A light ray emits from the origin making an angle 30° with the positive x-axis. After getting reflected by the line $x+y=1$, if this ray intersect x-axis at Q, then the abscissa of Q is

$\begin{array}{l}\text{ Incident ray is }\Rightarrow \mathrm{y}=\frac{1}{\sqrt{3}}\mathrm{x}\\ \mathrm{x}-\sqrt{3}\mathrm{y}=0\\ \text{ L.M : }\mathrm{x}+\mathrm{y}-1=0\\ \text{ Line }y=\frac{x}{\sqrt{3}}\text{ intersect }y+x=1\text{ at }\left(\frac{\sqrt{3}}{\sqrt{3}+1},\frac{1}{\sqrt{3}+1}\right)\\ \text{ Slope of reflected ray }=\tan {60}^{\circ }=\sqrt{3}\\ \text{ Reflected .Ray is : }y-\frac{1}{\sqrt{3}+1}=\sqrt{3}\left(x-\frac{\sqrt{3}}{\sqrt{3}+1}\right)\\ \\ \mathrm{substituting} \mathrm{y}=0 \mathrm{in} \mathrm{reflected} \mathrm{ray}\\ \\ \Rightarrow \boxed{x=\frac{2}{3+\sqrt{3}}}\end{array}$