1 markMCQSection A

An electromagnetic wave passes from vacuum into a dielectric medium with relative electrical permittivity $(3/2)$ and relative magnetic permeability $(8/3)$ . Then, its

Q: An electromagnetic wave passes from vacuum into a dielectric medium with relative electrical permittivity $(3/2)$ and relative magnetic permeability $(8/3)$. Then, its

**Correct option: C — wavelength is halved and frequency remains unchanged.** - Refractive index $n = \sqrt{\varepsilon_r \mu_r} = \sqrt{(3/2)(8/3)} = \sqrt{4}$ = **2**. - Frequency is set by the source and **does not change** on entering a medium. - Speed $v = c/n = c/2$, so wavelength $\lambda = v/f$ is **halved**.

Electromagnetic Waves

EM wave entering a dielectric: change of wavelength and frequency

Options

(A)wavelength is doubled and frequency remains unchanged.

(B)wavelength is doubled and frequency is halved.

(C)wavelength is halved and frequency remains unchanged.

(D)wavelength and frequency both will remain unchanged.

Official Answer

Correct option: C — wavelength is halved and frequency remains unchanged.

Refractive index $n = \sqrt{\varepsilon_r \mu_r} = \sqrt{(3/2)(8/3)} = \sqrt{4}$ = 2.
Frequency is set by the source and does not change on entering a medium.
Speed $v = c/n = c/2$ , so wavelength $\lambda = v/f$ is halved.

electromagnetic waverefractive indexrelative permittivityrelative permeabilityn = sqrt(εr μr)frequency unchangedwavelength in medium

Marking Scheme

11 mark: correct option C (wavelength halved, frequency unchanged).
2Reasoning credited: $n = \sqrt{\varepsilon_r \mu_r} = 2$ and constancy of frequency across the boundary.

Hint

Compute $n = \sqrt{\varepsilon_r \mu_r}$ . Frequency stays fixed; wavelength scales as $1/n$ .

Quick Oral Answer

The refractive index is $\sqrt{\varepsilon_r \mu_r} = \sqrt{3/2 \times 8/3} = 2$ , so the wave slows to $c/2$ ; frequency is fixed by the source, hence the wavelength is halved.

Analysis & Explanation

This question links the electromagnetic origin of the refractive index to how wavelength and frequency behave across a boundary.

Concept

The speed of an EM wave in a medium is $v = 1/\sqrt{\varepsilon\mu} = c/\sqrt{\varepsilon_r \mu_r}$ , so the refractive index is $n = \sqrt{\varepsilon_r \mu_r}$ .
Here $n = \sqrt{(3/2)(8/3)} = \sqrt{4} = 2$ .
Frequency is invariant: it is fixed by the oscillating source; the wave crests arrive at the boundary at the same rate they leave it, so f cannot change.
With $v = c/2$ and f constant, $\lambda = v/f$ is halved.

Why the distractors are wrong

A (wavelength doubled, f same): correct that f is unchanged, but with $n = 2$ the wave slows down, so wavelength shortens, not lengthens.
B (wavelength doubled, f halved): both parts wrong — f never halves and wavelength decreases.
D (both unchanged): ignores that a denser optical medium ( $n = 2$ ) alters speed and hence wavelength.

Exam trap

Many students forget that $\mu_r$ can differ from 1; both $\varepsilon_r$ and $\mu_r$ must be multiplied under the root. Also, only wavelength (and speed) change — frequency is always conserved across a boundary.

Common Mistakes

1Taking $\mu_r = 1$ and computing $n = \sqrt{\varepsilon_r}$ , giving a wrong refractive index.
2Assuming the frequency changes when a wave enters a medium (it does not).
3Thinking a larger n increases wavelength; a higher index reduces speed and shortens wavelength.

Interesting Facts

The relation $n = \sqrt{\varepsilon_r \mu_r}$ was one of Maxwell's triumphs — it connected optics to electricity and magnetism and predicted light's speed from purely electromagnetic constants.

Materials engineered with unusual $\varepsilon_r$ and $\mu_r$ (even negative values, called metamaterials) can bend light in ways impossible for natural media, enabling 'invisibility cloak' research.

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Frequently Asked Questions

Why does the frequency of the wave stay the same in the medium?

Frequency is determined by the source that generates the wave. At the boundary, the fields must be continuous, so wavefronts arrive and depart at the same rate. Only speed and wavelength adjust; frequency is conserved.

Do we need the relative permeability, or can we assume it is 1?

You must use the given value. The refractive index is $n = \sqrt{\varepsilon_r \mu_r}$ . Here $\mu_r = 8/3$ , not 1, and combining it with $\varepsilon_r = 3/2$ gives $n = 2$ . Ignoring μ_r would give the wrong answer.