The velocity of light in a medium with relative permittivity \( \epsilon_r \) and relative permeability \( \mu_r \) is given by: \[ v = \frac{c}{\sqrt{\epsilon_r \mu_r}} \] Where:
- \( c \) is the velocity of light in vacuum,
- \( \epsilon_r \) is the relative permittivity,
- \( \mu_r \) is the relative permeability. We are given that the relative permittivity \( \epsilon_r = 2 \) and the relative permeability \( \mu_r = 4.5 \). Substitute these values into the formula: \[ v = \frac{c}{\sqrt{2 \cdot 4.5}} \] Now, simplifying the expression: \[ v = \frac{c}{\sqrt{9}} = \frac{c}{3} \]
Thus, the velocity of light in this medium is \( \frac{c}{\sqrt{2 \cdot 4.5}} \). So the correct answer is (A) \( \frac{c}{\sqrt{2 \cdot 4.5}} \).
A transparent block A having refractive index $ \mu_2 = 1.25 $ is surrounded by another medium of refractive index $ \mu_1 = 1.0 $ as shown in figure. A light ray is incident on the flat face of the block with incident angle $ \theta $ as shown in figure. What is the maximum value of $ \theta $ for which light suffers total internal reflection at the top surface of the block ?