Question

A plane mirror $AB$, point object $P$ and two observers $O_1$ and $O_2$ are positioned as shown below. Find minimum length (in $m$) of the mirror $AB$ required for which both the observers are able to see the image of $P$ in the mirror.

Answer 1

2

Answer 2

The image of the point object $P(2, 0)$ in the plane mirror $AB$ (which is on the y-axis, i.e., $x=0$) is $P'(-2, 0)$.

For an observer to see the image of $P$, the line of sight from the observer to the image $P'$ must intersect the mirror.

1. Observer $O_1$ at $(4, 2)$: The line segment connecting $O_1(4, 2)$ and $P'(-2, 0)$ represents the line of sight. We need to find the point where this line intersects the mirror (the y-axis, where $x=0$). The slope of the line $O_1P'$ is $m_1 = \frac{0 - 2}{-2 - 4} = \frac{-2}{-6} = \frac{1}{3}$. The equation of the line passing through $P'(-2, 0)$ with slope $m_1$ is $y - 0 = \frac{1}{3}(x - (-2))$, which simplifies to $y = \frac{1}{3}(x + 2)$. To find the intersection with the y-axis, set $x=0$: $y_1 = \frac{1}{3}(0 + 2) = \frac{2}{3}$. So, the line of sight from $O_1$ to $P'$ intersects the mirror at $M_1(0, 2/3)$.

2. Observer $O_2$ at $(4, 8)$: Similarly, for observer $O_2(4, 8)$, the line segment connecting $O_2$ and $P'(-2, 0)$ represents the line of sight. The slope of the line $O_2P'$ is $m_2 = \frac{0 - 8}{-2 - 4} = \frac{-8}{-6} = \frac{4}{3}$. The equation of the line passing through $P'(-2, 0)$ with slope $m_2$ is $y - 0 = \frac{4}{3}(x - (-2))$, which simplifies to $y = \frac{4}{3}(x + 2)$. To find the intersection with the y-axis, set $x=0$: $y_2 = \frac{4}{3}(0 + 2) = \frac{8}{3}$. So, the line of sight from $O_2$ to $P'$ intersects the mirror at $M_2(0, 8/3)$.

For both observers to see the image, the mirror $AB$ must cover the portion of the y-axis that includes both intersection points $M_1$ and $M_2$. Since the mirror is a continuous segment on the y-axis, its minimum length will be the distance between the y-coordinates of $M_1$ and $M_2$.

The y-coordinates of the intersection points are $2/3$ and $8/3$. The minimum length of the mirror $AB$ is the absolute difference between these y-coordinates: Minimum length $= |y_2 - y_1| = |\frac{8}{3} - \frac{2}{3}| = |\frac{6}{3}| = 2$ meters.

The mirror must extend from $y = 2/3$ to $y = 8/3$ on the y-axis.