Apparent vs. Actual Location of an Underwater Mirror Viewed Obliquely: 04-BS-9, May 2017, Q8

Question 8: Apparent vs. Actual Location of an Underwater Mirror Viewed Obliquely (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
Angle of incidence (in air) $\theta_i$	$45^\circ$
Water depth $h$	$4$m
Index of refraction, water $n$	$1.33$

Find. The horizontal separation between the mirror's true location and the location the observer infers (by projecting the emerging $45^\circ$ ray straight down to the known depth, ignoring refraction).

Real path: refract in (45°→32.1°), travel to the mirror, reflect, refract out (32.1°→45°). The observer instead extends the exit ray straight down (dashed) at the observed 45°, placing the mirror closer than it really is.

Approach. Trace the real (refracted) round trip to find the mirror's true horizontal position, then separately trace the straight-line (unrefracted) backward extension of the emerging ray — the construction the observer implicitly uses when they "know the depth" but not the bending — to find the apparent position. Both are measured from the same entry point, at the same depth $h$.

Refraction angle in water (Snell's law). $$\sin\theta_r=\frac{\sin\theta_i}{n}=\frac{\sin45^\circ}{1.33}=0.5317\ \Rightarrow\ \theta_r=\boxed{32.12^\circ}$$
Actual mirror position. The ray descends from the entry point to depth $h$ at angle $\theta_r$ from vertical: $$x_{\text{actual}}=h\tan\theta_r=(4)\tan(32.12^\circ)=\boxed{2.511\ \text{m (from the entry point)}}$$
Exit point. By symmetry (horizontal mirror, equal angles), the reflected ray retraces the same horizontal displacement back to the surface, so the total round-trip horizontal travel is twice the one-way value: $$x_{\text{exit}}=2h\tan\theta_r=\boxed{5.022\ \text{m (from the entry point)}}$$ and by Snell's-law reversibility the ray re-emerges into air at the SAME $45^\circ$ angle it entered.
Apparent mirror position. The observer, knowing the depth but assuming straight-line propagation at the OBSERVED $45^\circ$ angle, projects the exit ray backward from the exit point down to depth $h$: $$x_{\text{apparent}}=x_{\text{exit}}-h\tan45^\circ=5.022-(4)(1)=\boxed{1.022\ \text{m (from the entry point)}}$$
Difference between apparent and actual location. $$\Delta x=x_{\text{actual}}-x_{\text{apparent}}=2.511-1.022$$ $$\boxed{\Delta x=1.489\ \text{m (apparent mirror appears displaced toward the entry point / observer)}}$$

Quantity	Result
Refraction angle $\theta_r$	$32.12^\circ$
Actual mirror position (from entry point)	$2.511$ m
Apparent mirror position (from entry point)	$1.022$ m
Difference (apparent vs. actual)	$1.489$ m

Check: "small horizontal mirror" is read as a flat reflector lying exactly on the ray's path at the bottom (so the geometry is a simple two-segment refracted path, symmetric about the mirror) — not a mirror at a fixed, separately-specified location that the ray must be aimed at. This is the natural reading given no second position is stated.

Concept & Theory

Topic. Apparent position of a submerged object viewed at an oblique angle — a generalization of the familiar "apparent depth" result (which only applies near-normal incidence) to a full lateral+vertical displacement at a stated oblique angle.

Key relations. Snell's law $n_1\sin\theta_1=n_2\sin\theta_2$; horizontal travel over a vertical drop $h$ at angle $\theta$ from vertical is $h\tan\theta$; the observer's naive straight-line reconstruction uses the OBSERVED (air-side) angle applied to the KNOWN depth.

Why this works. Refraction bends the light toward the normal on entry (since $n_{\text{water}}>n_{\text{air}}$, $\theta_r<\theta_i$), so the real underwater path is more "vertical" (less horizontal spread) than a straight extension of the observed exit angle would suggest; an observer who correctly measures the exit angle ($45^\circ$, same as entry by symmetry) but incorrectly assumes it held for the FULL depth places the object systematically closer to the surface point of exit than it truly is — exactly the classic apparent-depth illusion, generalized off-axis.

Common pitfall. Applying the small-angle apparent-depth formula $h'=h/n$ (valid only near-normal incidence, and purely vertical) to an explicitly oblique ($45^\circ$) problem instead of tracing the actual ray geometry; or forgetting that the round trip (down AND back up) doubles the refracted horizontal travel before the backward-extension comparison is made.

Source. Young & Freedman, University Physics with Modern Physics, Ch. 33–34 (Snell's law, apparent depth, image formation by refraction).

Question 8 of 8: Apparent vs. Actual Location of an Underwater Mirror Viewed Obliquely

Question 8: Apparent vs. Actual Location of an Underwater Mirror Viewed Obliquely (20 marks)