The electric field intensity produced by the radiations coming from a 100 W bulb at a 3 m distance is $E$. The electric field intensity produced by the radiations coming from a 50 W bulb at the same distance is (a) $\sqrt{2}\,E$. (b) $2E$. (c) $\dfrac{E}{2}$. (d) $\dfrac{E}{\sqrt{2}}$.

> [!TLDR] > The electric field intensity from a bulb scales as the square root of its power. Since power halves from 100 W to 50 W, the new field is $E/\sqrt{2}$. The key here is that the bulb radiates electromagnetic waves, and the electric field amplitude $E$ is not directly proportional to the power — it’s proportional to the **square root** of the power. This comes from the fact that the intensity (power per unit area) of an electromagnetic wave is proportional to $E^2$. Let’s walk through it. 1. **Intensity and power.** A bulb radiates its power $P$ uniformly in all directions (we assume it’s an isotropic source). At a distance $r$, the power spreads over a sphere of surface area $4\pi r^2$. The intensity $I$ (power per unit area) at that distance is $$ I = \frac{P}{4\pi r^2}. $$ 2. **Intensity and electric field.** For an electromagnetic wave in free space, the time‑averaged intensity is related to the peak electric field $E$ by $$ I = \frac{1}{2} \varepsilon_0 c \, E^2, $$ where $\varepsilon_0$ is the permittivity of free space and $c$ is the speed of light. This is a standard result — the factor of $1/2$ comes from averaging the square of a sinusoidal field. > [!FORMULA] > $$ I = \frac{1}{2} \varepsilon_0 c \, E^2 $$ 3. **Relating $E$ to $P$.** Equate the two expressions for $I$: $$ \frac{P}{4\pi r^2} = \frac{1}{2} \varepsilon_0 c \, E^2. $$ Solve for $E$: $$ E = \sqrt{ \frac{P}{2\pi \varepsilon_0 c \, r^2} }. $$ At a fixed distance $r$, all factors except $P$ are constant. So $$ E \propto \sqrt{P}. $$ 4. **Apply to the two bulbs.** For the 100 W bulb: $E \propto \sqrt{100} = 10$. …

The electric field intensity produced by the radiations coming from a 100 W bulb at a 3 m distance is E. The e

NCERT Exemplar · Q4

E

(a)

\sqrt{2}\,E

(b)

2E

(c)

\dfrac{E}{2}

(d)

\dfrac{E}{\sqrt{2}}

Rajasthan RbseMCQ· 1mImportance★★★★★

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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 electric field intensity from a bulb scales as the square root of its power. Since power halves from 100 W to 50 W, the new field is $E/\sqrt{2}$ .

The key here is that the bulb radiates electromagnetic waves, and the electric field amplitude $E$ is not directly proportional to the power — it’s proportional to the square root of the power. This comes from the fact that the intensity (power per unit area) of an electromagnetic wave is proportional to $E^2$ .

Let’s walk through it.

Intensity and power. A bulb radiates its power $P$ uniformly in all directions (we assume it’s an isotropic source). At a distance $r$ , the power spreads over a sphere of surface area $4\pi r^2$ . The intensity $I$ (power per unit area) at that distance is

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

Intensity and electric field. For an electromagnetic wave in free space, the time‑averaged intensity is related to the peak electric field $E$ by

I = \frac{1}{2} \varepsilon_0 c \, E^2,

where $\varepsilon_0$ is the permittivity of free space and $c$ is the speed of light. This is a standard result — the factor of $1/2$ comes from averaging the square of a sinusoidal field.

I = \frac{1}{2} \varepsilon_0 c \, E^2

Relating $E$ to $P$ . Equate the two expressions for $I$ :

\frac{P}{4\pi r^2} = \frac{1}{2} \varepsilon_0 c \, E^2.

Solve for $E$ :

E = \sqrt{ \frac{P}{2\pi \varepsilon_0 c \, r^2} }.

At a fixed distance $r$ , all factors except $P$ are constant. So

E \propto \sqrt{P}.

Apply to the two bulbs. For the 100 W bulb: $E \propto \sqrt{100} = 10$ . …

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

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 $\mathbf{B}$

Method: Relating Electric Field Amplitude to Source Power at a Fixed Distance

Steps

Q.The electric field intensity produced by the radiations coming from a 100 W bulb at a 3 m distance is EEE. The electric field intensity produced by the radiations coming from a 50 W bulb at the same distance is

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

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