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 intersects x-axis at Q, then the abscissa of Q is

Question

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 intersects x-axis at Q, then the abscissa of Q is (A) (2/(√3-1)) (B) (√3/2(√3+1)) (C) (2/3+√3) (D) (2/3-√3)

Tardigrade Admin · Accepted Answer

Slope of reflected ray = tan 60°=√3 Line y=(x/√3) intersect y+x=1 at ((√3/√3+1), (1/√3+1)) Equation of reflected ray is y-(1/√3+1)=√3(x-(√3/√3+1)) Put y =0 ⇒ x =(2/3+√3)

Q. 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

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

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

$\frac{2}{3 + 3}$

$\frac{2}{3 - 3}$

Slope of reflected ray $= tan 6 0^{\circ} = 3$
Line $y = \frac{x}{3}$ intersect $y + x = 1$ at $(\frac{3}{3 + 1}, \frac{1}{3 + 1})$
Equation of reflected ray is
$y - \frac{1}{3 + 1} = 3 (x - \frac{3}{3 + 1})$
Put $y = 0 \Rightarrow x = \frac{2}{3 + 3}$

Q. 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 intersects x-axis at Q, then the abscissa of Q is

(3​−1)2​

2(3​+1)3​​

3+3​2​

3−3​2​