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

(D) $\frac{\sqrt{3}}{2(\sqrt{3}+1)}$

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Straight Lines

A light ray emits from the origin making an angle

3 0^{\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

Answer: (B)

The incident ray

y = \frac{x}{3}

meets

x + y = 1

(\frac{3}{3 + 1}, \frac{1}{3 + 1})

. Reflecting the incident direction

(cos 3 0^{\circ}, sin 3 0^{\circ})

in the mirror

x + y = 1

, whose normal is along

(1, 1)

, gives the direction

(- \frac{1}{2}, - \frac{3}{2})

, so the reflected ray has slope

3

. Writing this ray through the point of incidence and setting

y = 0

- \frac{1}{3 + 1} = 3 (x - \frac{3}{3 + 1}) \Rightarrow 3 x = \frac{2}{3 + 1},

so the abscissa of

Q

\frac{2}{3 + 3}

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option