Apparent Displacement of a Submerged Object Viewed Obliquely: 04-BS-9, May 2016, Q8

Question 8: Apparent Displacement of a Submerged Object 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
Water depth $d$	$1$ m
Viewing angle in air (from vertical) $\theta_1$	$45^\circ$
Index of refraction of water $n$	$1.333$

Find. The horizontal distance between the object's true position and its apparent position, as perceived by the observer.

Ray leaves the object at $\theta_2$ (in water), refracts to $\theta_1=45^\circ$ at $P$; extending the air ray straight down to the true depth $d$ locates the apparent position, offset from the true position at the same depth.

Approach. Use Snell's law to find the ray's angle inside the water, $\theta_2$, then compare the true horizontal position of the object ($d\tan\theta_2$ from the point directly below $P$) with the apparent position obtained by extending the observed ray (at $\theta_1$) straight down, without bending, to the same depth $d$.

Angle of the ray inside the water, Snell's law. $n_{\text{air}}\sin\theta_1=n\sin\theta_2$ (with $n_{\text{air}}=1$): $$\sin\theta_2=\frac{\sin45^\circ}{1.333}=\frac{0.7071}{1.333}=0.5304$$ $$\theta_2=\boxed{32.04^\circ}$$
True horizontal offset of the object from the point below $P$. $$x_{\text{actual}}=d\tan\theta_2=(1)\tan(32.04^\circ)$$ $$x_{\text{actual}}=\boxed{0.6258\ \text{m}}$$
Apparent horizontal offset (straight-line extension of the air ray at $\theta_1$, to the same depth $d$). $$x_{\text{apparent}}=d\tan\theta_1=(1)\tan(45^\circ)$$ $$x_{\text{apparent}}=\boxed{1.000\ \text{m}}$$
Distance between the apparent and actual position. Both positions lie at the same depth $d$, so the separation is purely horizontal: $$\Delta x=x_{\text{apparent}}-x_{\text{actual}}=1.000-0.6258$$ $$\Delta x=\boxed{0.374\ \text{m, apparent position displaced away from the observer's foot-point, relative to the true position}}$$

An alternative convention, common in textbooks, places the apparent image on the vertical line through the object (where the extended air ray meets it), at apparent depth $x_{\text{actual}}/\tan\theta_1=0.6258$ m; the vertical distance to the true position is then $1-0.6258=0.374$ m. Because $\tan45^\circ=1$, both conventions give the same distance here, so the answer of $0.374$ m does not depend on which one is used.

Quantity	Result
Refraction angle in water $\theta_2$	$32.04^\circ$
True horizontal offset	$0.6258$ m
Apparent horizontal offset	$1.000$ m
Distance between apparent and actual position	$0.374$ m

Concept & Theory

Topic. Refraction at a plane interface and the apparent displacement of a submerged object viewed at an angle.

Key relations. Snell's law $n_1\sin\theta_1=n_2\sin\theta_2$; horizontal offset along a straight ray at angle $\theta$ from vertical over depth $d$, $x=d\tan\theta$.

Why this works. An observer who does not know the ray bent at the surface perceives the object as lying along the straight-line continuation of the ray that actually reaches their eye; comparing that straight-line path (at $\theta_1$) with the true bent path (at $\theta_2$), both referenced to the same known depth $d$, isolates exactly the horizontal displacement the refraction introduces.

Common pitfall. Applying the familiar near-vertical "apparent depth $=d/n$" formula, which only holds for near-normal viewing ($\theta_1\to0$) and does not apply at $45^\circ$; or comparing offsets at different reference depths instead of the same one, which mixes horizontal and vertical displacement into an ill-defined "distance."

Source. Young & Freedman, University Physics with Modern Physics, Ch. 33 (reflection and refraction at a plane interface).

Question 8 of 8: Apparent Displacement of a Submerged Object Viewed Obliquely

Question 8: Apparent Displacement of a Submerged Object Viewed Obliquely (20 marks)