Snell's Law and TIR: NEET Problem-Solving

1. Recapping the Core Concepts: Snell's Law and TIR

Before we dive into complex problems, let's quickly refresh the two fundamental principles we'll be using today.

Snell's Law: At any interface between two media, the relationship between the angle of incidence (${\theta }_1$) and the angle of refraction (${\theta }_2$) is given by:
$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Total Internal Reflection (TIR): This phenomenon occurs only when two conditions are met:

1. Light must travel from an optically denser medium to an optically rarer medium (i.e., $n_1>n_2$).
2. The angle of incidence ($i$) must be greater than the critical angle ($C$).

The critical angle is the specific angle of incidence for which the angle of refraction is 90°. It can be calculated using Snell's law:
$n_1\sin C=n_2\sin 90^{\circ }⟹\sin C=\frac{n_2}{n_1}$

To start, let's watch a short video that walks through the calculation of the critical angle in a few common scenarios. This will warm up our problem-solving mindset.

2. Solving Multi-Step Refraction Problems

NEET problems rarely involve a single, simple application of a formula. They often present scenarios with multiple layers or complex geometries. The key is to apply Snell's Law systematically at each interface.

Worked Example: A Classic NEET-style Problem

Let's apply these strategies to a typical problem that combines Snell's Law and TIR.

Strategy: Work backwards from the known information.

1. Analyze the water-air interface: The ray "emerges parallel to the surface," which means the angle of refraction in air is 90°. This tells us the angle of incidence in the water is the critical angle, $C$, for the water-air interface.

2. Calculate the critical angle $C$:
   Using Snell's Law for the water-air interface:
   ${\mu }_w\sin C={\mu }_{air}\sin 90^{\circ }$
   $\frac{4}{3}\sin C=1\times 1$
   $\sin C=\frac{3}{4}$

3. Use geometry to relate the angles: Look at the diagram. The refracted ray from the first surface becomes the incident ray for the second. The angle of refraction at the glass-water interface is $r$. From the right-angled triangle formed by the ray, the normal, and the horizontal surface, we can see that $r$ and $C$ are alternate interior angles. Therefore, $r=C$.

4. Apply Snell's Law at the glass-water interface:
   Now we can analyze the first boundary. The angle of incidence is given as $i$. The angle of refraction is $r$, which we now know is equal to $C$.
   ${\mu }_g\sin i={\mu }_w\sin r$
   The problem statement in the image implies the incident angle $i$ is such that $\sin i=3/4$. Let's assume the question meant the angle of incidence inside the glass onto the glass-water interface is i (this is a common ambiguity in diagrams, but let's follow the ray path shown). The ray enters glass, hits the glass-water interface at angle i, refracts at angle r into the water, and then hits the water-air interface at angle C. From the diagram's geometry, the angle of incidence at the glass-water interface i is the same as the critical angle C. This seems to be a slight mislabeling in the diagram text vs the drawing. Let's re-interpret based on a more standard problem type.

   Let's assume the angle of incidence at the first (air-glass, not shown) interface creates the ray inside the glass. This ray hits the glass-water interface at an angle we'll call ${\theta }_g$. It refracts into the water at an angle ${\theta }_w$. This ray then hits the water-air interface at the critical angle $C$. From the geometry with parallel interfaces, ${\theta }_w=C$.

   Let's re-examine the image text: "A ray of light is incident at an angle i on the glass-water interface...". This clarifies things. i is the angle in the glass.

   • Angle of incidence in glass = $i$
   • Angle of refraction in water = $r$ (from diagram geometry, $r=C$)

   Applying Snell's Law at the glass-water interface:
   ${\mu }_g\sin i={\mu }_w\sin r$
   ${\mu }_g\sin i=\frac{4}{3}\sin C$
   We already found $\sin C=3/4$.
   ${\mu }_g\sin i=\frac{4}{3}\times \frac{3}{4}=1$
   ${\mu }_g=\frac{1}{\sin i}$
   The question has a missing piece of information, which is the value of i. However, looking at the provided options, this structure is common. Often, the problem implies a specific value for i, such as 30°, 45°, or 60°. If we assume $i=45^{\circ }$, then ${\mu }_g=1/\sin 45^{\circ }=\sqrt{2}$. If $i=30^{\circ }$, ${\mu }_g=2$. Let's check the likely answer. Given the context of TIR problems, a common value for $\sin i$ is often related to the refractive indices. The solution in the image itself has (μg) sin i = (μw) sin r. If we assume the angle i in the diagram is what is intended, and from the geometry r = C, then (μg) sin i = (μw) sin C. Since μw sin C = μair sin 90 = 1, it simplifies to μg sin i = 1. This leads back to μg = 1/sin(i). A common setup is a right-angled prism, where i might be 45 degrees. Let's assume that was the implicit condition.

This exercise shows how crucial it is to dissect a problem: identify all interfaces, apply the correct laws, use geometry, and work step-by-step.

3. Vigorous Practice: NEET Problem Drill

As you requested, here is a vigorous drill with questions mirroring the style and difficulty of the NEET exam. Attempt to solve these problems on your own first.

I have selected them from the resources Problems on Ray Optics... (Disha Publication) and NEET Previous Year Questions... (Edurev).

Conclusion

Today, we've transitioned from learning concepts to applying them in an exam-oriented context. You've tackled a variety of problems on Snell's Law and Total Internal Reflection, sharpening the analytical and mathematical skills you'll need for NEET.

Key Takeaways:

• Systematic Approach: For any problem, first identify the media and interfaces. Apply Snell's Law at each boundary.
• Geometry is Key: Many problems require you to use geometry (e.g., properties of triangles, parallel lines) to determine angles of incidence and refraction.
• Check for TIR: Whenever light travels from a denser to a rarer medium, always check if the angle of incidence exceeds the critical angle.
• Wavelength Matters: Refractive index ($\mu$) depends on wavelength ($\lambda$). Specifically, ${\mu }_{violet}>{\mu }_{red}$. This means the critical angle also depends on color: $C_{violet}<C_{red}$. This is a frequent source of conceptual questions.

Preview of the Next Lesson:
Our intensive work with Snell's law and geometric interfaces today has perfectly prepared us for our next topic: prisms. In the next lesson, you will learn to trace the path of a ray through a prism and analyze the angle of deviation. You'll see that a prism is essentially a system with two non-parallel interfaces, and the skills you've honed today will be directly applicable. We will then formalize the relationship between the prism angle, deviation, and refractive index.