Question

If $\hat{e_1}$ represents unit vector along the incident ray, $\hat{e_2}$ represents unit vector along the refracted ray and $\hat{n}$ represents unit vector along the normal at the point of incidence on the plane separation two media 1 and 2 having refractive indices $\mu_1$ and $\mu_2$, then, the snells law may be represented as

Answer 1

$\mu_1 (\hat{e_1} \times \hat{n}) = \mu_2 (\hat{e_2} \times \hat{n})$

Answer 2

Snell's Law describes the relationship between the angles of incidence and refraction for a ray of light passing through the interface between two isotropic media. In its scalar form, it is given by:

$\mu_1 \sin i = \mu_2 \sin r$

where $\mu_1$ and $\mu_2$ are the refractive indices of medium 1 and medium 2, respectively, $i$ is the angle of incidence, and $r$ is the angle of refraction.

We are given:

• $\hat{e_1}$: unit vector along the incident ray.
• $\hat{e_2}$: unit vector along the refracted ray.
• $\hat{n}$: unit vector along the normal at the point of incidence on the plane separation.

The angle of incidence $i$ is the angle between the incident ray ($\hat{e_1}$) and the normal ($\hat{n}$). The angle of refraction $r$ is the angle between the refracted ray ($\hat{e_2}$) and the normal ($\hat{n}$).

From the definition of the cross product of two vectors, the magnitude of the cross product of two unit vectors is equal to the sine of the angle between them. So, the magnitude of the cross product of $\hat{e_1}$ and $\hat{n}$ is:

$|\hat{e_1} \times \hat{n}| = |\hat{e_1}| |\hat{n}| \sin i = (1)(1) \sin i = \sin i$

Similarly, the magnitude of the cross product of $\hat{e_2}$ and $\hat{n}$ is:

$|\hat{e_2} \times \hat{n}| = |\hat{e_2}| |\hat{n}| \sin r = (1)(1) \sin r = \sin r$

Substituting these into the scalar form of Snell's Law:

$\mu_1 |\hat{e_1} \times \hat{n}| = \mu_2 |\hat{e_2} \times \hat{n}|$

One of the fundamental properties of refraction is that the incident ray, the refracted ray, and the normal to the interface at the point of incidence all lie in the same plane. This is often referred to as the "plane of incidence."

Since $\hat{e_1}$, $\hat{e_2}$, and $\hat{n}$ are coplanar, the vectors $(\hat{e_1} \times \hat{n})$ and $(\hat{e_2} \times \hat{n})$ are both perpendicular to this common plane. Consequently, these two cross product vectors must be parallel to each other, pointing in the same direction (or opposite directions, depending on the orientation chosen for $\hat{n}$ and the definition of angles).

Let's use a standard coordinate system to confirm the direction. Assume the interface is the x-z plane, and the normal $\hat{n}$ is along the positive y-axis, i.e., $\hat{n} = (0, 1, 0)$. Let the incident ray $\hat{e_1}$ be in the x-y plane. If the incident ray comes from $y>0$ towards $y=0$, and the angle $i$ is measured from the normal, then $\hat{e_1}$ can be represented as $(\sin i, -\cos i, 0)$. The refracted ray $\hat{e_2}$ will also be in the x-y plane, going into $y<0$. It can be represented as $(\sin r, -\cos r, 0)$.

Now, calculate the cross products:

$\hat{e_1} \times \hat{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \sin i & -\cos i & 0 \\ 0 & 1 & 0 \end{vmatrix} = \hat{i}(0-0) - \hat{j}(0-0) + \hat{k}(\sin i - 0) = (\sin i) \hat{k}$

$\hat{e_2} \times \hat{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \sin r & -\cos r & 0 \\ 0 & 1 & 0 \end{vmatrix} = \hat{i}(0-0) - \hat{j}(0-0) + \hat{k}(\sin r - 0) = (\sin r) \hat{k}$

Since $\mu_1 \sin i = \mu_2 \sin r$, and the directions of $(\hat{e_1} \times \hat{n})$ and $(\hat{e_2} \times \hat{n})$ are the same (both along $\hat{k}$), we can write the vector form of Snell's Law as:

$\mu_1 (\hat{e_1} \times \hat{n}) = \mu_2 (\hat{e_2} \times \hat{n})$

This equation encapsulates both parts of Snell's Law:

1. The incident ray, refracted ray, and normal are coplanar (implied by the vectors $\hat{e_1} \times \hat{n}$ and $\hat{e_2} \times \hat{n}$ being parallel).
2. The relationship between the magnitudes, $\mu_1 \sin i = \mu_2 \sin r$.