Question

A ray of light along $x+\sqrt{3}y=\sqrt{3}$ gets reflected upon reaching $x-$axis, the equation of the reflected ray is $\sqrt{3}y=x-\sqrt{\lambda }$ then $\lambda =?$.

Answer 1

Given equation of the line is,

$x+\sqrt{3}y=\sqrt{3}$ …… (1)

$y=1-\frac{x}{\sqrt{3}}$

Slope of the incident ray is $-\frac{1}{\sqrt{3}}$.

Therefore, slope of the reflected ray will be $\frac{1}{\sqrt{3}}$ and the point of incident will be $(\sqrt{3},0-)$.

Therefore, equation of the reflected ray will be,

$y-0=\frac{1}{\sqrt{3}}(x-\sqrt{3})$

$\sqrt{3}y=x-\sqrt{3}$

Comparing this equation with the given equation, we have

$\Rightarrow \lambda =3$

Hence, this is the required result.