If relative permittivity and relative permeability of a medium are 3 and `4/3` respectively. the critical angle for this medium is.

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `15^(@)`

Answer

To find the critical angle for a medium with given relative permittivity (\( \epsilon_r \)) and relative permeability (\( \mu_r \)), we can follow these steps:

### Step 1: Understand the relationship between the speed of light in a medium and its permittivity and permeability.
The speed of light \( c \) in a vacuum is given by:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
In a medium, the speed of light \( v \) is given by:
\[ v = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}} \]
where \( \mu_0 \) and \( \epsilon_0 \) are the permeability and permittivity of free space, and \( \mu_r \) and \( \epsilon_r \) are the relative permeability and permittivity of the medium.

### Step 2: Calculate the refractive index \( n \) of the medium.
The refractive index \( n \) of the medium can be expressed as:
\[ n = \frac{c}{v} = \sqrt{\frac{\mu_r \epsilon_r}{1}} \]
Substituting the given values \( \epsilon_r = 3 \) and \( \mu_r = \frac{4}{3} \):
\[ n = \sqrt{\mu_r \epsilon_r} = \sqrt{\frac{4}{3} \cdot 3} = \sqrt{4} = 2 \]

### Step 3: Use Snell's Law to find the critical angle.
The critical angle \( \theta_c \) can be found using Snell's Law, which states:
\[ n_1 \sin(\theta_c) = n_2 \sin(90^\circ) \]
Here, \( n_1 = n \) (the refractive index of the denser medium) and \( n_2 = 1 \) (the refractive index of air/vacuum). Therefore:
\[ 2 \sin(\theta_c) = 1 \]
This simplifies to:
\[ \sin(\theta_c) = \frac{1}{2} \]

### Step 4: Solve for the critical angle \( \theta_c \).
To find \( \theta_c \):
\[ \theta_c = \arcsin\left(\frac{1}{2}\right) = 30^\circ \]

### Final Answer:
The critical angle for the medium is \( 30^\circ \).

---