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What is the formula for Real vs Apparent Depth, when not looking perpendicular to the interface?

I know the formula $$\frac{\text{real depth}}{\text{apparent depth}}=n$$ but I'm not looking for this.

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    $\begingroup$ You might want to give us at least a minimal description of what it is that you are talking about. $\endgroup$ – Pirx Dec 1 '16 at 19:43
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This question has been asked before but without a satisfactory answer :
Apparent depth and the viewing angle
Apparent depth is supposed to be independent of viewing angle but this seems inconsistent with critical angle refraction
Apparent depth and virtual image position

In the diagram below on the left the water surface is AB, the object is O and the image is I. One ray from object O strikes the water surface at angle of incidence $i$ and is refracted into the air at angle $r$ towards the observer.

The distances of object and image from the point at which the ray crosses the interface are $S, S'$ respectively, and their depths below surface AB are $y, y'$.

enter image description here

The exaggerated diagram on the right shows the formation of the image I from two rays which leave the object separated by an infinitesimally small angle $di$. These rays strike the surface of the water at points C and D with angles of incidence $i$ and $i+di$ respectively. They are then refracted into the air at angles of $r$ and $r+dr$. Extending the refracted rays backwards provides the location of the image I.

Angle ECD = $r$ and angle FCD = $i$. It follows that $$CD\cos r=S'dr$$ $$CD\cos i=S di$$

Differentiate Snell's Law : $$n\sin i =\sin r$$ $$n\cos i. di = \cos r. dr$$

Combining the above relations we get $$\frac{S}{S'}=n(\frac{\cos i}{\cos r})^2$$ $$\frac{y}{y'}=\frac{S\cos i}{S'\cos r}=n(\frac{\cos i}{\cos r})^3$$ $$\frac{x}{x'}=\frac{S\sin i}{S'\sin r}=(\frac{\cos i}{\cos r})^2$$

For small angles of refraction $r$ $$\frac{y}{y'}\approx n(1+\frac32 (1-\frac{1}{n^2})r^2)$$ When $r \approx 0$ we get the usual formula $$\frac{\text{real depth}}{\text{apparent depth}} \approx n$$ The above result also agrees with the solution given in Irodov Ex. 5.18 (Problems in General Physics), which is $$\frac{y'}{y}=\frac{n^2\cos^3 r}{(n^2-\sin^2 r)^{3/2}}$$

$r \ge i$ hence $\cos r \le \cos i$. So we always have $y \ge y'$ and $x \ge x'$. The image is only vertically above the object ($x=x'$) when viewed from above ($i=r=0$). Otherwise the image is always closer to the observer (not further away, as depicted in the above diagrams).

In the extreme case of $r \to 90^{\circ}$ the image is at the surface of the water ($y' \approx 0$) at a distance of $y\tan C$ horizontally from the object towards the observer, where $C$ is the Critical Angle - ie $n\sin C=1$.

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If the angle of incidence is important, one is no longer in the conditions of Gauss and the plane interface presents astigmatism. It's difficult to define the image of an object in this case.

https://en.wikipedia.org/wiki/Astigmatism_(optical_systems) :

"An optical system with astigmatism is one where it is propagated in two perpendicular planes have different foci, and it is an optical system with astigmatism is used to form an image of a cross, the vertical and horizontal lines will be in sharp focus at two different distances . "

The calculation provided in the previous answer assumes the rays contained in the plane of incidence. A different result would be found for rays perpendicular to the plane of incidence.

Sorry for my poor english !

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