Tilt Angle of a Ground Mirror Reflecting Light Between Two Towers: 04-BS-9, December 2016, Q8

Question 8: Tilt Angle of a Ground Mirror Reflecting Light Between Two Towers (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Tower A height	$300$ m, at $x=0$
Tower B height	$200$ m, at $x=600$ m
Mirror position	ground level, $x=300$ m (half-way)

Find. The mirror's tilt angle $\varphi$ from horizontal ground.

A flat, HORIZONTAL mirror at the midpoint would naturally reflect at $x\approx360$m (equal-angle ratio $h_A/h_B=300/200$), not at the stated $x=300$m; the mirror must therefore be tilted slightly to shift the reflection point to the midpoint.

Approach. Use the vector reflection law at the fixed point $P=(300,0)$: the mirror's unit tangent $\hat u=(\cos\varphi,\sin\varphi)$ defines a normal $\hat n=(-\sin\varphi,\cos\varphi)$; the incident ray $A\to P$ must reflect (about $\hat n$) into a ray parallel to $P\to B$. Solve the resulting trig equation for $\varphi$ numerically (there are two roots; pick the physically sensible near-horizontal one for a mirror "located on the ground").

Set up the reflection condition. With $A=(0,300)$, $P=(300,0)$, $B=(600,200)$, reflecting the incident direction $\vec d_{\text{in}}=P-A$ about the mirror's normal $\hat n(\varphi)$ must give a vector parallel to $\vec d_{\text{out}}=B-P$. Expanding the reflection formula $\vec d_{\text{in}}-2(\vec d_{\text{in}}\cdot\hat n)\hat n\parallel\vec d_{\text{out}}$ and simplifying (cross-product $=0$) reduces to $$\tan^2\varphi-10\tan\varphi-1=0$$
Solve the quadratic in $t=\tan\varphi$. $$t=\frac{10\pm\sqrt{100+4}}{2}=\frac{10\pm\sqrt{104}}{2}\ \Rightarrow\ t=10.099\ \text{or}\ t=-0.0990$$ giving $\varphi=84.35^\circ$ or $\varphi=-5.655^\circ$.
Select the physical root. A mirror "located on the ground" is a near-horizontal reflector, not an $84^\circ$ near-vertical plane (a spurious/extraneous root the reflection algebra also satisfies but which contradicts the problem's own description); the small-angle root is confirmed by direct substitution (the reflected ray from $P$ at $\varphi=-5.655^\circ$ is verified colinear with $B$ to within numerical precision): $$\boxed{\varphi=5.655^\circ\ \text{(tilted below horizontal, sloping down from the 300m-tower side toward the 200m-tower side)}}$$

Quantity	Result
Mirror tilt from horizontal, $\|\varphi\|$	$5.655^\circ$ (physical root; the algebra's other root, $84.35^\circ$, is extraneous)

Check: the reflection equation is quadratic in $\tan\varphi$ and has two mathematically valid roots; the near-horizontal root ($5.66^\circ$) is selected because the problem states the mirror is "located on the ground," which rules out the near-vertical ($84.3^\circ$) alternative. Both roots were confirmed by direct substitution back into the reflection law.

Concept & Theory

Topic. Law of reflection (angle of incidence = angle of reflection) applied to a TILTED plane mirror at a fixed, prescribed reflection point — solved via the vector reflection formula rather than the simpler equal-angle geometry that only applies to a horizontal mirror with a free reflection point.

Key relations. Vector reflection of a direction $\vec d$ about a unit normal $\hat n$: $\vec d'=\vec d-2(\vec d\cdot\hat n)\hat n$; for a mirror with tangent angle $\varphi$, $\hat n=(-\sin\varphi,\cos\varphi)$.

Why this works. For a horizontal mirror, the equal-angle condition alone determines a UNIQUE reflection point (via the similar-triangles ratio $h_A/h_B$) that need not coincide with any particular location; here the reflection point is instead FIXED (given as the midpoint), so the free parameter becomes the mirror's tilt, and satisfying the reflection law at that one fixed point is what pins down $\varphi$.

Common pitfall. Assuming a flat, horizontal mirror automatically satisfies equal angles at any specified point (true only at the ONE point the geometry naturally picks, here $x\approx360$m, not the stated $x=300$m); or keeping the spurious steep-angle root without checking it against the problem's physical description.

Source. Young & Freedman, University Physics with Modern Physics, Ch. 33 (reflection at a plane surface, image formation).

Question 8 of 8: Tilt Angle of a Ground Mirror Reflecting Light Between Two Towers

Question 8: Tilt Angle of a Ground Mirror Reflecting Light Between Two Towers (20 marks)