A ray of light along x+sqrt{3}y=sqrt{3} gets | vedclass

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(B) The given equation of the incident ray is $x+\sqrt{3}y=\sqrt{3}$.To find the point of incidence $A$ on the $x$-axis, set $y=0$ in the equation: $x

Vedclass · Accepted Answer

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

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(B) The given equation of the incident ray is $x+\sqrt{3}y=\sqrt{3}$.To find the point of incidence $A$ on the $x$-axis, set $y=0$ in the equation: $x+\sqrt{3}(0)=\sqrt{3} \Rightarrow x=\sqrt{3}$. Thus, $A \equiv (\sqrt{3}, 0)$.To find another point $B$ on the incident ray, set $x=0$: $0+\sqrt{3}y=\sqrt{3} \Rightarrow y=1$. Thus, $B \equiv (0, 1)$.The reflected ray passes through $A(\sqrt{3}, 0)$ and the reflection of point $B$ across the $x$-axis, which is $B' \equiv (0, -1)$.The equation of the line passing through $A(\sqrt{3}, 0)$ and $B'(0, -1)$ is given by the two-point form: $\frac{y-0}{x-\sqrt{3}} = \frac{-1-0}{0-\sqrt{3}}$.$\frac{y}{x-\sqrt{3}} = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}}$.$\sqrt{3}y = x-\sqrt{3}$.

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

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