Question

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

• 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.

Hint:

For electromagnetic waves entering a new medium: \[ f=\text{constant} \] Always calculate the new speed first using \[ v=\frac{c}{\sqrt{\mu_r\varepsilon_r}} \] and then use \[ \lambda=\frac{v}{f}. \]

Answer 1

The Correct Option is C

Step 1: Write the speed of the wave in the medium:
The refractive index is \(n=\sqrt{\mu_r\varepsilon_r}\).
Here \(\varepsilon_r=\dfrac{3}{2}\) and \(\mu_r=\dfrac{8}{3}\).
So \(\mu_r\varepsilon_r=\dfrac{3}{2}\times\dfrac{8}{3}=4\).
Hence \(n=\sqrt{4}=2\).
The speed in the medium is \(v=\dfrac{c}{n}=\dfrac{c}{2}\).

Step 2: Decide what happens to the frequency:
The frequency is fixed by the source of the wave.
It does not change when the wave crosses a boundary.
So \(f_{medium}=f_{vacuum}\).

Step 3: Find the new wavelength:
Use \(\lambda=\dfrac{v}{f}\).
\(\lambda_{medium}=\dfrac{c/2}{f}=\dfrac{1}{2}\dfrac{c}{f}=\dfrac{\lambda_{vacuum}}{2}\).
The wavelength becomes half of its vacuum value.

Step 4: Check each option:
(A) is wrong. It needs \(n=\dfrac{1}{2}\), but we found \(n=2\).
(B) is wrong. The frequency cannot be halved by a medium.
(C) is right. The wavelength is halved and the frequency is unchanged.
(D) is wrong. The wavelength changes because \(n\neq 1\).

Final Answer:
\(\lambda_{medium}=\dfrac{\lambda_{vacuum}}{2}\) and \(f\) is unchanged. The correct option is (C).