1. Given line of incident ray

The ray travels along $x + \sqrt{3}\,y = \sqrt{3}.$

Rewrite in slope-intercept form: $\sqrt{3}\,y = -x + \sqrt{3}$ $y = -\frac{1}{\sqrt{3}}x + 1.$

So the slope of the incident ray is $m_1 = -\frac{1}{\sqrt{3}}.$

2. Reflection from the $x$-axis

Reflection in the $x$-axis changes every point $(x,y)$ to $(x,-y)$.

For a line making angle $\theta$ with the positive $x$-axis, reflection in the $x$-axis changes the angle to $-\theta$, hence the slope changes from $m$ to $-m$.

Therefore the reflected ray has slope $m_2 = -m_1 = \frac{1}{\sqrt{3}}.$

3. Point where the ray hits the $x$-axis

On the $x$-axis, $y=0$. Substitute into the given line: $x + \sqrt{3}(0) = \sqrt{3}$ $x = \sqrt{3}.$

So the point of reflection is $(\sqrt{3},0).$

4. Equation of reflected ray

Using point-slope form with slope $\frac{1}{\sqrt{3}}$ through $(\sqrt{3},0)$: $y - 0 = \frac{1}{\sqrt{3}}(x-\sqrt{3}).$

Multiply by $\sqrt{3}$: $\sqrt{3}\,y = x - \sqrt{3}.$

5. Match with options

This is exactly Option B.

Correct answer: $\boxed{\sqrt{3}\,y = x - \sqrt{3}}$