Q61 / 79 · NEST (NISER) 2026 · Physics
An electromagnetic wave travels from vacuum into a non-magnetic dielectric medium with permittivity $\epsilon = 4\epsilon_0$. If the ratio $(E_0/B_0)$ of the electric field amplitude $E_0$ to the magnetic field amplitude $B_0$ in vacuum is equal to the speed of light $c$, then the corresponding ratio in the given medium is
- A$c/4$
- B$c$
- Correct answerC$c/2$
- D$c/\sqrt{2}$
Solution
The ratio of the electric field amplitude to the magnetic field amplitude in any medium is equal to the speed of the electromagnetic wave in that medium.
$v = \dfrac{E}{B}$
The speed of an electromagnetic wave in a medium is given by:
$v = \dfrac{1}{\sqrt{\mu \epsilon}}$
For a non-magnetic dielectric medium, $\mu = \mu_0$. Given $\epsilon = 4\epsilon_0$.
$v = \dfrac{1}{\sqrt{\mu_0 (4\epsilon_0)}} = \dfrac{1}{2\sqrt{\mu_0 \epsilon_0}}$
Since the speed of light in vacuum is $c = \dfrac{1}{\sqrt{\mu_0 \epsilon_0}}$, we get:
$v = \dfrac{c}{2}$
Thus, the corresponding ratio in the given medium is $c/2$.
Answer: $c/2$
$v = \dfrac{E}{B}$
The speed of an electromagnetic wave in a medium is given by:
$v = \dfrac{1}{\sqrt{\mu \epsilon}}$
For a non-magnetic dielectric medium, $\mu = \mu_0$. Given $\epsilon = 4\epsilon_0$.
$v = \dfrac{1}{\sqrt{\mu_0 (4\epsilon_0)}} = \dfrac{1}{2\sqrt{\mu_0 \epsilon_0}}$
Since the speed of light in vacuum is $c = \dfrac{1}{\sqrt{\mu_0 \epsilon_0}}$, we get:
$v = \dfrac{c}{2}$
Thus, the corresponding ratio in the given medium is $c/2$.
Answer: $c/2$
Hence, the correct answer is option: 'C'.