Monochromatic light is incident on a plane interface AB between two media of refractive indices n1 a

Q: Why does the problem state $\theta$ is "infinitesimally greater" than the critical angle?

This is a mathematical way of saying that $\theta$ is just large enough to cause Total Internal Reflection (TIR) at the original interface, giving us the strict inequality $n_2 \sin \theta > n_1$ without needing to worry about large deviations.

Q: In case (a), why doesn't the light enter the slab at all?

Because the refractive index of the slab ($n_3$) is even lower than medium I ($n_1$), the critical angle to enter the slab is smaller than the original critical angle. Since our incident angle already exceeds the original critical angle, it definitely exceeds this new, smaller critical angle, causing immediate TIR.

Q: Can we use Snell's law even if TIR occurs?

Snell's law ($n_i \sin \theta_i = n_r \sin \theta_r$) is universally valid for refraction. However, if the math gives $\sin \theta_r > 1$, it physically means refraction is impossible and TIR occurs instead. We use this exact logic to prove TIR at the upper boundary!

Q: What happens if $n_3$ is exactly equal to $n_1$?

If $n_3 = n_1$, the slab is optically identical to medium I. The light will undergo TIR at the boundary between medium II and the slab, exactly as it did before the slab was introduced.

Q: Why does the light emerge parallel to its original path if it refracts back into medium II?

Because the interfaces are parallel, the angle of incidence at the bottom of the slab (when returning) is equal to the angle of refraction when it entered. By the principle of reversibility and Snell's law, it must exit at the same angle $\theta$ it originally had.

Question

Monochromatic light is incident on a plane interface $AB$ between two media of refractive indices $n_1$ and $n_2$ ($n_2 > n_1$) at an angle of incidence $	heta$ as shown in the figure.
The angle $	heta$ is infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. Now if a transparent slab $DEFG$ of uniform thickness and of refractive index $n_3$ is introduced on the interface (as shown in the figure), show that for any value of $n_3$ all light will ultimately be reflected back again into medium II.
Consider separately the cases :
(a) $n_3 < n_1$ and (b) $n_3 > n_1$

Sigma Percentile · Accepted Answer

### The Inescapable Light: A Tale of Total Internal Reflection Imagine a beam of light trying to escape from a denser medium into a rarer one. If the angle is just right, the boundary acts like a perfect mirror, trapping the light inside. This phenomenon is known as Total Internal Reflection (TIR). In this fascinating problem, we explore what happens when we introduce a new transparent slab into the mix. Will the light finally escape, or is it doomed to return? ### The Master Condition Let's start by analyzing the initial setup. We are given that a light ray is incident from Medium II (with refractive index $n_2$) towards Medium I (with refractive index $n_1$). The angle of incidence, $ heta$, is infinitesimally greater than the critical angle for these two media. Mathematically, the critical angle $ heta_{c12}$ is defined by $\sin heta_{c12} = \frac{n_1}{n_2}$. Since our incident angle $ heta$ is greater than this critical angle, we can write our **master condition**: $$n_2 \sin heta > n_1$$ This simple inequality is the cornerstone of our entire proof. Keep it close! ### Case A: The Impenetrable Wall ($n_3 < n_1$) Now, let's introduce the transparent slab (Medium III) with refractive index $n_3$. In our first case, we assume $n_3$ is even smaller than $n_1$. This means $n_3 < n_1 < n_2$. What happens at the boundary between Medium II and the new slab? We need to check the new critical angle, $ heta_{c23}$, which is given by $\sin heta_{c23} = \frac{n_3}{n_2}$. Because $n_3 < n_1$, it naturally follows that $\frac{n_3}{n_2} < \frac{n_1}{n_2}$. This implies that the new critical angle is smaller than the original one: $$ heta_{c23} < heta_{c12} < heta$$ Since our incident angle $ heta$ is already greater than the original critical angle, it is definitely greater than this new, smaller critical angle. Therefore, the light ray doesn't even penetrate the slab! It undergoes Total Internal Reflection right at the first boundary and is immediately reflected back into Medium II. ### Case B: The Deceptive Entry ($n_3 > n_1$) What if the slab has a higher refractive index, such that $n_3 > n_1$? In this scenario, the light might actually enter the slab. Let's assume it does, and it refracts at an angle $i$. We can apply Snell's law at the boundary between Medium II and Medium III: $$n_2 \sin heta = n_3 \sin i$$ Once the light is inside the slab, it travels towards the upper boundary separating Medium III and Medium I. Will it escape here? For TIR to occur at this upper boundary, the angle of incidence (which is $i$) must be greater than the critical angle for these two media, $ heta_{c31}$. The condition for TIR at the upper boundary is: $$\sin i > \sin heta_{c31} = \frac{n_1}{n_3}$$ Which can be rewritten as: $$n_3 \sin i > n_1$$ ### The Mathematical Climax Now, let's bring all our pieces together. From Snell's law, we know that $n_3 \sin i$ is exactly equal to $n_2 \sin heta$. And remember our master condition from the very beginning? We established that $n_2 \sin heta > n_1$. By simply substituting $n_3 \sin i$ for $n_2 \sin heta$ in our master condition, we get: $$n_3 \sin i > n_1$$ This is exactly the condition required for Total Internal Reflection at the upper boundary! ### Conclusion Because the condition is perfectly satisfied, the light ray cannot escape into Medium I. It undergoes TIR at the upper boundary, bounces back down through the slab, and finally refracts back into Medium II. We have elegantly proven that whether the slab's refractive index is lower or higher than Medium I, the light is ultimately trapped. For **any** value of $n_3$, the light will always find its way back to Medium II. The physics of light, governed by the beautiful simplicity of Snell's law, ensures its inevitable return.

Visualized Solution

Initial Setup \& Critical Angle

Case (a): $n_{3} < n_{1}$

Conclusion for Case (a)

Case (b): $n_{3} > n_{1}$

Condition for TIR at III-I Interface

Proving TIR at III-I Interface

Final Conclusion

The Sigma Insight: Refraction and Total Internal Reflection

The Inescapable Light

The Master Condition

Case A

Case B

The Mathematical Climax

Conclusion

Similar Questions

A ray of light entering from air into a denser medium of refractive index $\frac{4}{3}$ , as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle $θ$ should be equal to

A ray of light passes through four transparent media with refractive indices $μ_{1}$ , $μ_{2}$ , $μ_{3}$ and $μ_{4}$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray $C D$ is parallel to the incident ray $A B$ , we must have

AB and CD are two slabs. The medium between the slabs has refractive index 2. Find the minimum angle of incidence at Q, so that the ray is totally reflected by both the slabs.