Question

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

Options

• wavelength is doubled and frequency remains unchanged.

• wavelength is doubled and frequency is halved.

• wavelength is halved and frequency remains unchanged.

• wavelength and frequency both will remain unchanged.

Answer 1

An electromagnetic wave passes from vacuum into a dielectric medium with relative electrical permittivity (3/2) and relative magnetic permeability (8/3). Then, its wavelength is halved and frequency remains unchanged.

Explanation:

When an electromagnetic (EM) wave transitions from one medium (like a vacuum) to another (like a dielectric), its frequency (f) remains unchanged. This is because the frequency is a characteristic of the wave source, not the medium through which it travels.

The speed of light in a vacuum is c. In a dielectric medium, the speed (v) is reduced and is given by the formula:

v = `c/n`

Where n is the refractive index. The refractive index can be calculated using the relative electrical permittivity (ε_r) and the relative magnetic permeability (μ_r) of the medium:

Given, ε_r = `3/2` 

μ_r = `8/3`

n = `sqrt(epsilon_r * mu_r)`

= `sqrt(3/2 xx 8/3)`

= `sqrt(24/6)`

= `sqrt 4`

= 2

Thus, the speed of the wave in this medium (v) = `c/2`

We know that,

v = f λ

`λ_"medium" = v/f`

= `(c/2)/f`

= `1/2(c/f)`

= `lambda_"vacuum"/2`

This calculation shows that the wavelength is halved in the dielectric medium.