnetic wave.

From Maxwell's equations in vacuum, the electric and magnetic fields each satisfy a wave equation with speed:

c = 1 / sqrt(mu0 epsilon0)

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{1}{\mu_0 \varepsilon_0}$
(a) $[L^{-1}T]$
(b) $[LT^{-1}]$
(c) $[L^{-2}T^2]$
(d) $[L^2T^{-2}]$
›Reveal solutionSolution

Since $1/\sqrt{\mu_0\varepsilon_0}=c$ (the speed of light), $1/(\mu_0\varepsilon_0)=c^2$ , whose dimension is $[L^2T^{-2}]$ .

Working

Maxwell's relation gives the speed of electromagnetic waves in vacuum as

$c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$

Squaring both sides:

$c^2 = \dfrac{1}{\mu_0\varepsilon_0}$

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set F1 markMCQ
$Q. Light year is equal to (A) 9.46 x 10^15 m (B) 9.46 x 10^12 m (C) 9.46 x 10^8 m (D) 9.46 x 10^10 m$
›Reveal solutionSolution

1 light year = c × (1 year) ≈ 3×10⁸ m/s × 3.15×10⁷ s ≈ 9.46 × 10¹⁵ m.

A light year is the distance travelled by light (an electromagnetic wave) in vacuum in one year.

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 mark
$Q. The velocity of electromagnetic waves in free space can be given by the relation ............. .$
›Reveal solutionSolution

Maxwell's equations predict that electromagnetic waves travel in free space at a speed fixed by the permeability and permittivity of free space.

From Maxwell's electromagnetic wave equation, the speed of an electromagnetic wave in free space (vacuum) is

$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ …

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. When medium of electromagnetic waves changes from air to water, their speed -$
(a) increases
(b) decreases
(c) remains same
(d) may increase or decrease
›Reveal solutionSolution

Speed of light in a medium is $v=c/n$ ; water has $n>1$ , so speed decreases relative to air.

The speed of an electromagnetic wave in a medium is $v = \dfrac{c}{n}$ , where $n$ is the refractive index of that medium. Water has a refractive index greater than that of air ( $n_{water}\approx1.33 > n_{air}\approx1$ ), so the speed of th …

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$Q. Relation between velocity of light (c), permeability of free space (μ₀), permittivity of free space (ε₀) is ____.$
(a) C = 1/(μ₀ε₀)
(b) C = 1/√(μ₀ε₀)
(c) C = μ₀ε₀
(d) C = √(μ₀ε₀)
›Reveal solutionSolution

Maxwell showed that the speed of light equals $1/\sqrt{\mu_0\varepsilon_0}$ , one of the great unifications of physics.

…

Start your 14-day free trial to see the full solution →
CBSE 2021Set A1 markMCQ
$Q. The unit of which physical quantity is light year? (A) Distance (B) Time (C) Energy (D) Intensity of light$
›Reveal solutionSolution

A light year is a unit of distance, not time.

Despite the word 'year', a light year is the distance travelled by light (speed c ≈ 3 × 10⁸ m/s) in one year:

$1\ \text{light year} = c \times (1\ \text{year}) \approx 9.46 \times 10^{15}\ \text{m}$

…

Start your 14-day free trial to see the full solution →
CBSE 2021Set ANNUAL1 markMCQ
$Q. The velocity of electromagnetic waves in free space is given by -$
(a) √(ε₀μ₀)
(b) 1/√(ε₀μ₀)
(c) ε₀/μ₀
(d) √(ε₀/μ₀)
›Reveal solutionSolution

The speed of an EM wave in free space comes directly from Maxwell's equations and equals $c = 1/\sqrt{\varepsilon_0\mu_0}$ .

Maxwell's equations predict electromagnetic waves travelling in vacuum with speed:

$v = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$ …

Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 markMCQ
$Q. If Vg, Vx and Vm are the speeds of gamma rays, X-rays and microwaves respectively in vacuum, then$
(a) Vg = Vx = Vm
(b) Vg > Vx > Vm
(c) Vg > Vx < Vm
(d) Vg < Vx < Vm
›Reveal solutionSolution

Gamma rays, X-rays and microwaves are all part of the same electromagnetic spectrum, and every electromagnetic wave travels at the speed of light c in vacuum, independent of its frequency.

Gamma rays, X-rays, and microwaves differ enormously in frequency and wavelength (gamma rays have the shortest wavelength/highest frequency, microwaves the longest wavelength/lowest frequency among these three), but they are all transverse electromagnetic waves consisting of oscillating electric and magnetic fields. A defining property of electromagnetic waves derived from Maxwell's equations is that in vacuum they all propagate at the same universal speed, …

Start your 14-day free trial to see the full solution →

←Previous questionQ29 — A plane electromagnetic wave propagating along x-direction can have th…→Next questionQ31 — The source of electromagnetic waves can be a charge (a) moving with a …

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

Concept understanding — Speed Of Light

Speed of Light

What It Is

Where the Value Comes From (Maxwell)

Key Properties

Speed of Light in a Medium

Why this formula?

Speed of Light: Why the Formula Holds

1. The Core Formula

2. Why This Formula? — The Derivation

Step 1: Maxwell's Equations

Step 2: The Wave Emerges

Step 3: The Wave Equation

3. Why Is It Constant?

The governing principle

Checking each option

Method: Relating an Oscillating Source's Frequency to the Emitted EM Wave's Frequency and Wavelength

Steps

Q.A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz10^9\ \text{Hz}109 Hz. The electromagnetic waves produced:

The governing principle

Checking each option

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days