Critical Angle and Total Internal Reflection

1. The Phenomenon of Total Internal Reflection

Let's begin, as you prefer, by visualizing what happens. As we increase the angle of incidence (${\theta }_1$) for a light ray traveling from a denser medium (refractive index $n_1$) to a rarer medium (refractive index $n_2$), the angle of refraction (${\theta }_2$) also increases, and faster than ${\theta }_1$.

There are three possible outcomes:

1. Refraction: For small angles of incidence, the ray refracts into the rarer medium, bending away from the normal.
2. Graze the Surface: At a specific angle of incidence, called the critical angle (${\theta }_c$), the refracted ray skims along the boundary between the two media. The angle of refraction is exactly 90°.
3. Total Internal Reflection (TIR): If the angle of incidence is greater than the critical angle, the ray no longer exits the denser medium. Instead, it is completely reflected back into it, obeying the laws of reflection.

The following video demonstrates this transition beautifully.

The Two Conditions for TIR

From the video and the underlying physics, we can establish two essential conditions that must be met for TIR to occur:

1. The light ray must be traveling from a denser medium to a rarer medium (i.e., $n_1>n_2$).
2. The angle of incidence in the denser medium must be greater than the critical angle (${\theta }_i>{\theta }_c$).

The image below provides a perfect summary of these three scenarios.

2. Calculating the Critical Angle

Now that we have the concept down, let's derive the formula to calculate the critical angle. The derivation is a straightforward application of Snell's Law.

We start with Snell's Law:
$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

By definition, at the critical angle of incidence (${\theta }_1={\theta }_c$), the angle of refraction is 90° (${\theta }_2=90^{\circ }$). Substituting this into the equation:
$n_1\sin {\theta }_c=n_2\sin (90^{\circ })$

Since $\sin (90^{\circ })=1$, the equation simplifies to:
$n_1\sin {\theta }_c=n_2$

Rearranging for the critical angle, we get the all-important formula:
$\sin {\theta }_c=\frac{n_2}{n_1}$
or
${\theta }_c={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

Notice that for the ${\sin }^{-1}$ function to be defined, the term $\frac{n_2}{n_1}$ must be less than or equal to 1. This mathematically confirms the first condition for TIR: $n_1$ must be greater than $n_2$.

Let's watch a clear, worked example of how to use this formula.

3. NEET-Style Problem Solving

Let's apply these concepts to the kind of problems you'll face in the NEET exam.

Problem 1: Calculating Velocity from Critical Angle (NEET 2020)

Question: If the critical angle for total internal reflection from a medium to vacuum is 45°, then the velocity of light in the medium is:

Solution Walkthrough:

1. Identify the media: Medium 1 is the unknown medium ($n_1=n$). Medium 2 is vacuum ($n_2=1$). The critical angle ${\theta }_c=45^{\circ }$.
2. Apply the critical angle formula:
   $\sin {\theta }_c=\frac{n_2}{n_1}⟹\sin (45^{\circ })=\frac{1}{n}$
3. Solve for the refractive index $n$:
   $\frac{1}{\sqrt{2}}=\frac{1}{n}⟹n=\sqrt{2}$
4. Relate refractive index to velocity: Recall the definition of refractive index: $n=\frac{c}{v}$, where $c$ is the speed of light in vacuum ($3\times 10^8$ m/s) and $v$ is the speed of light in the medium.
5. Solve for velocity $v$:
   $v=\frac{c}{n}=\frac{3\times 10^8\text{ m/s}}{\sqrt{2}}$

This is a classic two-step problem combining the critical angle formula with the definition of refractive index. You can find this question and its solution in the resource NEET Previous Year Questions (ID: LINK) under the 2020 section (Q2).

Problem 2: Calculating Critical Angle from Velocities (NEET 2022)

Question: Two transparent media A and B are separated by a plane boundary. The speed of light in those media are $1.5\times 10^8$ m/s and $2.0\times 10^8$ m/s, respectively. The critical angle for a ray of light for these two media is:

Solution Walkthrough:

1. Identify the denser and rarer media:
   • Speed in A: $v_A=1.5\times 10^8$ m/s.
   • Speed in B: $v_B=2.0\times 10^8$ m/s.
   • Since light is slower in medium A, A is the denser medium. So, $n_1=n_A$ and $n_2=n_B$.
2. Calculate refractive indices:
   • $n_A=\frac{c}{v_A}=\frac{3\times 10^8}{1.5\times 10^8}=2$
   • $n_B=\frac{c}{v_B}=\frac{3\times 10^8}{2.0\times 10^8}=1.5$
3. Apply the critical angle formula:
   $\sin {\theta }_c=\frac{n_B}{n_A}=\frac{1.5}{2}=0.75$
4. Express the final answer:
   ${\theta }_c={\sin }^{-1}(0.75)$

This problem tests your ability to first determine which medium is denser based on the speed of light. This problem is available in resource LINK as Q5 from 2022.

Problem 3: TIR in a Prism (NEET 2015)

This is a more advanced problem that combines TIR with dispersion.

Question: A beam of light consisting of red, green, and blue colours is incident on a right-angled prism. The refractive index of the material of the prism for the red, green, and blue wavelengths are 1.39, 1.44, and 1.47, respectively. The light is incident as shown. Which color(s) will be separated from the others?

This is the setup for the problem described, as seen in NEET 2015, Q3 in resource LINK.

Solution Walkthrough:

1. Analyze the geometry: The light enters the prism face AC perpendicularly, so it passes undeviated to the face AB. The prism is a right-angled isosceles prism, so the angle of incidence on face AB is 45°.
2. State the condition for TIR: The ray will undergo TIR if the angle of incidence (45°) is greater than the critical angle (${\theta }_c$).
3. Calculate the critical angle for each color: The light is trying to exit into air ($n_2=1$).
   • Red: $\sin {\theta }_{c,\text{red}}=\frac{1}{1.39}\approx 0.719$. ${\theta }_{c,\text{red}}={\sin }^{-1}(0.719)\approx 46^{\circ }$.
   • Green: $\sin {\theta }_{c,\text{green}}=\frac{1}{1.44}\approx 0.694$. ${\theta }_{c,\text{green}}={\sin }^{-1}(0.694)\approx 44^{\circ }$.
   • Blue: $\sin {\theta }_{c,\text{blue}}=\frac{1}{1.47}\approx 0.680$. ${\theta }_{c,\text{blue}}={\sin }^{-1}(0.680)\approx 42.8^{\circ }$.
4. Compare and conclude:
   • For red: ${\theta }_i(45^{\circ })<{\theta }_{c,\text{red}}(46^{\circ })$. Red light will refract out of the prism.
   • For green: ${\theta }_i(45^{\circ })>{\theta }_{c,\text{green}}(44^{\circ })$. Green light will undergo TIR.
   • For blue: ${\theta }_i(45^{\circ })>{\theta }_{c,\text{blue}}(42.8^{\circ })$. Blue light will undergo TIR.
   • Therefore, the prism separates the red light from the green and blue light.

4. Applications of Total Internal Reflection

Your interest in the sciences goes beyond just exam prep, so it's worth noting that TIR is not just an abstract concept; it's the basis for many technologies.

• Optical Fibers: These thin strands of glass or plastic guide light over long distances with minimal energy loss by ensuring the light always strikes the internal surface at an angle greater than the critical angle. This is the backbone of the internet and modern telecommunications.
• Endoscopy: Medical instruments use bundles of optical fibers to transmit light into the body and carry an image back out, allowing for minimally invasive diagnosis.
• Binoculars and Periscopes: High-quality optical instruments often use prisms instead of mirrors to redirect light. TIR in a prism is more efficient than reflection from a silvered mirror, leading to a brighter and clearer image.
• The Brilliance of Diamonds: A diamond's high refractive index ($n\approx 2.42$) gives it a very small critical angle ($\approx 24.4^{\circ }$). This means that most light entering the diamond is trapped inside, undergoing multiple internal reflections before exiting, which creates its characteristic sparkle.

Conclusion

Today, you've mastered the concept of Total Internal Reflection, a direct and powerful consequence of Snell's Law. We have followed your preferred learning path, starting with visual intuition, deriving the core mathematics, and then applying it vigorously to exam-style problems.

Key Takeaways:

• TIR occurs when light travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle.
• The critical angle ${\theta }_c$ is the angle of incidence that results in a 90° angle of refraction.
• The formula for the critical angle is derived directly from Snell's law: $\sin {\theta }_c=\frac{n_2}{n_1}$.
• You can solve a variety of problems by combining this formula with the definition of refractive index ($n=c/v$) and geometrical analysis.

Preview of the Next Lesson:
We have now thoroughly covered refraction at plane surfaces. The next logical step is to see how these rules apply when the boundary is curved. In our next lesson, we will apply the formula for refraction at a single spherical surface, which will lay the groundwork for understanding lenses.