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 (a) $36 \times 10^{-5}$ kg m/s. (b) $36 \times 10^{-4}$ kg m/s. (c) $108 \times 10^{4}$ kg m/s. (d) $1.08 \times 10^{7}$ kg m/s.

Light with an energy flux of 20W/cm² falls on a non-reflecting surface at normal incidence. If the surface has

NCERT Exemplar · Q3

20\ \text{W/cm}^2

(a)

36 \times 10^{-5}

kg m/s.

(b)

36 \times 10^{-4}

kg m/s.

(c)

108 \times 10^{4}

kg m/s.

(d)

1.08 \times 10^{7}

kg m/s.

Gujarat GsebMCQ· 1mImportance★★★★★

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✓ Free question

Concept understanding — Electromagnetic Wave Relation

Electromagnetic Wave Relation: From Intuition to Precision

Imagine you're standing at the beach. You see a wave coming in — it has a certain speed, a certain distance between crests (wavelength), and a certain number of crests passing you per second (frequency). The faster the wave, the more crests pass you in a given time. That's the basic idea: speed = frequency × wavelength.

Now, light is also a wave — an electromagnetic wave. It doesn't need water or air; it travels through empty space at a staggering speed. The relation that governs all waves, including light, is:

v = f \lambda

where $v$ is the wave speed, $f$ is the frequency (in hertz, Hz), and $\lambda$ (lambda) is the wavelength (in metres).

For electromagnetic waves in vacuum, this speed is a universal constant: $c = 3 \times 10^8$ m/s. So the relation becomes:

c = f \lambda

That's it. But let's unpack what this really means.

What is frequency? What is wavelength?

Frequency is how many complete wave cycles pass a fixed point in one second. A radio station broadcasting at 100 MHz means 100 million cycles per second. Higher frequency means more oscillations per second.

Wavelength is the distance between two consecutive crests (or troughs) of the wave. For visible light, wavelengths are tiny — around 400 to 700 nanometres (billionths of a metre).

The product $f \lambda$ always equals the wave speed. So if frequency goes up, wavelength must go down to keep the product constant. This is why:

Gamma rays have extremely high frequency and extremely short wavelength.
Radio waves have low frequency and very long wavelength (metres to kilometres).

Both travel at the same speed $c$ in vacuum.

Why does this matter for exams?

You'll use this relation in three main ways:

Given frequency, find wavelength (or vice versa) — just rearrange: $\lambda = \frac{c}{f}$ or $f = \frac{c}{\lambda}$ .
Compare different regions of the electromagnetic spectrum — know that as frequency increases, wavelength decreases proportionally.
Solve problems involving energy — because photon energy $E = hf$ (where $h$ is Planck's constant), the wave relation links energy to wavelength: $E = \frac{hc}{\lambda}$ .

Watch out

A common mistake: using $c = f \lambda$ for waves in a medium (like glass or water). In a medium, the speed is less than $c$ , so the wavelength changes but frequency stays the same. The relation $v = f \lambda$ still holds, but $v$ is now the speed in that medium.

A concrete example

A microwave oven operates at 2.45 GHz. What is its wavelength in vacuum?

$f = 2.45 \times 10^9$ Hz, $c = 3 \times 10^8$ m/s.

\lambda = \frac{c}{f} = \frac{3 \times 10^8}{2.45 \times 10^9} = 0.122 \text{ m} = 12.2 \text{ cm}

That's why the mesh on a microwave door has holes about 1–2 mm across — much smaller than 12 cm — so microwaves can't escape, but visible light (wavelength ~500 nm) passes through easily.

The big picture

The electromagnetic wave relation $c = f \lambda$ is not a deep law of nature — it's a definitional consequence of what frequency and wavelength mean. But it's the single most useful tool for navigating the electromagnetic spectrum. Memorise it, understand it, and you'll be able to connect wave properties to energy, to colour, to radiation types, and to countless exam problems.

$c = f \lambda$ — that's the relation. Everything else is just applying it.

The relation c = fλ connecting frequency and wavelength across the electromagnetic spectrum is introduced in the NCERT Class 12 Physics chapter on electromagnetic waves, tested in CBSE boards, JEE Main and NEET. Anyone searching "electromagnetic spectrum frequency wavelength relation class 12 physics" will find this wave-speed reasoning, including the medium-versus-vacuum distinction, matches the NCERT treatment.

Why this formula?

Electromagnetic Wave Relation: Why $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

Let's build this from first principles — not just memorising the formula, but understanding why light and all EM waves travel at this specific speed.

1. The Starting Point: Maxwell's Equations in Vacuum

In empty space (no charges, no currents), Maxwell's equations simplify to:

Gauss's law for electricity: $\nabla \cdot \mathbf{E} = 0$
Gauss's law for magnetism: $\nabla \cdot \mathbf{B} = 0$
Faraday's law: $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$
Ampère-Maxwell law: $\nabla \times \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$

The key insight: a changing electric field creates a magnetic field, and a changing magnetic field creates an electric field. This mutual induction is what sustains the wave.

2. Deriving the Wave Equation for $\mathbf{E}$

Take the curl of Faraday's law:

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

Now use the vector identity: $\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}$

Since $\nabla \cdot \mathbf{E} = 0$ in vacuum, this becomes:

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

Substitute $\nabla \times \mathbf{B}$ from Ampère-Maxwell:

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

Result: The electric field satisfies the wave equation:

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

3. Identifying the Wave Speed

Compare with the standard wave equation for any wave travelling at speed $v$ :

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

Matching terms:

\frac{1}{v^2} = \mu_0 \varepsilon_0 \quad \Rightarrow \quad v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}

This $v$ is the speed of electromagnetic waves in vacuum — denoted $c$ .

Why this is profound: The constants $\mu_0$ (permeability of free space) and $\varepsilon_0$ (permittivity of free space) come from static electricity and magnetism. Yet their combination gives the speed of light — showing light is an electromagnetic wave.

4. The Magnetic Field Follows Suit

Exactly the same derivation starting from Ampère-Maxwell law gives:

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

So both $\mathbf{E}$ and $\mathbf{B}$ propagate at the same speed $c$ .

5. The Crucial Relationship Between $\mathbf{E}$ and $\mathbf{B}$

For a plane wave travelling in the $x$ -direction:

$\mathbf{E}$ oscillates along $y$ : $E_y = E_0 \sin(kx - \omega t)$
$\mathbf{B}$ oscillates along $z$ : $B_z = B_0 \sin(kx - \omega t)$

From Faraday's law: $\frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{\partial t}$

Differentiating the wave forms:

k E_0 \cos(kx - \omega t) = \omega B_0 \cos(kx - \omega t)

Since $\omega = ck$ , we get:

\frac{E_0}{B_0} = \frac{\omega}{k} = c

Key result: In vacuum, the magnitudes are related by:

\boxed{E = cB}

This means:

$\mathbf{E}$ and $\mathbf{B}$ are perpendicular to each other and to the direction of propagation
They are in phase (peaks and zeros occur together)
The electric field is $c$ times stronger than the magnetic field in SI units

6. Physical Intuition: Why This Speed?

Think of it this way:

$\mu_0$ measures how strongly a current creates a magnetic field
$\varepsilon_0$ measures how strongly a charge creates an electric field
Their product in the denominator means: the more "reluctant" space is to create fields, the slower the wave

If space were more "magnetic" (larger $\mu_0$ ) or more "electric" (larger $\varepsilon_0$ ), EM waves would travel slower. The actual value $c \approx 3 \times 10^8$ m/s emerges from the measured values of these constants.

Summary: The Core Relations

Quantity	Formula	Why
Wave speed	$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$	From wave equation derived from Maxwell's equations
Field ratio	$E = cB$	From Faraday's law applied to plane waves
Direction	$\mathbf{E} \perp \mathbf{B} \perp$ propagation	From cross-product structure of Maxwell's equations

Exam tip: Never just quote $c = 1/\sqrt{\mu_0\varepsilon_0}$ — be ready to show it comes from taking curls of Maxwell's equations and identifying the wave equation form.

The momentum delivered by an electromagnetic wave is found using $p = \frac{U}{c}$ , where $U$ is the total energy absorbed. Here, $U = \text{power} \times \text{time} = (20\ \text{W/cm}^2 \times 30\ \text{cm}^2) \times (30 \times 60\ \text{s}) = 1.08 \times 10^6\ \text{J}$ , so $p = \frac{1.08 \times 10^6}{3 \times 10^8} = 3.6 \times 10^{-3}\ \text{kg m/s}$ . The final answer is $\boxed{3.6 \times 10^{-3}\ \text{kg m/s}}$ .

The key idea is that light carries momentum, and when it is completely absorbed by a surface, that momentum is transferred to the surface. For electromagnetic waves, the momentum per unit energy is a fixed constant: $p = \frac{U}{c}$ for complete absorption. This relation comes directly from Maxwell’s equations and the fact that the energy flux (Poynting vector) and momentum density are linked by $c$ .

Why does this work without needing to know the wavelength or frequency? Because the momentum of a photon is $p = \frac{h}{\lambda}$ , and its energy is $E = \frac{hc}{\lambda}$ , so $p = \frac{E}{c}$ for each photon. Summing over all photons gives the same relation for the total energy and total momentum. So the problem reduces to finding the total energy absorbed over the given time.

Let’s go step by step.

Find the power incident on the surface. The energy flux (intensity) is $20\ \text{W/cm}^2$ . This means each square centimetre receives $20\ \text{J}$ of energy per second. The surface area is $30\ \text{cm}^2$ , so the total power (energy per second) incident is:

P = \text{intensity} \times \text{area} = 20\ \frac{\text{W}}{\text{cm}^2} \times 30\ \text{cm}^2 = 600\ \text{W}.

(Recall $1\ \text{W} = 1\ \text{J/s}$ .)

Calculate the total energy absorbed over 30 minutes. Time $t = 30\ \text{minutes} = 30 \times 60\ \text{s} = 1800\ \text{s}$ . Since the surface is non-reflecting (complete absorption), all incident energy is absorbed:

U = P \times t = 600\ \text{W} \times 1800\ \text{s} = 1.08 \times 10^6\ \text{J}.

Apply the momentum-energy relation for complete absorption. For an electromagnetic wave that is completely absorbed, the momentum delivered is:

p = \frac{U}{c},

where $c = 3 \times 10^8\ \text{m/s}$ is the speed of light.

Substituting:

p = \frac{1.08 \times 10^6}{3 \times 10^8} = 3.6 \times 10^{-3}\ \text{kg m/s}.

Tip

If the surface were perfectly reflecting, the momentum delivered would be twice as large ( $p = 2U/c$ ) because the wave reverses direction, giving a change in momentum of $2p_{\text{photon}}$ per photon. Here, since it’s non-reflecting (absorbing), we use the single factor.

Watch out

A common mistake is to forget to convert minutes to seconds, or to use the area incorrectly (e.g., multiplying intensity by time directly without area). Always check units: flux is per cm² per second, so area and time must both be accounted for.

✓Final answer

\boxed{3.6 \times 10^{-3}\ \text{kg m/s}}

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

GUJCET 2026Set x1 markMCQ
$8 \times 10^9$
›Reveal solutionSolution
[!TLDR]

Using the NCERT/CBSE list of insect-resistant crop varieties: $X$ = flat bean, $Y$ = Pusa A-4, $Z$ = Aphids.

Concept

Under 'Strategies for Enhancement in Food Production', plant breeders release crop varieties resistant to specific insect pests. A standard table pairs each crop with its released variety and the pest it resists. This GSEB/NCERT-aligned question tests recall of three of those pairings.

Solution

Row (i): Rape-seed mustard, variety Pusa Gaurav. This mustard variety is bred for resistance to the mustard aphid, so $Z$ = Aphids.

Row (ii): variety Pusa Sem 2, resisting Jassids. 'Pusa Sem' varieties belong to flat bean, so $X$ = flat bean.

Row (iii): Okra (bhindi) resisting shoot borer. The released okra variety here is Pusa A-4 (resistant to shoot and fruit borer), so $Y$ = Pusa A-4.

Combining: $X$ = flat bean, $Y$ = Pusa A-4, $Z$ = Aphids.

[!ANSWER]

(A) $X$ - flat bean, $Y$ - Pusa A-4, $Z$ - Aphids

Note

This solution was worked out by our team and independently cross-checked by a second solve. The official answer key on record for this question could not be confirmed, so please cross-verify with the official paper where possible.
GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. The amplitude of the magnetic field of Electromagnetic wave is B_0 = 510 nT, then amplitude of electric field of Electromagnetic wave is E_0 = ___.$
(a) 143 V/m
(b) 153 V/m
(c) 135 V/m
(d) 170 V/m
›Reveal solutionSolution

In an electromagnetic wave, the electric and magnetic field amplitudes are related by E_0 = c B_0, where c is the speed of light.

Given B_0 = 510 nT = 510 x 10^-9 T, c = 3 x 10^8 m/s.

E_0 = c B_0 = 3 x 10^8 x 510 x 10^-9 = 1530 x 10^-1 = 153 V/m.

✓Final answer
$(b) 153 V/m.$
GUJCET 2023Set 091 markMCQ
$\vec{E}$
›Reveal solutionSolution

The Poynting direction $\vec E\times\vec B$ gives the wave's propagation direction.

Concept: In an electromagnetic wave, $\vec E$ , $\vec B$ , and the propagation direction form a right-handed triad, with propagation along $\vec E\times\vec B$ .

✓Final answer
$\vec E\times\vec B$

ANSWER: (D)
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. For a given electromagnetic waves the magnitude of electric field is 6.6 V/m at a point in space. The magnitude of magnetic field at this point is ___ T.$
(a) 2.1 x 10^-8
(b) 6.6 x 10^-8
(c) 19.8 x 10^-8
(d) 2.2 x 10^-8
›Reveal solutionSolution

In an EM wave the field magnitudes obey E = cB, so B = E/c = 6.6/(3x10^8) = 2.2 x 10^-8 T.

In an electromagnetic wave the electric and magnetic field amplitudes are related by E = c B.

So B = E/c = 6.6/(3 x 10^8) = 2.2 x 10^-8 T.

✓Final answer
$(d) 2.2 x 10^-8 T.$
GUJCET 2022Set 171 markMCQ
$c = 3 \times 10^8$
›Reveal solutionSolution
$\lambda = c/f$ ; higher frequency gives shorter wavelength.

Steps.

At $f = 6$ MHz: $\lambda = \dfrac{3\times10^8}{6\times10^6} = 50$ m.

At $f = 12$ MHz: $\lambda = \dfrac{3\times10^8}{12\times10^6} = 25$ m.

Band: 25 m to 50 m.

✓Final answer
$(B) 25 m to 50 m$

ANSWER: (B)
GUJCET 2022Set 171 markMCQ
$10^9$
›Reveal solutionSolution
An oscillating charge radiates EM waves at exactly its own oscillation frequency.

Concept. An accelerating/oscillating charge produces electromagnetic waves whose frequency equals the frequency of oscillation of the charge.

Given oscillation frequency $= 10^9$ Hz $\Rightarrow$ EM wave frequency $= 10^9$ Hz.

✓Final answer
$10^9$

ANSWER: (B)
GUJCET 2021Set 151 markMCQ
$\vec{B} = 2.1 \times 10^{-8}\,\hat{k}$
›Reveal solutionSolution

For an EM wave $E=cB$ , with $\vec E,\vec B$ and the propagation direction mutually perpendicular ( $\vec E\times\vec B$ points along propagation).

Concept. $E=cB$ and $\hat E\times\hat B=\hat{\text{propagation}}$ .

Solution. $E=cB=(3\times10^{8})(2.1\times10^{-8})=6.3$ V/m. Wave goes along $+\hat i$ ; with $\vec B\parallel\hat k$ , we need $\hat E\times\hat k=\hat i$ , i.e. $\hat E=\hat j$ . So $\vec E=6.3\,\hat j$ V/m.

✓Final answer
$6.3\,\hat{j}\,\frac{V}{m}$

ANSWER: (B)
GUJCET 2019Set 131 markMCQ
$E$
›Reveal solutionSolution
Far from the source, radiated $E$ and $B$ are in phase and decrease as $\dfrac{1}{r}$ .

Concept: The radiation (far) field of an accelerating charge dominates at large distances because it decays only as $1/r$ , unlike the static ( $1/r^2$ ) or induction ( $1/r^3$ ) terms. This slow decay is why radiated energy reaches far away.

Steps:

Near-field terms $\propto r^{-2}, r^{-3}$ die rapidly.

The radiation term $\propto r^{-1}$ survives, so $E,B\propto r^{-1}$ .

✓Final answer
$r^{-1}$

ANSWER: (D)
GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. The maximum value of E in an electromagnetic wave is equal to 1.8 Vm^-1. Thus the maximum value of B is ___.$
(a) 6 x 10^-8 T
(b) 3 x 10^-6 T
(c) 6 x 10^-9 T
(d) 2 x 10^-10 T
›Reveal solutionSolution

In an electromagnetic wave, the peak electric and magnetic fields are related by $B_0=E_0/c$ .

Given $E_0=1.8$ V/m, $c=3\times10^8$ m/s.

$B_0=\dfrac{E_0}{c}=\dfrac{1.8}{3\times10^8}=0.6\times10^{-8}=6\times10^{-9}$ T.

✓Final answer
$6\times10^{-9}$
GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. For a radiation of 6 GHz passing through air, the wave number (number of waves) per 1 m length is ___ (1 GHz = 10^9 Hz).$
(a) 5
(b) 3
(c) 20
(d) 30
›Reveal solutionSolution

The number of complete waves per unit length (wave number in this sense) equals $f/c$ , the reciprocal of wavelength.

Given $f=6$ GHz $=6\times10^9$ Hz, $c=3\times10^8$ m/s.

$\lambda=c/f=\dfrac{3\times10^8}{6\times10^9}=0.05$ m.

Number of waves in 1 m $=1/\lambda=1/0.05=20$ .

✓Final answer
$(c) 20.$
GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. In the region closer to the oscillating charges, the phase difference between E (vector) and B (vector) fields is ___ and their magnitude quickly decreases as ___ with distance r from the source.$
(a) 0, r^-1
(b) pi/2, r^-3
(c) pi/2, r^-1
(d) 0, r^-3
›Reveal solutionSolution
Close to the oscillating charges (the near field), E and B are pi/2 out of phase and their amplitudes decrease steeply, as r^-3.

Near an oscillating charge (the induction/near-field zone, distances small compared with the wavelength):

The electric and magnetic fields are out of phase by pi/2 (90 degree).

Their magnitudes fall off quickly with distance, as r^-3.

(Only in the far radiation zone do E and B become in phase and decrease as 1/r, carrying energy away as the radiated wave.)

✓Final answer
$(b) pi/2, r^-3.$
GUJCET 2015Set C1 markMCQ
$Q. To transmit a signal of 3 KHz frequency, the minimum length of antenna is _____ km (A) 25 (B) 20 (C) 50 (D) 75$
›Reveal solutionSolution
[!TLDR]

$\lambda = c/f = 100$ km; minimum antenna length $= \lambda/4 = 25$ km. Answer: (A).

Concept

To radiate a signal efficiently, an antenna should have a length of at least about a quarter of the signal wavelength, $L_{min} = \lambda/4$ , where $\lambda = c/f$ (NCERT/CBSE communication systems).

Solution

Wavelength of the 3 kHz signal:
$\lambda = \frac{c}{f} = \frac{3\times10^8}{3\times10^3} = 10^5\,\text{m} = 100\,\text{km}$
Minimum antenna length:
$L_{min} = \frac{\lambda}{4} = \frac{100}{4} = 25\,\text{km}$
[!ANSWER]

(A) 25.

←Previous questionQ2 — A linearly polarized electromagnetic wave given as mathbfE = E₀haticos…→Next questionQ4 — The electric field intensity produced by the radiations coming from a …

Q.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

Concept understanding — Electromagnetic Wave Relation

Electromagnetic Wave Relation: From Intuition to Precision

What is frequency? What is wavelength?

Why does this matter for exams?

A concrete example

The big picture

Why this formula?

Electromagnetic Wave Relation: Why $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

1. The Starting Point: Maxwell's Equations in Vacuum

2. Deriving the Wave Equation for $\mathbf{E}$

3. Identifying the Wave Speed

4. The Magnetic Field Follows Suit

5. The Crucial Relationship Between $\mathbf{E}$ and $\mathbf{B}$

6. Physical Intuition: Why This Speed?

Summary: The Core Relations

Method: Momentum Delivered by an Absorbed or Reflected EM Wave

Steps

Concept

Solution

Concept

Solution

Q.Light with an energy flux of 20 W/cm220\ \text{W/cm}^220 W/cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm230\ \text{cm}^230 cm2, the total momentum delivered (for complete absorption) during 30 minutes is

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Q.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

Unlock everything free for 14 days