Is the image of an object submerged in water (at the bottom of a pool say) directly above the object? Take a ball bearing resting at the bottom of a swimming pool.
My textbook shows an figure (not the figure below) illustrating apparent depth, where the observer is viewing the ball from an angle. In the figure, the image of the ball is directly above the object.
I derived a formula for apparent depth A in terms of incident angle i and real depth R as:
$$A(i)=\frac{R \cos{i} \sqrt{n^2- \sin^2{i}}}{n^2}$$
It can be seen that as i goes to 0 (looking straight into the water) we get the expression:
$$A=\frac{R}{n}$$
Then I was looking the wikipedia page for Refraction and noticed a figure that showed this situation but the image is not directly above the object.

This figure took me by surprise as it hadn't occurred to me that this could be the case.
If this is indeed the case my formula is not correct.
So my question is whether the image of the object is indeed directly above the object or not ?
And how do we know this ?
 A: To locate the image of the point object, O, we consider two rays from O, refracted at the water surface to emerge at angles $\phi_1$ and $\phi_2$. We extrapolate these refracted rays back to their intersection. However, the position of the intersection depends not just on the mean refracted angle, $\frac 12(\phi_1+\phi_2)$, but also on $|\phi_2-\phi_1|$. The position stabilises for small $|\phi_2-\phi_1|$, so it is reasonable to take the image, I, of O to be at the intersection when $|\phi_2-\phi_1|\to 0$.
Let N be the point on the water surface directly above O, and put ON = $h$. Consider rays from the object that reach the water surface at points P and Q, at angles of incidence of $\theta$ and $(\theta + \Delta\theta)$ and emerge into the air at angles of refraction of $\phi$ and $(\phi + \Delta\phi)$. Put NP = $\xi$.
At this stage, drawing a diagram is recommended! From it we see that:
$$\xi=h\tan\theta$$
and that
$$\lim_{\Delta\theta\to0}\frac{\text{PQ}}{\Delta\theta}=\frac{d\xi}{d\theta}=h\sec^2\theta$$
We now apply the sine formula to triangle IPQ...
$$\frac{\text{IP}}{\sin\left[\frac{\pi}2-(\phi+\Delta\phi)\right]}=\frac{\text{PQ}}{\sin(\Delta\phi)}$$
So as $\Delta\phi \to 0$,
$$\text{IP}=\cos\phi\frac{d\xi}{d\phi}$$
The sideways displacement of I from the vertical line ON is
$$x=\text{NP}-\text{IP}\sin\phi=\xi-\sin\phi\cos\phi \frac{d\xi}{d\phi}$$
that is:
$$x=h\tan\theta-\sin\phi \cos\phi \frac{d\xi}{d\theta}\frac{d\theta}{d\phi}$$
We know that $\frac{d\xi}{d\theta}=h\sec^2 \theta$ and we can use Snell's law to find $\frac{d\theta}{d\phi}$ ...
$$\sin\phi=n\sin\theta$$
$$\text{So}\ \ \ \frac{d\theta}{d\phi}=\frac 1n \frac{\cos\phi}{\cos\theta}$$
Substituting for $\frac{d\xi}{d\theta}$ and $\frac{d\theta}{d\phi}$ in our expression for $x$,
$$x=h\tan\theta-\sin\phi\cos\phi\ h\sec^2\theta \frac 1n \frac{\cos\phi}{\cos\theta}$$
With more use of $\sin\phi=n\sin\theta$, this can be tidied up to yield the rather neat...
$$x=h\ (n^2-1)\tan^3 \theta.$$
It's re-assuring to note that when $n=1$, that is the water has been replaced by air, $x=0$. Also, for any value of $n$ greater than 1, when $\theta=0$ (that is we view from directly above) $x=0$. The other limiting case to consider is the critical ray, for which $\sin\theta=\frac 1n$; we find that $x=\text{NP}$.
For completeness, the upward displacement, $y$ of the image is
$$y=h-\text{IP}\cos\phi=h-\frac hn \left[1-(n^2-1)\tan^2\theta\right]^{3/2}.$$
