Apparent depth

Look straight down at a coin resting on the floor of a swimming pool and it seems to float just below the surface, closer than your fingers can actually reach. Nothing about the coin has moved — only the light carrying its image has bent. As each ray leaves the water and enters the air above, it refracts away from the normal, and your eye, which assumes light has travelled in a straight line, projects the coin's image to a shallower point than where it truly sits. This is apparent depth: a precise, calculable consequence of refraction at a plane interface. It explains why fish look nearer than they are to a heron, why a ruler dipped in a beaker seems to bend and shorten, and why a thick glass slab over print makes the page seem to rise toward you. The same idea, applied layer by layer, handles stacked media — water under glass, or any pool with a transparent lid. This article builds the near-normal-viewing formula from Snell's law and works the standard single- and multi-layer numericals.

The core insight

For near-normal viewing, the refractive index of a medium equals real depth divided by apparent depth, so a denser medium viewed from a rarer one always looks shallower than it truly is.

The core idea

Why the object appears to shift. Light from a submerged object travels from a denser medium (water) into a rarer one (air) to reach your eye. At the water-air interface it bends away from the normal, since it is moving into the less dense medium. When your brain traces these diverging rays backward in straight lines — the only way it knows how to interpret light — it locates their apparent source higher up than the object's real position. The denser the medium relative to air, the more the rays bend, and the greater the apparent rise.

The near-normal formula. The full geometry of refraction depends on the actual viewing angle, but ISC problems restrict attention to near-normal viewing: the eye positioned almost directly above the object, so every angle involved is small. Under this restriction, applying Snell's law together with the small-angle approximation sin⁡θ≈tan⁡θ\sin\theta \approx \tan\theta gives a clean ratio between the object's true position and its perceived one:

n=Real depthApparent depthn=\frac{\text{Real depth}}{\text{Apparent depth}}

where: nn is the refractive index of the medium containing the object, taken relative to the medium the observer stands in (n=ndenser/nrarern = n_{\text{denser}}/n_{\text{rarer}} when looking from a rarer medium into a denser one). Rearranged, this gives the apparent depth directly:

Apparent depth=Real depthn\text{Apparent depth}=\frac{\text{Real depth}}{n}

Because n>1n>1 whenever you view a denser medium from a rarer one, the apparent depth is always smaller than the real depth — submerged objects never look deeper than they are, only shallower. This single inequality is a fast sanity check on any answer you compute.

Apparent shift. The amount by which the object appears to have risen is simply the difference between the two depths:

s=Real depth−Apparent depth=Real depth(1−1n)s=\text{Real depth}-\text{Apparent depth}=\text{Real depth}\left(1-\frac1n\right)

Stacked layers. Real problems often place several transparent media on top of one another — a glass slab floating on water, both viewed from air. Since each layer produces its own independent apparent-depth reduction under near-normal viewing, the total apparent depth is the sum of each layer's own contribution:

Total apparent depth=∑itini\text{Total apparent depth}=\sum_i \frac{t_i}{n_i}

where: tit_i is the true thickness of layer ii and nin_i is that layer's own refractive index. You never divide the combined thickness by a single averaged index — each layer is treated on its own terms, then the pieces are added.

Figure
Rays from a submerged coin bend away from the normal at the water surface and appear to an observer above to come from a shallower point marked apparent depth than the coin true position marked real depth
Apparent depth: refraction lifts the image toward the surface air water coin real depth apparent depth eye Solid lines are real refracted rays. Dashed lines show where the eye traces them back to.
Derivation AReal depth over apparent depth from Snell's law

The goal is to show, starting only from Snell's law at the water-air interface and the small-angle approximation valid for near-normal viewing, that the refractive index of water equals the object's real depth divided by its apparent depth as seen from directly above — the single relation every apparent-depth numerical rests on.

Step 1. Set up the geometry. A point object O sits at real depth hh below a plane water surface. A ray leaves O, travels up through the water, and strikes the surface at small angle rr to the normal, then refracts into air at angle ii, obeying Snell's law nwatersin⁡r=nairsin⁡in_{\text{water}}\sin r = n_{\text{air}}\sin i, or n=sin⁡i/sin⁡rn = \sin i/\sin r with nn the refractive index of water relative to air.

Step 2. Locate the horizontal displacement of the ray at the surface. Let it strike the surface at horizontal distance xx from the point directly above O. Since O is at depth hh, right-triangle geometry gives tan⁡r=x/h\tan r = x/h.

Step 3. Trace the refracted ray back to find the perceived position. The observer, almost directly above O, extends the refracted ray backward until it crosses the vertical below the eye; this crossing point is the apparent position, at depth h′h'. The same xx now gives tan⁡i=x/h′\tan i = x/h' along this backward-extended ray.

Step 4. Apply the small-angle approximation. Because viewing is near-normal, both rr and ii are small, so sin⁡r≈tan⁡r\sin r \approx \tan r and sin⁡i≈tan⁡i\sin i \approx \tan i. Substituting into Snell's law: n=sin⁡isin⁡r≈tan⁡itan⁡r=x/h′x/h=hh′n = \dfrac{\sin i}{\sin r} \approx \dfrac{\tan i}{\tan r} = \dfrac{x/h'}{x/h} = \dfrac{h}{h'}. The distance xx cancels, leaving a relation independent of where the ray struck the surface.

Step 5. Read off the result. Since hh is the real depth and h′h' the apparent depth, this is exactly n=Real depth/Apparent depthn = \text{Real depth}/\text{Apparent depth}. Because n>1n>1 for water viewed from air, h′<hh' < h always — the object looks shallower than it is.

Worked example

A fish swims at a real depth of 2.4 m below the surface of a pond of refractive index 4/3. A bird looks straight down from directly above the surface. How deep does the fish appear, and by how much has its image shifted?

Step 1. Identify the quantities. Real depth =2.4= 2.4 m, n=4/3n = 4/3 (pond water relative to air), and the viewing is near-normal (bird looking straight down), so the formula from the derivation applies directly.

Step 2. Apply Apparent depth=Real depth/n\text{Apparent depth} = \text{Real depth}/n: Apparent depth=2.44/3=2.4×34=1.8 m\text{Apparent depth} = \dfrac{2.4}{4/3} = 2.4\times\dfrac{3}{4} = 1.8\ \text{m}.

Step 3. Find the apparent shift using s=Real depth−Apparent depths = \text{Real depth} - \text{Apparent depth}: s=2.4−1.8=0.6 ms = 2.4 - 1.8 = 0.6\ \text{m}.

Answer: the fish appears to be at a depth of 1.8 m, having shifted upward by 0.6 m from its true position.

Competitive stretch

Now suppose a glass slab of thickness 6 cm (n=1.5n = 1.5) is floated on top of the same pond, and the observer looks vertically down through slab and water together at a coin resting where the fish was, at the pond's original real depth. Treat the coin's depth in water as 2.4 m and add the glass layer above it.

Step 1. Compute each layer's own apparent-depth contribution using ti/nit_i/n_i: for the glass, 6/1.5=4 cm6/1.5 = 4\ \text{cm}; for the water, converting 2.4 m to 240 cm, 240/(4/3)=240×34=180 cm240/(4/3) = 240\times\dfrac34 = 180\ \text{cm}.

Step 2. Add the two contributions: total apparent depth =4+180=184 cm= 4 + 180 = 184\ \text{cm}, measured from the top of the glass slab.

Step 3. Compare with the true total thickness, 6+240=246 cm6 + 240 = 246\ \text{cm}: the layered stack shifts the coin's image upward by 246−184=62 cm246 - 184 = 62\ \text{cm}, more than water alone would, because the glass layer adds its own, smaller reduction on top of the water's.

Answer: total apparent depth 184 cm, an upward shift of 62 cm compared to the true 246 cm stack.

Common traps

  • Inverting the formula. It is n=real depth/apparent depthn = \text{real depth}/\text{apparent depth}, not the reverse. Since n>1n>1, apparent depth must always come out smaller than real depth — check this before submitting an answer.
  • Applying the formula to oblique viewing. The simple ratio holds only for near-normal (nearly vertical) viewing; it breaks down once the line of sight makes a large angle with the normal, where full geometric refraction is needed instead.
  • Using one averaged index for a layered stack. Each layer's own thickness must be divided by that layer's own refractive index; do not apply a single overall nn to the combined thickness of several different media.
  • Mixing up real and apparent totals. In multi-layer problems, keep the sum of true thicknesses and the sum of ti/nit_i/n_i apparent depths as two separate running totals — conflating them is the most common arithmetic slip.

Ask the tutor: want to walk through the Snell's-law derivation with an air gap sandwiched between two liquid layers, or try a numerical where the apparent shift itself is given and you must find the real depth?

Why does the object look shallower and not deeper?

Because light always bends away from the normal when it passes from a denser medium into a rarer one, such as water into air. The refracted rays spread out more than they would if travelling in a single straight line, and when your eye extends them backward, it finds their apparent source higher up, closer to the surface, than the object's true position. Since this bending only ever reduces the apparent depth for a denser-to-rarer transition, apparent depth is always less than real depth, never more.

Does the near-normal formula work if I look at the pool from an angle, standing at the edge?

No, not accurately. The relation n=real depth/apparent depthn = \text{real depth}/\text{apparent depth} was derived assuming both the angle of incidence and the angle of refraction are small, which lets sin⁡θ≈tan⁡θ\sin\theta \approx \tan\theta. When you view from a steep angle at the pool's edge, that approximation fails, and the apparent position depends on the actual viewing angle in a more complicated way that a full geometric ray trace is needed to capture. ISC numericals always specify or imply near-normal viewing, so you can use the simple formula with confidence there.

In a layered problem, why do I add the individual t over n terms instead of using one refractive index for the whole stack?

Each layer refracts the rays independently as they cross its own boundary, and under near-normal viewing each layer's contribution to the total apparent-depth reduction is exactly its own thickness divided by its own refractive index, regardless of what lies above or below it. There is no single "combined refractive index" for a stack of different media, so the only correct method is to compute each layer's apparent depth separately and sum them, exactly as in the two-layer glass-and-water example.

Before this

Where this leads