Question

A meter stick (object) is placed in an upright position in front of a plane mirror as shown in the diagram. The image of the meter stick is equidistant from the mirror. Suppose that the meter stick is equipped with a working eyeball capable of viewing the top and the bottom of its image. The eyeball is located at the $90\text{cm}$ mark on the meter stick. Determine the location of the intersection of the eye's line of sight with the mirror as the eyeball sights at the top of the image.

• A. $50\text{cm}$
• B. $45\text{cm}$
• C. $90\text{cm}$
• D. $95\text{cm}$

Answer 1

The correct option is D $95\text{cm}$

Let us show the given situation geometrically as :


From the diagram it is clear that the point of intersection of the eye's line of sight with the mirror is $B$ if, eyeball sights at the top of the image.
Therefore, to find its location we need to find the length $BD$.
From the geometry $\Delta ABO\simeq \Delta CBO$
$\Rightarrow AO=OC=\frac{AC}{2}=\frac{10}{2}=5\text{cm}$
Hence, $BD=OE=OC+CE$
$\Rightarrow BD=5+90=95\text{cm}$