What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam t

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Q. What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of a room, its intensity essentially remains constant. What geometrical characteristic of LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?

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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. …

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:

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

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The intensity of a bulb’s light falls to one-fourth when distance is doubled because it spreads spherically; a laser beam’s intensity stays nearly constant because it is highly collimated (very low divergence), so its cross‑sectional area hardly changes over room‑scale distances.

1. The core idea: intensity depends on how the power spreads

Intensity $I$ is power per unit area:

I = \frac{P}{A}.

For a given source, if the same power $P$ is spread over a larger area $A$ , the intensity drops. The key difference between a bulb and a laser lies in how the beam’s cross‑sectional area changes with distance.

2. For a bulb: spherical spreading

A bulb emits light in all directions. At a distance $r$ , the light power is uniformly distributed over the surface of a sphere of radius $r$ :

A = 4\pi r^2.

So the intensity at distance $r$ is

I(r) = \frac{P}{4\pi r^2}.

If you double the distance to $2r$ :

I(2r) = \frac{P}{4\pi (2r)^2} = \frac{P}{16\pi r^2} = \frac{1}{4} \cdot \frac{P}{4\pi r^2} = \frac{I(r)}{4}.

Watch out

A common mistake is to think intensity halves when distance doubles. Because area depends on $r^2$ , the intensity actually falls by a factor of $4$ — the inverse‑square law.

3. For a laser: collimation keeps the area nearly constant

A laser beam is designed to be highly collimated — its rays are almost perfectly parallel. This means the beam’s diameter changes very little over ordinary distances (like across a room).

If the beam has a constant cross‑sectional area $A_0$ , then at any distance $r$ within the collimated region:

I(r) \approx \frac{P}{A_0} = \text{constant}.

That’s why the intensity of a laser beam does not noticeably drop as it travels across a room. …

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

Showing the 12 most recent of 15 on this concept.

COMEDK 2025Set 2025-E1 markMCQ
$2 \times 10^{-8} T$
›Reveal solutionSolution
In free space, the electric and magnetic fields of an electromagnetic wave are related by $E = cB$ . Given $B = 2 \times 10^{-8}\,\mathrm{T}$ and $c = 3 \times 10^8\,\mathrm{m/s}$ , the electric field is $6\,\mathrm{V/m}$ , so the correct option is (B).

The key concept here is the intrinsic relationship between the electric and magnetic fields in an electromagnetic wave in free space. Unlike in circuits or static fields, where $E$ and $B$ are independent, in a traveling EM wave they are locked together: their magnitudes are proportional, and the constant of proportionality is the speed of light $c$ . This comes directly from Maxwell’s equations — specifically, Faraday’s law and Ampère’s law — which show that a changing magnetic field creates an electric field, and vice versa, with the ratio fixed by $c = 1/\sqrt{\mu_0 \varepsilon_0}$ .

The frequency given (40 MHz) is a red herring here: it tells us the wave is in the radio band, but the instantaneous relation $E = cB$ holds at every point and time for a plane wave in vacuum, regardless of frequency. So we don’t need the frequency at all.

Let’s work it through:

Recall the fundamental relation for a plane electromagnetic wave in free space:

$\frac{E}{B} = c$
where $c = 3 \times 10^8\,\mathrm{m/s}$ is the speed of light. This holds because the wave’s energy is equally shared between the fields, and the wave equation forces the ratio.

Plug in the given magnetic field:

$B = 2 \times 10^{-8}\,\mathrm{T}$
So
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COMEDK 2025Set 2025-M1 markMCQ
$E=E_0 i, \quad B=B_0 j$
›Reveal solutionSolution
For an electromagnetic wave propagating along the +z axis, the electric and magnetic fields must be perpendicular to each other and to the direction of propagation. The only option satisfying this is (A): $E = E_0 \hat{i}$ , $B = B_0 \hat{j}$ .

The key concept here is the mutual perpendicularity of $\vec{E}$ , $\vec{B}$ , and the direction of propagation in an electromagnetic wave. In free space, electromagnetic waves are transverse: the electric field, magnetic field, and wave vector $\vec{k}$ (pointing in the propagation direction) form a right-handed orthogonal set. This means:

$\vec{E} \perp \vec{k}$

$\vec{B} \perp \vec{k}$

$\vec{E} \perp \vec{B}$

And the direction of $\vec{E} \times \vec{B}$ gives the direction of propagation.

Let’s check each option step by step.

Identify the propagation direction. The wave propagates along the +z axis, so the wave vector $\vec{k}$ is along $\hat{k}$ (the unit vector in the z-direction).

Check option (A): $E = E_0 \hat{i}$ , $B = B_0 \hat{j}$ .

$\vec{E}$ is along x-axis ( $\hat{i}$ ), $\vec{B}$ is along y-axis ( $\hat{j}$ ).

Both are perpendicular to $\hat{k}$ (z-axis) because $\hat{i} \cdot \hat{k} = 0$ and $\hat{j} \cdot \hat{k} = 0$ .

$\vec{E} \times \vec{B} = E_0 B_0 (\hat{i} \times \hat{j}) = E_0 B_0 \hat{k}$ , which points along +z. This matches the propagation direction.

So (A) is valid.

Check option (B): $E = E_0 \hat{i}$ , $B = B_0 \hat{k}$ .

$\vec{E}$ is along x-axis, $\vec{B}$ is along z-axis.

$\vec{B}$ is parallel to the propagation direction, not perpendicular. This violates the transverse nature of EM waves in free space.

Invalid.

Check option (C): $E = E_0 \hat{k}$ , $B = B_0 \hat{j}$ .

$\vec{E}$ is along z-axis, parallel to propagation. Again, not transverse. …
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COMEDK 2024Set 2024-E1 markMCQ
$10^{18} \text { to } 10^{20} \mathrm{~Hz}$
›Reveal solutionSolution
A-R, B-S, C-Q, D-P by increasing frequency.

Match each band by frequency:

Radiowaves (A): lowest frequency $10^4$ – $10^8$ Hz = R.

Microwaves (B): $10^9$ – $10^{12}$ Hz = S.

Infrared (C): $10^{11}$ – $5\times10^{14}$ Hz = Q (the entry misprinted as a second P). …
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COMEDK 2023Set 2023-E1 markMCQ
$Q. What feature of the infrared waves make it useful for the haze photography? (A) Since it is invisible (B) Since it has large wave length (C) Since it is absorbed by the medium (D) Since it has high frequency$
›Reveal solutionSolution

Infrared's long wavelength means little scattering by haze particles, so it passes through and produces clearer distant images.

Scattering of light by small particles (Rayleigh scattering) falls off strongly with increasing wavelength ( $\propto 1/\lambda^4$ ). Infrared radiation has a much larger wavelength than visible light, so it is scattered far less by the fine particles of haze, mist and smoke. It therefore penetrates the haze and reaches the camera, allowing clear photograp …
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COMEDK 2023Set 2023-M1 markMCQ
$Q. The ratio of amplitude of magnetic field to the amplitude of electric field of an electromagnetic wave propagating in vacuum is (A) reciprocal of speed of light in vacuum (B) the speed of light in vacuum (C) proportional to frequency of the electromagnetic wave (D) inversely proportional to the frequency of the electromagnetic wave$
›Reveal solutionSolution

Maxwell's relation $E_0=cB_0$ makes $B_0/E_0=1/c$ , i.e. the reciprocal of the speed of light, independent of frequency.

For a plane electromagnetic wave in vacuum, the field amplitudes satisfy:

$\frac{E_0}{B_0}=c\quad\Rightarrow\quad \frac{B_0}{E_0}=\frac{1}{c}.$ …
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COMEDK 2023Set 2023-M1 markMCQ
$20 \mathrm{~MHz}$
›Reveal solutionSolution

In an EM wave $E$ and $B$ are related by $B=E/c$ , giving $B=6/(3\times10^8)=2\times10^{-8}\,$ T.

For a plane electromagnetic wave, the field magnitudes satisfy

$B=\frac{E}{c}=\frac{6\ \text{V m}^{-1}}{3\times10^{8}\ \text{m s}^{-1}}=2\times10^{-8}\ \text{T}.$ …
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KCET 2022Set B-31 markMCQ
$\pi \sqrt{LC}$
›Reveal solutionSolution
In an LC oscillation, energy sloshes between capacitor and inductor. Equal sharing happens when the charge on the capacitor is $q_0/\sqrt{2}$ , which occurs at $t = \frac{\pi}{4}\sqrt{LC}$ .

The problem is about an ideal LC circuit — a capacitor charged to $q_0$ connected to an inductor at $t=0$ . No resistance, so total energy is conserved. At any instant, the capacitor stores electric energy $U_E = \frac{q^2}{2C}$ and the inductor stores magnetic energy $U_B = \frac{1}{2} L i^2$ . Their sum is constant: $\frac{q_0^2}{2C}$ .

The question asks: when are these two equal? That means $U_E = U_B$ , so each is half the total energy. That gives $\frac{q^2}{2C} = \frac{1}{2} \cdot \frac{q_0^2}{2C}$ , i.e. $q^2 = \frac{q_0^2}{2}$ , so $q = \pm \frac{q_0}{\sqrt{2}}$ .

Now we need the time when the charge first reaches this value. The charge on the capacitor in an LC circuit oscillates sinusoidally. Since the capacitor is fully charged at $t=0$ and then begins to discharge, the charge follows a cosine function:
$q(t) = q_0 \cos(\omega t)$
where $\omega = \frac{1}{\sqrt{LC}}$ is the angular frequency of the LC oscillation.

Set $q(t) = \frac{q_0}{\sqrt{2}}$ .

$q_0 \cos(\omega t) = \frac{q_0}{\sqrt{2}} \quad\Rightarrow\quad \cos(\omega t) = \frac{1}{\sqrt{2}}$

The smallest positive angle whose cosine is $1/\sqrt{2}$ is $\pi/4$ radians. So: $\omega t = \frac{\pi}{4}$ …
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COMEDK 2022Set 20221 markMCQ
$\varepsilon_r$
›Reveal solutionSolution

(Option A is dimensionally wrong — it is a pure number, not a speed; the refractive index is n = √(μ_r ε_r), and v = c/n.)

Concept: Speed of an EM wave in a medium, v = 1/√(με), where μ = μ₀μ_r and ε = ε₀ε_r.

v = 1/√(μ₀μ_r ε₀ε_r) = [1/√(μ₀ε₀)] × 1/√(μ_r ε_r) = c/√(μ_r ε_r). …
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COMEDK 2022Set 20221 markMCQ
$\varepsilon_0=9\times10^{-12}$
›Reveal solutionSolution

Solve for E0: E0 = sqrt( 2I / (epsilon0 c) ) = sqrt( 2 x 2000 / (9 x 10^-12 x 3 x 10^8) ) = sqrt( 4000 / (2.7 x 10^-3) ) = sqrt( 1.481 x 10^6 ) = 1.22 x 10^3 V/m = 1.22 kV/m

Concept: intensity of an EM wave in terms of the peak electric field:

I = (1/2) epsilon0 c E0^2

Intensity of the beam:

I = P/A = 30 x 10^-3 W / (15 x 10^-6 m^2) = 2000 W/m^2

Solve for E0:

E0 = sqrt( 2I / (epsilon0 c) ) …
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COMEDK 2021Set 20211 markMCQ
$Q. The correct arrangement in increasing order of wavelength of X-rays, UV rays, microwave is (A) microwave, X-rays, UV rays (B) UV rays, X-rays, microwave (C) X-rays, UV rays, microwave (D) microwave, UV rays, X-rays$
›Reveal solutionSolution
Increasing wavelength: X-rays < UV rays < microwave.

Concept: electromagnetic spectrum ordering.

Typical wavelengths:

X-rays: ~1e-10 m (0.01-10 nm)

Ultraviolet: ~1e-8 m (10-400 nm)

Microwave: ~1e-2 m (1 mm - 1 m) …
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COMEDK 2021Set 20211 markMCQ
$Q. Which of the following waves are used to treatment of muscles ache? (A) Ultraviolet (B) Infrared (C) Microwave (D) X-rays$
›Reveal solutionSolution

Infrared radiation has a heating effect (it excites molecular vibrations) and is used in heat lamps / IR therapy to relieve muscular aches and sprains. UV is used for sterilisation, microwaves for cooking and radar, X-rays for imaging.

Concept: uses of electromagnetic waves. …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. The radiations used in treatment of muscles ache are (A) Microwave (B) Ultraviolet (C) Infrared (D) X-rays$
›Reveal solutionSolution

Infrared radiation is the heat radiation used to treat muscle aches.

Infrared waves are absorbed by tissue and produce a warming effect, which relaxes muscles, dilates blood vessels and eases pain. This is the basis of infrared physiotherapy lamps used for muscular aches. Microwaves and ultraviolet are not used this way f …
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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

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 $\mathbf{B}$

1. The core idea: intensity depends on how the power spreads

2. For a bulb: spherical spreading

3. For a laser: collimation keeps the area nearly constant

Method: Comparing Intensity Fall-off — Point Source vs Collimated Beam

Steps

1. The core idea: intensity depends on how the power spreads

2. For a bulb: spherical spreading

3. For a laser: collimation keeps the area nearly constant

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