Let the zx-plane be the boundary between two transparent media. Medium 1 in z = 0 has a refractive

Q: The question says the zx-plane is the boundary. Why did we use the xy-plane?

Great observation! There's a slight contradiction in the problem statement. It says 'zx-plane' but then defines the media using $z \ge 0$ and $z < 0$. The mathematical inequalities ($z \ge 0$ and $z < 0$) strictly define the boundary as the plane where $z=0$, which is the xy-plane. In physics problems, always trust the mathematical constraints!

Q: Why do we only use the z-component to find the angle of incidence?

The angle of incidence is defined as the angle between the incident ray and the normal to the surface. Since our boundary is the xy-plane, the normal is the z-axis. The angle a vector makes with the z-axis can be found using its z-component: $\cos \theta = \frac{A_z}{|\mathbf{A}|}$.

Q: What if the ray was moving upwards instead of downwards?

If the ray was moving from $z 0$, its z-component would be positive. The process remains exactly the same. We take the absolute value of the z-component to find the acute angle with the normal.

Q: Can we use the dot product to find the angle of incidence?

Absolutely! The formula $\cos i = \frac{|A_z|}{|\mathbf{A}|}$ is actually derived from the dot product. $\mathbf{A} \cdot \hat{\mathbf{k}} = |\mathbf{A}| |\hat{\mathbf{k}}| \cos i$. Since $\hat{\mathbf{k}}$ is a unit vector, it simplifies to our formula.

Q: Why did the $\sqrt{3}$ cancel out so perfectly in Snell's Law?

That's the beauty of well-designed JEE problems! The examiners often choose values that lead to elegant mathematical cancellations, rewarding students who set up the equations correctly rather than getting bogged down in messy decimals.

Question

Let the zx-plane be the boundary between two transparent media. Medium 1 in $z \ge 0$ has a refractive index of $\sqrt{2}$ and medium 2 with $z < 0$ has a refractive index of $\sqrt{3}$. A ray of light in medium 1 given by the vector $\mathbf{A} = 6\sqrt{3}\hat{\mathbf{i}} + 8\sqrt{3}\hat{\mathbf{j}} - 10\hat{\mathbf{k}}$ is incident on the plane of separation. The angle of refraction in medium 2 is

Options:

(A) $45^\circ$

(B) $60^\circ$

(C) $75^\circ$

(D) $30^\circ$

Sigma Percentile · Accepted Answer

The journey of mastering ray optics often takes us from simple 2D diagrams into the fascinating world of 3D vectors. This problem is a brilliant example of how vector geometry and Snell's Law intertwine to describe the physical reality of light refraction. Let's dive into the elegant mechanics of this problem.

### Decoding the Boundary

The very first step in any optics problem is to establish our geometry. The problem statement presents a classic trap: it mentions the "zx-plane" as the boundary, but immediately follows up by defining the two media as $z \ge 0$ and $z < 0$.

Which one do we trust? In physics, mathematical constraints always overrule descriptive text. The inequalities $z \ge 0$ and $z < 0$ unequivocally define the boundary as the plane where $z = 0$. This is the **xy-plane**.

Because the boundary is the xy-plane, the normal to this surface is simply the **z-axis**, represented by the unit vector $\hat{\mathbf{k}}$. This realization is the crucial key that unlocks the rest of the problem.

### The Geometry of the Incident Ray

We are given the incident ray as a vector: 
$\mathbf{A} = 6\sqrt{3}\hat{\mathbf{i}} + 8\sqrt{3}\hat{\mathbf{j}} - 10\hat{\mathbf{k}}$

To use Snell's Law, we need the angle of incidence, $i$. By definition, this is the angle between the incident ray and the normal. Since our normal is the z-axis, we can find this angle using the fundamental property of vectors—the direction cosine.

First, let's find the magnitude of our incident ray vector:
$|\mathbf{A}| = \sqrt{(6\sqrt{3})^2 + (8\sqrt{3})^2 + (-10)^2}$
$|\mathbf{A}| = \sqrt{108 + 192 + 100} = \sqrt{400} = 20$

Now, the cosine of the angle with the z-axis is simply the absolute value of the z-component divided by the total magnitude:
$\cos i = \frac{|A_z|}{|\mathbf{A}|} = \frac{|-10|}{20} = \frac{1}{2}$

What angle gives a cosine of $\frac{1}{2}$? Exactly, $i = 60^\circ$. The vector geometry has beautifully collapsed into a simple, familiar angle.

### The Magic of Snell's Law

With our angle of incidence secured, we cross the boundary using Snell's Law. The law states that the product of the refractive index and the sine of the angle is conserved across the interface:
$\mu_1 \sin i = \mu_2 \sin r$

We plug in our known values: $\mu_1 = \sqrt{2}$, $i = 60^\circ$, and $\mu_2 = \sqrt{3}$.
$\sqrt{2} \sin 60^\circ = \sqrt{3} \sin r$

Substituting the value of $\sin 60^\circ$:
$\sqrt{2} \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} \sin r$

Notice the elegant design of the problem—the $\sqrt{3}$ terms on both sides cancel out perfectly!
$\frac{\sqrt{3}}{\sqrt{2}} = \sqrt{3} \sin r$
$\sin r = \frac{1}{\sqrt{2}}$

This leaves us with a fundamental trigonometric value. The angle whose sine is $\frac{1}{\sqrt{2}}$ is $45^\circ$.

**Therefore, the angle of refraction in the second medium is $45^\circ$.**

This problem beautifully demonstrates that no matter how complex a 3D vector might look, the underlying physical principles like Snell's Law remain elegantly simple and universally applicable.

Select Answer:

Visualized Solution

$Identifying the Boundary and Normal$

$Angle of Incidence Formula$

$Magnitude of Incident Ray Vector$

$Calculating Angle of Incidence$

$Applying Snell’s Law$

$Calculating Angle of Refraction$

The Sigma Insight: Refraction and Total Internal Reflection

Decoding the Boundary

The Geometry of the Incident Ray

The Magic of Snell's Law

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