The x-y plane is the boundary between two transparent media. Medium-1 with z = 0 has a refractive i

Q: Why do we separate the z-component from the x and y components?

Because the x-y plane is the boundary between the two media. The z-axis acts as the normal to this boundary. Separating them allows us to easily find the angle of incidence with the normal.

Q: How did we find the angle of incidence using $\tan i$?

In the plane of incidence, the incident ray forms a right-angled triangle with the normal (z-axis) and the x-y plane. The perpendicular is the magnitude of the x-y component, and the base is the magnitude of the z-component.

Q: Why is the z-component of the refracted ray negative?

The incident ray comes from $z \ge 0$ and travels into medium-2 where $z \le 0$. Therefore, the refracted ray must point downwards, in the negative z-direction.

Q: What is the significance of the unit vector $\hat{\mathbf{p}}$?

$\hat{\mathbf{p}}$ represents the direction of the incident ray's projection on the x-y plane. Since the refracted ray lies in the same plane of incidence, its horizontal component will also point in the direction of $\hat{\mathbf{p}}$.

Q: Can we use the vector form of Snell's law directly?

Yes, you can use $\mu_1 (\hat{\mathbf{i}} \times \hat{\mathbf{n}}) = \mu_2 (\hat{\mathbf{r}} \times \hat{\mathbf{n}})$, but decomposing the vectors geometrically provides a much clearer physical intuition of the ray's path.

Question

The $x$-$y$ plane is the boundary between two transparent media. Medium-1 with $z \ge 0$ has a refractive index $\sqrt{2}$ and medium-2 with $z \le 0$ has a refractive index $\sqrt{3}$. A ray of light in medium-1 given by 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. Find the unit vector in the direction of the refracted ray in medium-2.

Sigma Percentile · Accepted Answer

The problem of finding the path of a refracted ray in 3D space can seem daunting at first glance. However, by breaking down the vectors into their geometric components, we can transform a complex 3D problem into a simple 2D trigonometry exercise. Let's embark on this journey!

### The 3D Geometry of Light

Imagine you are standing at the boundary of two worlds. The $x$-$y$ plane acts as the floor beneath you. Above this floor is Medium-1 ($z \ge 0$) with a refractive index of $\mu_1 = \sqrt{2}$. Below the floor is Medium-2 ($z \le 0$) with a refractive index of $\mu_2 = \sqrt{3}$.

A ray of light shoots down from Medium-1, represented by the vector $\mathbf{A} = 6\sqrt{3}\hat{\mathbf{i}} + 8\sqrt{3}\hat{\mathbf{j}} - 10\hat{\mathbf{k}}$. Our mission is to find the exact direction this ray takes after it pierces through the floor into Medium-2.

### Deconstructing the Incident Ray

To understand how the ray hits the boundary, we need to find its angle of incidence. The angle of incidence is always measured with respect to the normal. Since the boundary is the $x$-$y$ plane, the normal is simply the $z$-axis.

Let's break our incident vector $\mathbf{A}$ into two distinct parts:
1. A component lying flat on the $x$-$y$ plane: $\mathbf{A}_1 = 6\sqrt{3}\hat{\mathbf{i}} + 8\sqrt{3}\hat{\mathbf{j}}$
2. A component pointing straight down along the $z$-axis: $\mathbf{A}_2 = -10\hat{\mathbf{k}}$

Now, let's find the lengths (magnitudes) of these components. 
The length of the horizontal component is $|\mathbf{A}_1| = \sqrt{(6\sqrt{3})^2 + (8\sqrt{3})^2} = \sqrt{108 + 192} = \sqrt{300} = 10\sqrt{3}$.
The length of the vertical component is $|\mathbf{A}_2| = \sqrt{(-10)^2} = 10$.

If we visualize this, the incident ray forms a right-angled triangle with the $z$-axis. The angle of incidence $i$ is the angle between the ray and the $z$-axis. Using basic trigonometry, we can find this angle:
$$ 	an i = \frac{|\mathbf{A}_1|}{|\mathbf{A}_2|} = \frac{10\sqrt{3}}{10} = \sqrt{3} $$
This tells us that the angle of incidence is exactly $i = 60^\circ$.

### Snell's Law in Action

Now that we know how the ray hits the boundary, we can use Snell's Law to find out how it bends. Snell's Law states:
$$ \mu_1 \sin i = \mu_2 \sin r $$

Let's plug in our known values:
$$ \sqrt{2} \sin 60^\circ = \sqrt{3} \sin r $$
$$ \sqrt{2} \left(\frac{\sqrt{3}}{2}ight) = \sqrt{3} \sin r $$

The $\sqrt{3}$ cancels out on both sides, leaving us with:
$$ \sin r = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} $$
This beautiful result means the angle of refraction is $r = 45^\circ$.

### Constructing the Refracted Vector

We know the refracted ray bends to an angle of $45^\circ$, but in which direction? The laws of optics tell us that the incident ray, the normal, and the refracted ray all lie in the same plane (the plane of incidence).

This means the horizontal direction of the refracted ray will be exactly the same as the horizontal direction of the incident ray. Let's find the unit vector $\hat{\mathbf{p}}$ that points in this horizontal direction:
$$ \hat{\mathbf{p}} = \frac{\mathbf{A}_1}{|\mathbf{A}_1|} = \frac{6\sqrt{3}\hat{\mathbf{i}} + 8\sqrt{3}\hat{\mathbf{j}}}{10\sqrt{3}} = \frac{3\hat{\mathbf{i}} + 4\hat{\mathbf{j}}}{5} $$

Finally, we can construct the unit vector $\hat{\mathbf{n}}$ for the refracted ray. It will have a horizontal component of $\sin r$ in the direction of $\hat{\mathbf{p}}$, and a vertical component of $\cos r$ pointing downwards (in the $-\hat{\mathbf{k}}$ direction):
$$ \hat{\mathbf{n}} = (\sin r)\hat{\mathbf{p}} - (\cos r)\hat{\mathbf{k}} $$

Substituting $r = 45^\circ$:
$$ \hat{\mathbf{n}} = \frac{1}{\sqrt{2}} \left( \frac{3\hat{\mathbf{i}} + 4\hat{\mathbf{j}}}{5} ight) - \frac{1}{\sqrt{2}}\hat{\mathbf{k}} $$

Factoring out the common terms, we arrive at our final, elegant answer:
$$ \hat{\mathbf{n}} = \frac{1}{5\sqrt{2}} (3\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}}) $$

By breaking the 3D problem into a 2D plane of incidence, we bypassed complex vector cross-products and solved the problem with pure geometric intuition!

Visualized Solution

$Visualizing the Setup$

$Decomposing the Incident Ray$

$Magnitudes of Components$

$Angle of Incidence (i)$

$Applying Snell’s Law$

$Unit Vector in x - y Plane (\hat{p})$

$Refracted Ray Unit Vector (\hat{n})$

$Conclusion$

The Sigma Insight: Refraction and Total Internal Reflection

The 3D Geometry of Light

Deconstructing the Incident Ray

Snell's Law in Action

Constructing the Refracted Vector

Similar Questions

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

A ray of light $A O$ in vacuum is incident on a glass slab at angle $6 0^{\circ}$ and refracted at angle $3 0^{\circ}$ along $O B$ as shown in the figure. The optical path length of light ray from $A$ to $B$ is

A transparent solid cylinder rod has a refractive index of $\frac{2}{3}$ . It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle $θ$ for which the light ray grazes along the wall of the rod is

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decomposing the Incident Ray

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnitudes of Components

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle of Incidence (i)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Snell’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Unit Vector in x-y Plane (p^​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Refracted Ray Unit Vector (n^)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion