Horizontal Offset Between the Apparent and Real Position of an Insect Seen From Underwater: 04-BS-9, May 2018, Q8

Question 8: Horizontal Offset Between the Apparent and Real Position of an Insect Seen From Underwater (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
Fish-eye depth $h_f$	0.50 m below the surface
Insect height $h_i$	0.20 m above the surface (known to the fish)
Observed angle from vertical (water side) $\theta_2$	$30^\circ$
Index of refraction of water $n$	1.33

Find. The horizontal distance between the insect's TRUE position and where the fish — not applying Snell's law — perceives it to be.

The TRUE ray bends at the surface (P): steeper θ₁ in air, shallower θ₂ in water, both measured from vertical. A fish that does not know Snell's law instead extrapolates its OWN observed water-side angle θ₂ in a straight dashed line all the way up to the insect's known height — landing short of the real position.

Approach. First locate the refraction point $P$ on the surface using the fish's own observed water-side angle $\theta_2$ over its depth $h_f$. Use Snell's law to find the TRUE, larger air-side angle $\theta_1$ and hence the insect's real horizontal offset beyond $P$. Separately, compute the NAIVE straight-line extension the fish would compute — the SAME angle $\theta_2$, but carried the full vertical distance $h_f+h_i$, since a fish ignorant of refraction assumes no bending occurs at the surface.

Horizontal position of the refraction point $P$. Using the fish's observed angle $\theta_2=30^\circ$ over its depth $h_f=0.50$ m: $$x_P=h_f\tan\theta_2=(0.50)\tan30^\circ=0.2887\ \text{m}$$
True air-side angle via Snell's law. Light bends toward the normal entering water, so it bends AWAY from the normal on the way back out into air — $\theta_1$ is the larger angle: $$\sin\theta_1=n\sin\theta_2=1.33\sin30^\circ=0.665$$ $$\boxed{\theta_1=41.68^\circ}$$
Real horizontal position of the insect. Over its height $h_i=0.20$ m above $P$, using the TRUE (larger) air-side angle: $$x_i=h_i\tan\theta_1=(0.20)\tan41.68^\circ=0.1781\ \text{m}$$ $$x_{\text{real}}=x_P+x_i=0.2887+0.1781=0.4668\ \text{m}$$
Apparent position (fish's naive straight-line guess). Not knowing about refraction, the fish extends its OWN observed angle $\theta_2$ in a single straight line over the FULL vertical distance $h_f+h_i=0.70$ m: $$x_{\text{apparent}}=(h_f+h_i)\tan\theta_2=(0.70)\tan30^\circ=0.4041\ \text{m}$$
Horizontal distance between the two. $$\Delta x=x_{\text{real}}-x_{\text{apparent}}=0.4668-0.4041$$ $$\boxed{\Delta x=6.261\times10^{-2}\ \text{m}=6.261\ \text{cm (the true position lies further out)}}$$

Check: $\theta_2=30^\circ$ is read as the WATER-side viewing angle the fish actually measures (the fish's eye is underwater, looking up), the natural reading of "it sees an insect... at 30° angle away from vertical."

Quantity	Result
Refraction point offset $x_P$	0.2887 m
True air-side angle $\theta_1$	$41.68^\circ$
Real horizontal position $x_{\text{real}}$	0.4668 m
Apparent (naive) horizontal position $x_{\text{apparent}}$	0.4041 m
Horizontal distance between them $\Delta x$	6.261 cm

Concept & Theory

Topic. Comparing the true refracted position of an object to the position inferred by an observer who wrongly assumes light travels in a straight line across a refracting interface.

Key relations. Snell's law $\sin\theta_1=n\sin\theta_2$ (water $\to$ air); horizontal offset $=$ height $\times\tan(\text{angle from vertical})$.

Why this works. An observer below a refracting surface who does not correct for refraction effectively assumes the LOCAL angle it measures continues unchanged all the way to the object. But the true ray actually bends AWAY from the normal as it exits into the less-dense medium ($\theta_1>\theta_2$), so the real object sits further from directly overhead than the naive straight-line extrapolation predicts — the mirror image of the everyday "apparent depth" illusion, applied looking up instead of down.

Common pitfall. Applying the near-normal apparent-depth shortcut $d_{\text{apparent}}=d/n$ (valid only for small viewing angles) at this fairly oblique $30^\circ$ angle instead of the full offset-geometry calculation; or swapping which angle ($\theta_1$ vs. $\theta_2$) belongs to which leg of the path.

Source. Young & Freedman, University Physics with Modern Physics, Ch. 33 (Snell's law, refraction at a plane interface).

Question 8 of 8: Horizontal Offset Between the Apparent and Real Position of an Insect Seen From Underwater

Question 8: Horizontal Offset Between the Apparent and Real Position of an Insect Seen From Underwater (20 marks)