A transparent cube of side d, made of a material of refractive index mu_2, is immersed in a liquid o

Q: Why is the angle of incidence at the top face $90^\circ - r$?

The normal to the vertical face AB is horizontal, and the normal to the horizontal face BC is vertical. These two normals form a $90^\circ$ angle. The refracted ray forms a right-angled triangle with these normals, so the two acute angles $r$ and $i$ must add up to $90^\circ$.

Q: Why do we use $\cos r > \sin C$ instead of $\sin(90^\circ - r) > \sin C$?

They are exactly the same thing! $\sin(90^\circ - r)$ is a standard trigonometric identity that equals $\cos r$. Using $\cos r$ makes it easier to substitute our expression for $\sin r$ from Snell's law later.

Q: What happens if $\theta$ is exactly equal to the value we found?

If $\theta$ is exactly equal to that value, the ray will hit the top face exactly at the critical angle. The refracted ray will graze the surface of the face BC (angle of refraction will be $90^\circ$).

Q: Why did the inequality sign flip when we rearranged the terms?

When we moved the negative term $-\frac{\mu_1^2}{\mu_2^2} \sin^2 \theta$ to the other side, it became positive. We then read the inequality from right to left, which naturally flips the direction of the symbol relative to the terms.

Q: Can this formula be used if the cube is in air?

Absolutely! If the cube is in air, the refractive index of the surrounding medium $\mu_1$ is simply 1. You can substitute $\mu_1 = 1$ into the final formula to find the condition for a cube in air.

Question

A transparent cube of side $d$, made of a material of refractive index $\mu_2$, is immersed in a liquid of refractive index $\mu_1 (\mu_1 < \mu_2)$. A ray is incident on the face $AB$ at an angle $ heta$ (shown in the figure). Total internal reflection takes place at point $E$ on the face $BC$. Then, $ heta$ must satisfy Options: (A) $ heta < \sin^{-1} \frac{\mu_1}{\mu_2}$ (B) $ heta > \sin^{-1} \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$ (C) $ heta < \sin^{-1} \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$ (D) $ heta > \sin^{-1} \frac{\mu_1}{\mu_2}$

Sigma Percentile · Accepted Answer

### The Geometry of Trapped Light: Mastering Total Internal Reflection in a Cube Imagine you are a photon of light, swimming through a liquid of refractive index $\mu_1$. Suddenly, you encounter a dense, transparent cube made of a material with a higher refractive index, $\mu_2$. You strike the vertical face of this cube at an angle $ heta$. What happens next is a beautiful dance of optics and geometry. Our goal in this problem is to find the exact condition for $ heta$ such that when the light ray reaches the top face of the cube, it doesn't escape back into the liquid. Instead, it must undergo **Total Internal Reflection (TIR)**, bouncing back entirely into the cube. Let's break down this journey step by step. ### The Boundary Condition: Total Internal Reflection Let's fast-forward to the critical moment: the ray hitting the top face $BC$ at point $E$. For the ray to be trapped inside the cube, the angle at which it strikes this top boundary—let's call it the angle of incidence $i$—must be strictly greater than the critical angle $C$ for the $\mu_2$ to $\mu_1$ interface. Mathematically, this is our absolute requirement: $$i > C$$ We know from the principles of optics that the critical angle is defined by the ratio of the refractive indices: $$\sin C = \frac{\mu_1}{\mu_2}$$ ### The Geometric Link Now, let's rewind and look at the path the ray took inside the cube. The ray entered through the vertical face $AB$ and refracted at an angle $r$. It then traveled in a straight line to hit the horizontal top face $BC$ at an angle $i$. Here is where the geometry of the cube becomes our greatest tool. The normal to the vertical face $AB$ is perfectly horizontal, while the normal to the horizontal face $BC$ is perfectly vertical. These two normals intersect at a perfect $90^\circ$ angle. If you trace the path of the refracted ray between these two normals, you'll see it forms a right-angled triangle. Because the sum of angles in a triangle is $180^\circ$, and one angle is $90^\circ$, the two acute angles must add up to $90^\circ$. These two acute angles are exactly our angle of refraction $r$ and our angle of incidence $i$! $$r + i = 90^\circ \implies i = 90^\circ - r$$ Let's substitute this geometric truth back into our TIR condition: $$90^\circ - r > C$$ Taking the sine of both sides gives us: $$\sin(90^\circ - r) > \sin C$$ Using the fundamental trigonometric identity $\sin(90^\circ - x) = \cos x$, we arrive at a powerful inequality: $$\cos r > \frac{\mu_1}{\mu_2}$$ ### Tracing Back to the Source: Snell's Law We have a condition for $r$, but the problem asks for a condition on our initial angle $ heta$. To bridge this gap, we must apply **Snell's Law** at the very first interface, face $AB$, where the ray entered the cube. $$\mu_1 \sin heta = \mu_2 \sin r$$ Rearranging this to isolate $\sin r$, we get: $$\sin r = \frac{\mu_1}{\mu_2} \sin heta$$ ### The Mathematical Synthesis We now have an inequality involving $\cos r$ and an equation for $\sin r$. To combine them, we use the Pythagorean identity $\cos r = \sqrt{1 - \sin^2 r}$. Substituting our expression from Snell's law into this identity yields: $$\cos r = \sqrt{1 - \left(\frac{\mu_1}{\mu_2} \sin heta ight)^2}$$ Now, we plug this massive expression back into our TIR inequality: $$\sqrt{1 - \frac{\mu_1^2}{\mu_2^2} \sin^2 heta} > \frac{\mu_1}{\mu_2}$$ To solve for $ heta$, we must carefully unravel this inequality. First, square both sides to eliminate the square root: $$1 - \frac{\mu_1^2}{\mu_2^2} \sin^2 heta > \frac{\mu_1^2}{\mu_2^2}$$ Next, rearrange the terms to isolate the term containing $ heta$: $$1 - \frac{\mu_1^2}{\mu_2^2} > \frac{\mu_1^2}{\mu_2^2} \sin^2 heta$$ Find a common denominator for the left side: $$\frac{\mu_2^2 - \mu_1^2}{\mu_2^2} > \frac{\mu_1^2}{\mu_2^2} \sin^2 heta$$ Notice how beautifully the $\mu_2^2$ denominators cancel out on both sides. This leaves us with: $$\mu_2^2 - \mu_1^2 > \mu_1^2 \sin^2 heta$$ Divide by $\mu_1^2$ to completely isolate $\sin^2 heta$: $$\frac{\mu_2^2 - \mu_1^2}{\mu_1^2} > \sin^2 heta$$ Which can be rewritten as: $$\sin^2 heta < \frac{\mu_2^2}{\mu_1^2} - 1$$ Taking the square root of both sides gives us the final constraint on the sine of our initial angle: $$\sin heta < \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$$ Finally, taking the inverse sine reveals the ultimate condition that $ heta$ must satisfy to ensure the light ray is trapped by total internal reflection: $$ heta < \sin^{-1} \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$$ And there we have it! By weaving together Snell's Law, the geometry of a cube, and the condition for Total Internal Reflection, we've successfully decoded the path of the light ray.

Select Answer:

Visualized Solution

Visualizing the Setup

Condition for Total Internal Reflection

Geometric Relationship of Angles

Substituting into TIR Condition

Applying Snell's Law

Trigonometric Substitution

Solving the Inequality

Final Simplification

The Way Forward

The Sigma Insight: Refraction and Total Internal Reflection

The Geometry of Trapped Light

The Boundary Condition

The Geometric Link

Tracing Back to the Source

The Mathematical Synthesis

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