OPTICS BYU 3

Objective: Find the minimum distance $d$ from the mirror such that the boy can see his full image.

Logic:

• Since the mirror is tilted backward with its bottom edge on the floor, the field of view is limited by the bottom edge.
• To see the full image, the boy must be able to see his lowest point (his feet).
• A light ray traveling from his feet to the bottom edge of the mirror must reflect and reach his eyes.

Let’s analyze the angles at the point of incidence (Origin $O$):

1. The mirror makes an angle $\theta$ with the vertical. Therefore, the normal to the mirror makes an angle $\theta$ with the horizontal.
2. The incident ray from the feet travels horizontally along the floor. Thus, the angle of incidence is $i = \theta$.
3. According to the law of reflection ($i = r$), the angle of reflection is also $r = \theta$.
4. The total angle between the incident ray (horizontal) and the reflected ray is $i + r = 2\theta$.

Now, consider the right-angled triangle formed by the boy’s position and the origin:

• Base: Distance from mirror $d$.
• Height: Height of eyes $h$.
• Angle of elevation of the reflected ray from the floor: $2\theta$.

Using trigonometry in this triangle:

$$ \tan(2\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{d} $$

Solving for the minimum distance $d$:

$$ d = \frac{h}{\tan(2\theta)} $$

Using the identity $\frac{1}{\tan \alpha} = \cot \alpha$: