A light ray emerging from a point source at A | vedclass

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(B) Let the point of reflection $B$ be $(0, y)$. By the law of reflection,the angle of incidence equals the angle of reflection. Let the normal at $B$

Vedclass · Accepted Answer

$(0,5)$

Vedclass · Answer

(B) Let the point of reflection $B$ be $(0, y)$. By the law of reflection,the angle of incidence equals the angle of reflection. Let the normal at $B$ be the horizontal line $y = y_B$. The slope of line $AB$ is $m_1 = \frac{y-3}{0-2} = \frac{y-3}{-2}$. The slope of line $BC$ is $m_2 = \frac{10-y}{5-0} = \frac{10-y}{5}$. Since the angle of incidence equals the angle of reflection with respect to the normal (horizontal line),the slopes must be negatives of each other: $m_1 = -m_2$. $\frac{y-3}{-2} = -\left(\frac{10-y}{5}\right)$ $\frac{y-3}{-2} = \frac{y-10}{5}$ $5(y-3) = -2(y-10)$ $5y - 15 = -2y + 20$ $7y = 35$ $y = 5$ Thus,the coordinates of $B$ are $(0, 5)$.

$A$ light ray emerging from a point source at $A(2,3)$ is reflected on the $Y$-axis at point $B$ and passes through point $C(5,10)$. Then the coordinates of $B$ are:

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