The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3

Q: Why do we use relative permittivity and permeability instead of absolute values?

Using relative values simplifies the calculation because the absolute values contain constants like $\epsilon_0$ and $\mu_0$. The ratio $\frac{c}{v}$ naturally isolates the relative terms as $\sqrt{\epsilon_r \mu_r}$.

Q: What happens if the angle of incidence is exactly the critical angle?

At exactly the critical angle, the refracted ray grazes the boundary between the two media, meaning the angle of refraction is exactly $90^\circ$.

Q: Can the critical angle exist if light travels from a rarer to a denser medium?

No, the critical angle only exists when light travels from a denser medium (higher refractive index) to a rarer medium (lower refractive index). Otherwise, $\sin i_c$ would be greater than 1, which is mathematically impossible.

Q: Why did we take the refractive index of the second medium as 1?

The problem asks for the critical angle of the medium, which implicitly assumes the second medium is vacuum or air, both of which have a refractive index of approximately 1.

Q: What is the physical significance of the refractive index being 2?

A refractive index of 2 means that light travels exactly half as fast in this medium compared to its speed in a vacuum. It's a measure of how much the medium slows down the light.

Question

The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability $\frac{4}{3}$ for this wavelength, will be

Options:

(A) $45^\circ$

(B) $60^\circ$

(C) $15^\circ$

(D) $30^\circ$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a light ray traveling from a denser medium to a rarer medium. The critical angle is the angle of incidence for which the refracted ray grazes the surface. Look closely at the diagram. When the angle of incidence reaches the critical angle $i_c$, the angle of refraction $r$ becomes exactly $90^\circ$.

### The Electromagnetic Connection

Now, the speed of light in any medium depends on its permittivity and permeability. Remember the formula? The speed of light $v$ is given by:

$$v = \frac{1}{\sqrt{\epsilon \mu}}$$

We can express the absolute permittivity $\epsilon$ and permeability $\mu$ in terms of their relative counterparts and the constants for free space:

$$v = \frac{1}{\sqrt{\epsilon_0 \mu_0}} 	imes \frac{1}{\sqrt{\epsilon_r \mu_r}} = \frac{c}{\sqrt{\epsilon_r \mu_r}}$$

### Finding the Refractive Index

The refractive index, denoted by $n$, is the ratio of the speed of light in a vacuum $c$ to the speed of light in the medium $v$. So, $n$ becomes the square root of $\epsilon_r$ times $\mu_r$:

$$n = \frac{c}{v} = \sqrt{\epsilon_r \mu_r}$$

In the question, we are given the relative permittivity as $3$ and relative permeability as $\frac{4}{3}$. Let's substitute these values into our equation:

$$n = \sqrt{3 	imes \frac{4}{3}}$$

The $3$s cancel out perfectly. We are left with the square root of $4$, which is $2$. So, the refractive index of the medium is $2$.

### The Critical Angle

Now let's apply the formula for the critical angle. From Snell's Law, when the angle of refraction is $90^\circ$, we get:

$$\sin i_c = \frac{1}{n}$$

### Final Calculation

Putting the value of $n$ as $2$, $\sin i_c$ becomes $\frac{1}{2}$. And when is sine equal to $\frac{1}{2}$? At $30^\circ$!

$$i_c = \sin^{-1}\left(\frac{1}{2}ight) = 30^\circ$$

So our answer is **$30^\circ$**.

### The Way Forward

Think about what would happen if the incident angle is greater than $30^\circ$? The light won't escape into the rarer medium; instead, **Total Internal Reflection (TIR)** will occur, and the light will bounce entirely back into the denser medium. This beautiful phenomenon is the core principle behind optical fibers and mirages!

Animated Solution for Physics - Optics: The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability $\frac{4}{3}$ for this wavelength, will be

Select Answer:

Visualized Solution

Visual Anchor

Critical Angle Condition

Speed of Light in Medium

Refractive Index

Substitution

Calculation

Snell's Law at Critical Angle

Final Answer

The Way Forward

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Setup

The Electromagnetic Connection

Finding the Refractive Index