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Step-by-step Solution
Step 1: Identify the normal to the boundary
The boundary between the two media is given as the plane $z = 0$.
The normal to this plane points along the $+ \hat{k}$ direction.
Since the incident ray travels from the region $z \ge 0$ into $z < 0$,
it moves in the negative $z$-direction.
Step 2: Express the incident ray vector and find its magnitude
The incident ray in medium 1 is given by the direction vector
$ \vec{A} = 6\sqrt{3}\,\hat{i} + 8\sqrt{3}\,\hat{j} - 10\,\hat{k} $.
The magnitude of $ \vec{A} $ is:
$$
|\vec{A}|
= \sqrt{(6\sqrt{3})^2 + (8\sqrt{3})^2 + (-10)^2}
= \sqrt{108 + 192 + 100}
= \sqrt{400}
= 20.
$$
Step 3: Determine the angle of incidence
The angle of incidence $i$ is measured between the direction of the ray
and the outward normal to the plane. The outward normal to the $z=0$ plane (on the $z\ge0$ side)
is $+\hat{k}$.
Compute the dot product:
$$
\vec{A} \cdot \hat{k}
= (6\sqrt{3})(0) + (8\sqrt{3})(0) + (-10)(1)
= -10.
$$
Since $|\hat{k}| = 1$ and $|\vec{A}| = 20$,
$$
\cos \theta = \frac{\vec{A} \cdot \hat{k}}{|\vec{A}|\cdot|\hat{k}|}
= \frac{-10}{20 \times 1}
= -\frac{1}{2}.
$$
Therefore, $\theta = 120^\circ$ is the angle between $\vec{A}$ and $+\hat{k}$.
However, the angle of incidence $i$ is conventionally taken as the acute angle with the normal.
Thus,
$$
i = 180^\circ - 120^\circ = 60^\circ.
$$
Step 4: Apply Snell's law
Snell's law states:
$$
n_1 \,\sin i \;=\; n_2 \,\sin r,
$$
where
$n_1 = \sqrt{2}$ (refractive index of medium 1),
$n_2 = \sqrt{3}$ (refractive index of medium 2),
$i = 60^\circ$ (angle of incidence),
and $r$ is the angle of refraction.
Step 5: Solve for the angle of refraction
Substitute the known values into Snell’s law:
$$
\sqrt{2} \,\sin 60^\circ \;=\; \sqrt{3} \,\sin r.
$$
Recall that $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Hence,
$$
\sqrt{2} \times \frac{\sqrt{3}}{2}
= \sqrt{3} \,\sin r
\quad \Longrightarrow \quad
\frac{\sqrt{6}}{2}
= \sqrt{3}\,\sin r.
$$
Divide both sides by $\sqrt{3}$:
$$
\sin r
= \frac{\sqrt{6}}{2 \,\sqrt{3}}
= \frac{\sqrt{6}}{2\sqrt{3}}
= \frac{\sqrt{2}}{2}
= \frac{1}{\sqrt{2}}.
$$
Therefore,
$$
r = 45^\circ.
$$
Step 6: Final Answer
The angle of refraction in medium 2 is $45^\circ$.
