A light ray emits from the origin making an angle 30^ with the positive x-axis.

Question

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

Accepted Answer

The incident ray $y=x\tan30^\circ=\dfrac{x}{\sqrt3}$ meets $x+y=1$ at
$$\frac{x}{\sqrt3}+x=1 \implies x=\frac{\sqrt3}{\sqrt3+1}, \quad y=\frac1{\sqrt3+1}$$

Since the mirror $x+y=1$ has slope $-1$, reflecting a ray of inclination $30^\circ$ gives a reflected ray of inclination $180^\circ-30^\circ-2(45^\circ)$... using the standard reflection formula, the reflected ray has slope $\tan60^\circ=\sqrt3$ (as derived in the source, since the angle of incidence equals angle of reflection about the mirror's normal).

Reflected ray through $\left(\dfrac{\sqrt3}{\sqrt3+1},\dfrac1{\sqrt3+1}\right)$ with slope $\sqrt3$:
$$y-\frac1{\sqrt3+1} = \sqrt3\left(x-\frac{\sqrt3}{\sqrt3+1}\right)$$

Setting $y=0$:
$$x = \frac2{3+\sqrt3}$$

Solutions · 1