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

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.