Question

The relative permittivity of distilled water is 81. The velocity of light in it will be : (Given $\mu_r = 1$)

• A. $3.33 \times 10^7$ m/s
• B. $4.33 \times 10^7$ m/s
• C. $5.33 \times 10^7$ m/s
• D. $2.33 \times 10^7$ m/s

Hint:

The refractive index 'n' connects mechanics/optics ($v=c/n$) with electromagnetism ($n=\sqrt{\epsilon_r \mu_r}$). For non-magnetic media, a common case in exams, $\mu_r \approx 1$, so $n \approx \sqrt{\epsilon_r}$.

Answer 1

The Correct Option is A

The refractive index (n) of a medium is related to its relative permittivity ($\epsilon_r$) and relative permeability ($\mu_r$) by the formula:
$n = \sqrt{\epsilon_r \mu_r}$
Given the values for distilled water: $\epsilon_r = 81$ and $\mu_r = 1$.
$n = \sqrt{81 \times 1} = \sqrt{81} = 9$
The velocity of light (v) in a medium is related to the speed of light in vacuum (c) and the refractive index (n) of the medium:
$v = \frac{c}{n}$
Using the speed of light in vacuum, $c \approx 3 \times 10^8$ m/s, and the calculated refractive index, n = 9:
$v = \frac{3 \times 10^8 \text{ m/s}}{9}$
$v = \frac{1}{3} \times 10^8 \text{ m/s}$
$v \approx 0.333 \times 10^8 \text{ m/s} = 3.33 \times 10^7 \text{ m/s}$