Question 8 of 8: Time Interval Between a Directly Reflected and a Refracted-and-Reflected Light Beam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2017 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\varepsilon_0=8.85\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.6\times10^{-19}$ C. Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — Coulomb's law and superposition for discrete point charges, Gauss's law for cylindrical charge distributions and coaxial capacitors, the Biot–Savart/Ampère force between parallel currents, Biot–Savart on-axis loop fields, Faraday's law for a moving loop in a spatially varying field, and the point (differential) form of Gauss's law for recovering a charge distribution from a given field; Young & Freedman, University Physics with Modern Physics — Snell's law and refracted-ray time-of-flight geometry.
Question 8: Time Interval Between a Directly Reflected and a Refracted-and-Reflected Light Beam (20 marks)
Find. The time delay between the surface-reflected beam and the beam that refracts into the water, reflects off the bottom, and refracts back out, both arriving at the same detector.
Two paths from source to detector: a direct surface reflection (red), and a path that refracts into the water, reflects off the flat bottom, and refracts back out (green). Both are symmetric about the horizontal midpoint (x = 50 m) because the two towers are equal height.
Approach. Both paths are symmetric about the horizontal midpoint by equal tower heights. For the direct path, the surface reflection point is the geometric midpoint (angle of incidence = angle of reflection). For the refracted path, Snell's law at entry/exit AND the fixed horizontal span (100 m total, split between two air legs and two water legs) together fix the refraction angle; solve that constraint numerically, then convert each path's geometric length to a travel time using $c$ in air and $c/n$ in water.
Direct (surface-reflected) path length and time. By symmetry the reflection point is at the midpoint (50 m out); each leg is the hypotenuse of a 30 m × 50 m right triangle:
$$L_{\text{direct}}=2\sqrt{30^2+50^2}=2\sqrt{3400}=\boxed{116.62\ \text{m}}$$
$$t_{\text{direct}}=\frac{L_{\text{direct}}}{c}=\frac{116.62}{3\times10^8}=\boxed{3.887\times10^{-7}\ \text{s}}$$
Refraction geometry. Let $\theta_1$ be the air-side angle from vertical and $\theta_2$ the water-side angle. By symmetry the bottom reflection is at the horizontal midpoint (50 m), so the horizontal air-leg distance $x_1=30\tan\theta_1$ and the horizontal water-leg distance $50-x_1=2\tan\theta_2$ must hold together with Snell's law $\sin\theta_1=n\sin\theta_2$. Solving this pair simultaneously (numerically):
$$\boxed{\theta_1=58.18^\circ,\qquad\theta_2=39.71^\circ}$$
(check: $x_1=30\tan58.18^\circ=48.34$ m $=50-2\tan39.71^\circ$ — consistent.)
Refracted-path length (one full round trip, both legs doubled by symmetry) and its travel time. Each air leg has length $30/\cos\theta_1$, each water leg $2/\cos\theta_2$; light travels at $c$ in air and $c/n$ in water:
$$L_{\text{air,1-way}}=\frac{30}{\cos58.18^\circ}=56.89\ \text{m},\qquad L_{\text{water,1-way}}=\frac{2}{\cos39.71^\circ}=2.600\ \text{m}$$
$$t_{\text{refr}}=2\left(\frac{L_{\text{air,1-way}}}{c}\right)+2\left(\frac{nL_{\text{water,1-way}}}{c}\right)=\frac{2(56.89)}{3\times10^8}+\frac{2(1.33)(2.600)}{3\times10^8}$$
$$\boxed{t_{\text{refr}}=4.023\times10^{-7}\ \text{s}}$$
Time interval between the two detections.
$$\Delta t=t_{\text{refr}}-t_{\text{direct}}=4.023\times10^{-7}-3.887\times10^{-7}$$
$$\boxed{\Delta t=1.360\times10^{-8}\ \text{s}=13.60\ \text{ns}\ \text{(refracted beam arrives later)}}$$
$1.360\times10^{-8}$ s (13.60 ns); refracted beam is later
Check: the refraction angle pair $(\theta_1,\theta_2)$ is the unique physical solution of Snell's law combined with the fixed 100 m span and 2 m depth (found numerically); it is NOT a free/assumed choice — the bottom reflection point lands at the horizontal midpoint only because the two tower heights are equal, which is what lets both the direct and refracted geometries be treated as mirror-symmetric about x = 50 m.
