A light ray along the line x + sqrt{3}y = sqr | vedclass

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(B) The given line is $x + \sqrt{3}y = \sqrt{3}$,which can be written as $y = -\frac{1}{\sqrt{3}}x + 1$.The slope of this line is $m_1 = -\frac{1}{\sq

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The given line is $x + \sqrt{3}y = \sqrt{3}$,which can be written as $y = -\frac{1}{\sqrt{3}}x + 1$.The slope of this line is $m_1 = -\frac{1}{\sqrt{3}}$,so the angle it makes with the $x$-axis is $	heta = 150^{\circ}$ or $-30^{\circ}$.The ray hits the $x$-axis at the point where $y = 0$,which is $x = \sqrt{3}$. So the point of incidence is $(\sqrt{3}, 0)$.According to the law of reflection,the reflected ray makes an angle of $+30^{\circ}$ with the $x$-axis. Thus,its slope is $m_2 = 	an(30^{\circ}) = \frac{1}{\sqrt{3}}$.The equation of the reflected line passing through $(\sqrt{3}, 0)$ with slope $m_2 = \frac{1}{\sqrt{3}}$ is:$y - 0 = \frac{1}{\sqrt{3}}(x - \sqrt{3})$$\sqrt{3}y = x - \sqrt{3}$.

$A$ light ray along the line $x + \sqrt{3}y = \sqrt{3}$ reaches the $x$-axis and gets reflected. Find the equation of the reflected ray.

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