QJEEM13Ma516f: JEE Main 2013 Mathematics - Class 11 JEEM13Ma516f Solution: A ray of light along gets reflected upon reaching -axis, the…

$x + \sqrt 3 y = \sqrt 3$ or $y = - {1 \over {\sqrt 3 }}x + 1$

Let $\theta$ be the angle which the line makes with the positive x-axis.

$\therefore$ $\tan \theta = - {1 \over {\sqrt 3 }} = \tan \left( {\pi - {\pi \over 6}} \right)$ or $\theta = \pi - {\pi \over 6}$

$\therefore$ $\angle ABC = {\pi \over 6}$ ; $\therefore$ $\angle DBE = {\pi \over 6}$

$\therefore$ the equation of the line BD is,

$y = \tan {\pi \over 6}x + c$ or $y = {x \over {\sqrt 3 }} + c$ ..... (1)

The line $x + \sqrt 3 y = \sqrt 3$ intersects the x-axis at $B(\sqrt 3 ,0)$ and, the line (1) passes through $B(\sqrt 3 ,0)$ .

$\therefore$ $0 = {{\sqrt 3 } \over {\sqrt 3 }} + c$ or, c = $-$ 1

Hence, the equation of the reflected ray is,

$y = {x \over {\sqrt 3 }} - 1$ or $y\sqrt 3 = x - \sqrt 3$