Real and apparent depth

Look down at a swimming pool from the poolside and the bottom always seems closer than it really is — step in and you find yourself in deeper water than your eyes promised. A coin resting on the floor of a glass of water seems to have "risen" toward the surface, and a straw dipped in a tumbler appears to snap sharply at the waterline. All three illusions come from the same piece of physics: light refracting as it leaves a denser medium for a rarer one. This article stays inside the ICSE Class 10 syllabus topic "Refraction of Light at Plane Surfaces," building directly on refractive index and Snell's law. We build the relation between real depth, apparent depth and refractive index for near-normal viewing, work through the standard pool-and-fish numericals in both directions, and explain the bent-stick illusion using the same idea. Nothing here goes beyond what a Class 10 student needs — oblique viewing and the full ray-tracing treatment are left for later study.

The core insight

For near-normal viewing, a denser medium's refractive index equals the ratio of an object's real depth to its apparent depth, $\mu = \text{real depth}/\text{apparent depth}$ , so any object viewed from a rarer medium always looks shallower than it truly is.

The core idea

Why the object seems to move. Light from an object at the bottom of water (or glass) travels up through the denser medium and bends away from the normal as it crosses into the rarer medium, air, on its way to your eye. Your eye and brain do not know the ray bent — they simply trace it back as a straight line, and that straight line meets the vertical at a point higher up than the object's true position. The true position is the real depth; the traced-back, seen position is the apparent depth, and it is always smaller.

The governing relation. For near-normal viewing — looking almost straight down, the standard assumption at this level — the refractive index of the denser medium links the two depths simply:

\mu = \frac{\text{real depth}}{\text{apparent depth}}

where: $\mu$ is the refractive index of the medium containing the object (relative to air), real depth is the object's true vertical distance below the surface, and apparent depth is the distance at which it appears to lie, both measured in the same unit. Because $\mu > 1$ for every medium denser than air, apparent depth is always less than real depth: the object looks nearer the surface than it is, whichever denser medium you view it through.

The vertical shift. The amount the object seems to have risen is the apparent shift, $\text{real depth} - \text{apparent depth}$ . Since apparent depth $= \text{real depth}/\mu$ , a larger $\mu$ shrinks apparent depth further and so produces a bigger shift for the same real depth — an object at the bottom of a glass block (denser than water) appears to rise more than the same object at the same depth in water.

The bent-stick illusion follows from the same rule. A stick dipped partly into water looks broken at the waterline because the submerged part is seen at a smaller apparent depth than its real depth, while the part above water is seen exactly where it is (air to air, no bending). The two portions no longer line up, so the stick appears to kink sharply right where it enters the water.

Figure

A coin at the true bottom of a water-filled beaker sends a ray bending away from the normal as it leaves the water surface, and the eye traces that ray back in a straight line to a shallower apparent position marked above the real coin

Derivation ADeriving mu = real depth over apparent depth for near-normal viewing

The goal is to show, starting from Snell's law applied to a ray leaving a denser medium for a rarer one at a small angle to the normal, that the refractive index of the denser medium equals the ratio of real depth to apparent depth, and to see exactly why this simple ratio only holds for near-normal (almost vertical) viewing.

Step 1. Set up the ray. Let an object O sit at real depth $h_r$ below the flat surface of a denser medium (refractive index $\mu$ relative to air). A ray from O travels upward, striking the surface at a small angle $i$ to the normal, and refracts into air at a larger angle $r$ , since light bends away from the normal on leaving the denser medium. Let the point where the ray meets the surface be P, and let the horizontal distance from the normal through O to P be $x$ .

Step 2. Apply Snell's law at the surface. Going from the denser medium into air, $\mu \sin i = \sin r$ , so $\mu = \dfrac{\sin r}{\sin i}$ .

Step 3. Use the small-angle, near-normal approximation. For nearly vertical viewing, both $i$ and $r$ are small, so $\sin i \approx \tan i$ and $\sin r \approx \tan r$ . From the geometry, $\tan i = x/h_r$ (using the real depth as the vertical side), and $\tan r = x/h_a$ , where $h_a$ is the apparent depth at which the eye, tracing the emergent ray straight back, perceives O to lie. The horizontal distance $x$ is the same for both triangles because it is measured to the same point P on the surface.

Step 4. Substitute and simplify. Since $\mu = \sin r/\sin i \approx \tan r/\tan i = (x/h_a)/(x/h_r)$ , the common factor $x$ cancels, leaving $\mu = h_r/h_a$ , that is,

\mu = \frac{\text{real depth}}{\text{apparent depth}}.

Because this result relied on replacing sines with tangents, it is only valid when $i$ and $r$ are small — near-normal viewing. At large, oblique angles the approximation breaks down and the simple ratio no longer applies. Verified: for real depth $24\ \text{cm}$ and $\mu = 1.33$ , apparent depth $= 24/1.33 = 18.0\ \text{cm}$ ; for apparent depth $2.0\ \text{m}$ and $\mu = 1.33$ , real depth $= 1.33 \times 2.0 = 2.66\ \text{m}$ — both recomputed directly from $\mu = h_r/h_a$ .

Worked example

A coin lies at the bottom of a beaker filled with water to a depth of $24\ \text{cm}$ . The refractive index of water is $1.33$ . Find the apparent depth of the coin as seen from directly above, and the apparent shift.

Step 1. Write the governing relation and identify the knowns: $\mu = \text{real depth}/\text{apparent depth}$ , with real depth $= 24\ \text{cm}$ and $\mu = 1.33$ , apparent depth unknown.

Step 2. Rearrange for apparent depth: $\text{apparent depth} = \dfrac{\text{real depth}}{\mu} = \dfrac{24}{1.33}$ .

Step 3. Compute: $\text{apparent depth} \approx 18.0\ \text{cm}$ .

Step 4. Find the apparent shift: $\text{shift} = \text{real depth} - \text{apparent depth} = 24 - 18.0 = 6.0\ \text{cm}$ . The coin appears to be about $18.0\ \text{cm}$ below the surface, an apparent rise of about $6\ \text{cm}$ from its true $24\ \text{cm}$ depth.

Competitive stretch

A fish appears to be at a depth of $2.0\ \text{m}$ when a person on the bank looks straight down into a pond of refractive index $1.33$ . Find the true depth of the fish, and state how much deeper it actually is than it looks.

Step 1. Here apparent depth is given and real depth is unknown, so rearrange $\mu = \text{real depth}/\text{apparent depth}$ the other way: $\text{real depth} = \mu \times \text{apparent depth}$ .

Step 2. Substitute: $\text{real depth} = 1.33 \times 2.0$ .

Step 3. Compute: $\text{real depth} = 2.66\ \text{m}$ .

Step 4. Compare with the apparent depth: the fish is actually $2.66 - 2.0 = 0.66\ \text{m}$ deeper than it looks. The fish's true depth is about $2.66\ \text{m}$ — a useful safety fact, since someone judging a pond as $2.0\ \text{m}$ deep by eye would misjudge the real depth by more than half a metre.

Common traps

Dividing the wrong way — apparent depth is always smaller than real depth, so real depth $= \mu \times$ apparent depth, and apparent depth $=$ real depth $/\mu$ ; check which quantity is given before rearranging.
Applying the formula at large, oblique viewing angles — $\mu = \text{real depth}/\text{apparent depth}$ holds only for near-normal (almost vertical) viewing, the assumption ICSE numericals make unless stated otherwise.
Using the wrong medium's refractive index — always use the refractive index of the medium containing the object, the denser medium being looked into from air, never the reverse.
Thinking the object physically moves — nothing actually shifts position; only the apparent position, reconstructed by the eye from the bent ray, changes, and the same logic explains why a dipped stick looks bent at the waterline.

Ask the tutor: Want to see how real and apparent depth connects to the critical angle and total internal reflection later in this chapter, with more practice problems?

Why does apparent depth use the refractive index of the medium the object is in, not the medium the eye is in?

Because refraction happens as light leaves the medium containing the object and enters the medium containing the eye, and Snell's law measures the bending relative to the denser medium's index (with air taken as the reference, index $1$ ). When you view a coin in water from air, it is water's refractive index that appears in $\mu = \text{real depth}/\text{apparent depth}$ — the eye's medium (air) does not get its own separate factor because its index is already built in as the reference value of $1$ .

Does the real-apparent depth formula work if I look at the pool from the side at a shallow angle?

No, not with this simple ratio. The derivation from Snell's law replaces $\sin i$ and $\sin r$ with $\tan i$ and $\tan r$ , which is only accurate when both angles are small — that is, when you are looking almost straight down. At large, oblique angles that approximation fails, and the apparent position depends on the viewing angle in a more complicated way that goes beyond the Class 10 syllabus. ICSE numericals always assume near-normal viewing unless a diagram states otherwise.

If a glass block has a higher refractive index than water, does an object under it look even shallower?

Yes. Since apparent depth $= \text{real depth}/\mu$ , a larger $\mu$ divides the real depth by a bigger number, giving a smaller apparent depth and hence a larger apparent shift. An object at the same real depth appears to rise more when viewed through glass ( $\mu \approx 1.5$ ) than through water ( $\mu \approx 1.33$ ), because glass bends light away from the normal more strongly as it exits.

Real and apparent depth

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads