Question

If an optical medium possesses a relative permeability of $ \frac{10}{\pi} $ and relative permittivity of $ \frac{1}{0.0885} $, then the velocity of light is greater in vacuum than in that medium by ________ times. $ (\mu_0 = 4\pi \times 10^{-7} \, \text{H/m}, \quad \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}, \quad c = 3 \times 10^8 \, \text{m/s}) $

Hint:

The velocity of light in any medium is determined by its relative permeability and permittivity. It is slower than in a vacuum unless both \( \mu \) and \( \epsilon \) are 1.

Answer 1

Correct Answer: 6

The velocity of light in any medium is given by the formula: \[ V = \frac{C}{\sqrt{\mu \epsilon}} \] where: 
- \( C \) is the speed of light in vacuum, 
- \( \mu \) is the permeability of the medium, 
- \( \epsilon \) is the permittivity of the medium. 
In this case: 
- The relative permeability \( \mu_r = \frac{10}{\pi} \), 
- The relative permittivity \( \epsilon_r = \frac{1}{0.0885} \), 
- The permeability \( \mu = \mu_0 \mu_r \), 
- The permittivity \( \epsilon = \epsilon_0 \epsilon_r \). 
Substituting the values: \[ \mu = (4\pi \times 10^{-7}) \times \frac{10}{\pi} = 4 \times 10^{-6} \, \text{H/m} \] \[ \epsilon = 8.85 \times 10^{-12} \times \frac{1}{0.0885} = 1 \times 10^{-10} \, \text{F/m} \] Now, substitute these values into the formula for velocity: \[ V = \frac{3 \times 10^8}{\sqrt{(4 \times 10^{-6})(1 \times 10^{-10})}} = \frac{3 \times 10^8}{\sqrt{4 \times 10^{-16}}} = \frac{3 \times 10^8}{2 \times 10^{-8}} = 1.5 \times 10^{16} \, \text{m/s} \] 
Thus, the velocity of light in the medium is \( \frac{1}{6} \) times the velocity in a vacuum, so the correct answer is 6.