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 A

Concept: The speed of an electromagnetic wave in a medium is given by \[ v=\frac{c}{\sqrt{\mu_r\varepsilon_r}}, \] where \[ \mu_r=\text{relative permeability}, \qquad \varepsilon_r=\text{relative permittivity}. \] When an electromagnetic wave enters another medium, its frequency remains unchanged because the frequency is determined by the source of the wave. The wavelength changes according to \[ \lambda=\frac{v}{f}. \] Therefore, once the new speed is known, the new wavelength can be determined.

Step 1: Calculate the speed of the wave in the medium.
Given, \[ \varepsilon_r=\frac32, \qquad \mu_r=\frac83. \] Hence, \[ v=\frac{c}{\sqrt{\frac32\times\frac83}} \] \[ v=\frac{c}{\sqrt{4}} \] \[ v=\frac{c}{2} \] Thus, the speed becomes half of its value in vacuum.

Step 2: Determine the change in frequency and wavelength.
Frequency does not change: \[ f'=f. \] Using \[ \lambda=\frac{v}{f}, \] we get \[ \lambda'=\frac{v'}{f'} = \frac{c/2}{f} = \frac12\left(\frac{c}{f}\right) = \frac{\lambda}{2}. \] Therefore, the wavelength becomes half of its original value while the frequency remains unchanged. \[ \boxed{\lambda'=\frac{\lambda}{2},\qquad f'=f} \] Hence, the correct option is \[ \boxed{\text{(C) wavelength is halved and frequency remains unchanged}} \]