Mirror Location and Time Delay for a Reflected Light Pulse: 04-BS-9, May 2015, Q8

Question 8: Mirror Location and Time Delay for a Reflected Light Pulse (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
Source height $h_s$	$10$ m above the ground plane
Detector height $h_d$	$30$ m above the ground plane
Horizontal separation $L$	$40$ m
Speed of light $c$	$3\times10^8$ m/s

Find. (i) The mirror's location on the ground; (ii) the time delay between the direct and reflected pulses.

Source S and detector D above the ground plane; the reflected ray obeys angle-of-incidence = angle-of-reflection, equivalent to a straight line from S's mirror image (below ground) to D.

Approach. Use the image-source method: the law of reflection (angle of incidence = angle of reflection off a plane mirror) is exactly reproduced by drawing a straight line from the source's mirror image (reflected through the ground plane) to the detector; where that line crosses the ground is the true mirror point, and its length equals the true reflected path length. Compute the direct path length by ordinary distance, and the time delay from the path-length difference.

Locate the mirror point via the image source. Place the image source at depth $h_s$ below the ground, directly under S. The straight line from the image $(0,-h_s)$ to the detector $(L,h_d)$ crosses the ground ($y=0$) at: $$x_m=\frac{h_s}{h_s+h_d}\,L=\frac{10}{10+30}(40)$$ $$x_m=\boxed{10.0\ \text{m from the source (horizontally)}}$$
Direct path length. $$d_{\text{direct}}=\sqrt{L^2+(h_d-h_s)^2}=\sqrt{40^2+(30-10)^2}$$ $$d_{\text{direct}}=\boxed{44.72\ \text{m}}$$
Reflected path length (source–mirror–detector). $$d_1=\sqrt{x_m^2+h_s^2}=\sqrt{10^2+10^2}=14.14\ \text{m},\qquad d_2=\sqrt{(L-x_m)^2+h_d^2}=\sqrt{30^2+30^2}=42.43\ \text{m}$$ $$d_{\text{reflected}}=d_1+d_2=\boxed{56.57\ \text{m}}$$ (this matches the direct image-to-detector distance $\sqrt{L^2+(h_d+h_s)^2}=\sqrt{40^2+40^2}=56.57$ m, confirming the construction.)
Time delay between the two arrivals. $$\Delta t=\frac{d_{\text{reflected}}-d_{\text{direct}}}{c}=\frac{56.57-44.72}{3\times10^8}$$ $$\Delta t=\boxed{39.5\ \text{ns}}$$, with the reflected pulse arriving later.

Quantity	Result
Mirror location $x_m$	$10.0$ m from the source, on the ground
$d_{\text{direct}}$	$44.72$ m
$d_{\text{reflected}}$	$56.57$ m
$\Delta t$	$39.5$ ns (reflected pulse arrives later)

Concept & Theory

Topic. Geometric reflection off a plane mirror, solved by the method of images.

Key relations. Angle of incidence = angle of reflection; the image-source construction $x_m=\dfrac{h_s}{h_s+h_d}L$; time delay $\Delta t=(d_{\text{reflected}}-d_{\text{direct}})/c$.

Why this works. Reflecting the source through the mirror plane turns the two-segment reflected ray (which must obey the law of reflection at an unknown point) into a single straight line problem: any straight line from the image to the detector automatically satisfies equal angles at the crossing point, because the image construction preserves path length and angle by symmetry. This avoids setting up and solving the reflection condition trigonometrically.

Common pitfall. Averaging the two heights or assuming the mirror sits at the horizontal midpoint — the mirror point divides $L$ in the ratio $h_s:h_d$ of the two heights, not evenly, unless the heights happen to be equal.

Source. Young & Freedman, University Physics with Modern Physics, Ch. 33 (reflection at a plane surface); Sadiku, Elements of Electromagnetics, Ch. 10 (plane-wave propagation, used for the constant-$c$ travel-time relation).

Question 8 of 8: Mirror Location and Time Delay for a Reflected Light Pulse

Question 8: Mirror Location and Time Delay for a Reflected Light Pulse (20 marks)