The image of the stone is at depth h′ = n² h cos³θ / (n² − sin² θ)^(3/2) below the surface. For n = 1.33, h = 2 m and θ = 40°, h′ = 1.007 m. Looking straight down, h′ = h/n.
Problem statement
A man standing at the edge of a swimming pool looks at a stone lying on the bottom. The depth of the pool is h. At what distance from the surface of the water is the image of the stone formed if the line of vision makes an angle θ with the normal to the surface?
We assume the water has refractive index n (1.33 for water) and the air has index 1. This means Snell's law reads n sin α = sin θ, where α is the angle in the water.
We assume the surface is flat and still. This means the normal is the same everywhere on the surface.
We assume the eye receives a narrow bunch of rays from the stone. This means the image is the point where neighbouring rays seem to cross when traced backward.
Symbols: h is the real depth of the stone. h′ is the depth of its image, measured down from the surface. α is the angle of the ray in the water. θ is the angle of the line of vision in the air.
The big idea in plain words
Rays leave the stone in many directions. The eye catches only a few of them, so it uses neighbouring rays. Each ray bends at the surface, so traced backward the rays do not all come from the stone. Neighbouring rays seem to cross at one point I, a bit higher than the stone. That point is the image. Its depth below the surface is h′. The depth changes with the angle of view, so the stone seems to rise as you look more sideways.
The solution in each frame
Lab frame
The pool and the observer are at rest. The full solution is in the next section.
Other frames
No other frame is needed. Nothing moves, so one frame is enough.
The solution step by step
A ray from the stone O leaves at angle α to the vertical. It reaches the surface at a distance x = h tan α from the point straight above O. Snell's law gives the angle θ in the air:
In air the ray goes on at angle θ. Traced backward below the surface, it is the line
where z is the depth below the surface. Now take a neighbouring ray with a slightly different α and θ. The two backward lines cross at the image point I. At that point the value of x is the same for both rays, so x does not change when θ changes a little. Set the change of x with θ to zero:
To find dα/dθ, differentiate Snell's law: n cos α dα = cos θ dθ. So
Put this in. The depth z of the crossing point is the image depth h′:
From Snell's law, cos α = √(n² − sin²θ) / n. Then cos³α = (n² − sin²θ)^(3/2) / n³, so
For looking straight down, θ = 0, this becomes h′ = h/n. For water this is 0.75 h. Your current values give h′ = 1.007 m, so the stone seems 0.993 m higher than it is.
A separate check takes two rays only 10⁻⁵ radian apart and finds where their backward lines cross. The crossing is at depth 1.00748 m, the same as the formula, h′ = 1.007 m.
Five methods
1. Force method
Not needed. This is a ray-optics problem about directions, not forces.
2. Energy method
Not needed. Only the directions of the rays matter.
3. Graph method
The first graph plots h′/h against θ. It starts at 1/n and falls to zero as the line of vision becomes nearly flat. The second graph shows h′/h against n, and the third shows h′ rising in a straight line with h.
4. Calculus method
This is the method used above: the image is the crossing of neighbouring rays, found by setting dx/dθ = 0.
5. Check method
Limits: at θ = 0, h′ = h/n, the usual apparent depth. At n = 1 the formula gives h′ = h for every θ, because there is no bending. As θ tends to 90°, h′ tends to zero. Units: h′ has the same unit as h.
Fastest: the closed formula. Most reliable: the numerical crossing of two close rays.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Refractive index n | Yes | α, h′, h′/h, the drawing and all graphs |
| Line of vision θ | Yes | α, h′, the rays, the arcs, the image point I and the graph dots |
| Pool depth h | Yes | The pool height, h′, the position of I and the third graph |
| Wanted h′/h (given-answer mode) | Sets θ | The model solves θ for the wanted ratio |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Refractive index | n | 1.33 |
| Pool depth | h | 2 m |
| Line of vision | θ | 40° |
| Angle in the water | α | 28.901° |
| Image depth | h′ | 1.007 m |
Exam traps
- Using h′ = h/n for every angle. That holds only for looking straight down.
- Using the single ray from the stone to find the image. The image needs the crossing of two neighbouring rays.
- Swapping the angles in Snell's law: the line of vision θ is in the air, so it is sin θ = n sin α.
- Measuring h′ along the ray instead of straight down from the surface.
- Forgetting that the image is also shifted sideways, not only upward.