Why Submerged Objects Appear Shallower: Refraction Explained

A coin sits at the bottom of a swimming pool. You look down through the water, and it seems much closer to the surface than it really is. If you reached towards its apparent position, your hand would stop short.

Why does this happen? Light travelling from the coin through water bends away from the normal when it enters the air. Your eye receives that bent light and traces its direction backwards, making the coin appear to be at a shallower position than its actual depth.

Before working through the physics, consider a prediction: if a ray travels straight upwards from the coin and doesn’t bend at all, shouldn’t the coin appear exactly where it is? That sounds reasonable, but one ray alone cannot tell your eye how far away an object is. The answer comes from comparing several rays.

01The experiment: Making a hidden coin reappear

You can reveal the effect of refraction without measuring a single angle.

Place a coin at the bottom of an opaque cup. Look into the cup from one side, then lower your head until the near rim just hides the coin.

Keep your head, the cup, and the coin in the same positions. Now slowly fill the cup with water.

The coin becomes visible again.

What would the straight-line model predict?

With the cup empty, the rim blocks the light travelling directly from the coin towards your eye.

If light continued along straight lines through the water-air boundary, adding water would not remove that obstruction. The coin would remain hidden.

Instead, light reflected from the coin travels upwards through the water, reaches the water-air surface, and changes direction. Some rays that previously couldn’t reach your eye can now pass above the rim.

The observation supports the idea that the light’s path changes at the boundary.

It doesn’t yet tell us the angle of bending or how much shallower the coin appears. We need a more controlled arrangement for that.

02Follow the ray from water into air

Imagine one small point on the coin reflecting light towards your eye.

The light travels through two media, in this order:

1. Water, where the ray begins its journey towards the surface.
2. Air, where the refracted ray travels towards your eye.

For our calculations, use the following illustrative refractive indices for visible light.

A refractive index, \(n\), describes how much more slowly light travels in a medium than in a vacuum. It has no units.

Step 1: Draw the normal

At the point where the ray meets the water surface, draw a line perpendicular to that surface. This line is called the normal.

For a flat, horizontal water surface, the normal is vertical.

All angles of incidence and refraction are measured between the ray and the normal, not between the ray and the water surface.

Step 2: Identify the incident and refracted angles

For light travelling from the coin towards an observer in air:

• The incident angle, \(i\), is the angle between the ray in water and the normal.
• The refracted angle, \(r\), is the angle between the ray in air and the normal.

Because light passes from water into air, it bends away from the normal.

Therefore, for a ray that crosses the surface at a non-zero angle,

\[ r>i \]

Step 3: Calculate the bending with Snell’s law

Snell’s law gives the relationship between the two angles:

\[ n_{\mathrm w}\sin i=n_{\mathrm a}\sin r \]

Suppose a ray leaves the coin and reaches the water surface at \(20^\circ\) to the normal.

Substituting:

\[ \begin{aligned} 1.33\sin20^\circ&=1.00\sin r\\ \sin r&=0.455\\ r&=27.1^\circ \end{aligned} \]

The ray travels at \(20^\circ\) to the normal in water and \(27.1^\circ\) to the normal in air.

Notice what has changed: the emerging ray is further away from the normal than the underwater ray.

The coin doesn’t move. Only the direction of the light changes.

03A better experiment: Measuring the ray rather than judging the coin

The cup experiment tells us something has changed, but it isn’t a good way to measure refraction. The rim obstructs our view, and small changes in eye position or curvature of the water surface can affect what we see.

We can improve the investigation using a transparent, straight-sided tank, a narrow light beam, and a protractor.

Arrange the apparatus in this order:

1. Fill a rectangular transparent tank with water, allowing the surface to settle.
2. Position a ray box in the air above the tank, directing a narrow beam diagonally downwards towards the water surface.
3. Mark the point where the beam enters the water.
4. Draw a normal perpendicular to the surface at that point.
5. Measure the beam’s angle in air and its refracted angle in water using a protractor.

A ray box directed from air into water reverses the coin-to-eye path. This works because light paths through these transparent media are reversible: a ray following the same path in the opposite direction obeys the same refraction law.

What should the measurements show?

For an air-to-water ray, the angle in water should be smaller than the angle in air.

The following values are theoretical predictions calculated using Snell’s law with the illustrative refractive indices above. They are not recorded experimental measurements.

If the measured angles follow this pattern, they support the refraction model.

For a fair test, the interface should remain flat, and the tank should not move between readings. When observing the ray through a tank wall, view approximately perpendicular to that wall to minimise additional refraction.

This revised setup also helps distinguish refraction from reflection. A reflected ray remains in the original medium; a refracted ray crosses into the second medium.

You could use several measured angle pairs to estimate the refractive index rather than assuming its value. The full graph method is covered in How to Find Refractive Index from a Sine-Sine Graph.

04Why does the coin appear higher than its real position?

We know which way the light bends. Now we need to connect that bending to the position you see.

Your eye receives light travelling along the refracted rays in air. It interprets their directions as though the light has travelled towards you in straight lines.

Imagine extending those emerging rays backwards into the water without allowing them to bend again.

For rays close to the normal, their backward extensions meet approximately at a point above the real coin.

That apparent point is called a virtual image.

The dashed backward extensions are not real light paths. They show where the observer interprets the light as coming from.

This also explains our opening puzzle about the undeviated ray.

A ray travelling exactly along the normal doesn’t bend, but it only identifies a straight line on which the coin could lie. Neighbouring angled rays provide the additional directional information needed to estimate its distance.

Deriving the apparent-depth relationship

Call the real depth of the coin \(d\) and its apparent depth \(d_{\mathrm{app}}\).

Both are positive vertical distances measured downwards from the flat water surface, in metres.

Consider an angled ray leaving the coin and emerging at a point horizontally displaced by \(x\) from the vertical line directly above the coin.

Let:

• \(\theta_{\mathrm w}\) be the angle the ray makes with the normal in water.
• \(\theta_{\mathrm a}\) be the angle the refracted ray makes with the normal in air.
• \(x\) be the horizontal displacement, measured in metres.

The underwater triangle gives:

\[ x=d\tan\theta_{\mathrm w} \]

If we extend the air ray backwards until it crosses the vertical line through the coin, the second triangle gives:

\[ x=d_{\mathrm{app}}\tan\theta_{\mathrm a} \]

Combining them:

\[ d_{\mathrm{app}} =d\frac{\tan\theta_{\mathrm w}} {\tan\theta_{\mathrm a}} \]

For rays close to the normal, the angles are small. Their sines and tangents are approximately equal.

Snell’s law then allows us to write:

\[ d_{\mathrm{app}}\approx d\frac{n_{\mathrm a}}{n_{\mathrm w}} \]

For water viewed from air:

\[ d_{\mathrm{app}}\approx\frac{d}{1.33} \]

So an underwater object appears to be at approximately 75% of its real depth when viewed nearly vertically from air.

This relationship is an approximation, not an exact rule for every viewing angle.

It assumes a flat, stationary surface, uniform transparent media, and rays travelling close to the normal. At larger viewing angles, the back-projected position depends on ray direction. A strongly curved or rippled surface makes the simple model less reliable.

05The contrasting case: What if the observer is underwater?

It’s tempting to remember one rule: objects viewed through water look closer.

But that rule is incomplete. You must identify which medium contains the object and which contains the observer.

Imagine an idealised underwater observer looking upwards at an object suspended in air.

The light now travels from air into water.

It bends towards the normal rather than away from it. The underwater observer traces the received rays backwards and, near the normal, sees the object apparently further above the surface than its real position.

For a flat interface, the general near-normal relationship is:

\[ d_{\mathrm{app}}\approx d\frac{n_{\mathrm{observer}}} {n_{\mathrm{object}}} \]

Here, \(d\) and \(d_{\mathrm{app}}\) are positive distance magnitudes from the boundary on the object’s side. An object above the surface therefore remains apparently above it.

If an object is \(0.50\,\mathrm m\) above the water, an ideal observer in water would estimate its height as:

\[ d_{\mathrm{app}}\approx 0.50\times\frac{1.33}{1.00} =0.665\,\mathrm m \]

It appears approximately \(0.67\,\mathrm m\) above the surface, not \(0.50\,\mathrm m\).

This assumes the observer is optically in water without an additional air gap, mask or glass window.

An object in water viewed by another observer in the same uniform water doesn’t undergo this boundary refraction at all, provided the light travels directly between them without crossing the surface.

The useful decision is not simply whether water is present. It’s whether the light crosses a boundary on its way to the observer, and in which direction.

06Questions and solutions

The following questions use idealised, flat interfaces and the supplied refractive indices. Unless a question states otherwise, assume the observer looks using rays close to the normal, and ignore additional glass surfaces. All numerical values are illustrative.

Question 1

A small stone rests \(0.80\,\mathrm m\) below the surface of a still pond. An observer looks almost vertically down from air.

Use \(n_{\mathrm w}=1.33\) and \(n_{\mathrm a}=1.00\).

a) Calculate the apparent depth of the stone.

b) Calculate how far above its real position it appears.

c) State whether the refracted ray bends towards or away from the normal as it leaves the water.

Solution 1

The stone appears approximately \(0.60\,\mathrm m\) below the surface, which is about \(0.20\,\mathrm m\) above its real position. The light bends away from the normal when it enters air.

a) Apparent depth

\[ \begin{aligned} d_{\mathrm{app}} &\approx d\frac{n_{\mathrm a}}{n_{\mathrm w}}\\ &=0.80\times\frac{1.00}{1.33}\\ &=0.602\,\mathrm m \end{aligned} \]

b) Apparent upward displacement

\[ \begin{aligned} \Delta d&=0.80-0.602\\ &=0.198\,\mathrm m \end{aligned} \]

c) Direction of refraction

Light travels from water, with refractive index \(1.33\), into air, with refractive index \(1.00\). Snell’s law requires a larger angle to the normal in air.

The common mistake is to draw the refracted ray bending towards the normal simply because it is entering a new medium. The correct direction depends on the refractive indices of both media and the direction of travel.

Question 2

Two small illuminated markers are positioned on the same vertical line through a flat water surface.

• Marker A is \(0.60\,\mathrm m\) above the surface in air.
• Marker B is \(0.80\,\mathrm m\) below the surface in water.

An observer in air looks nearly vertically down and estimates that the apparent vertical separation between the markers is \(1.20\,\mathrm m\).

a) Determine the refractive index of the water from this observation. Take the refractive index of air as \(1.00\).

b) A second idealised observer is fully underwater and looks upwards along nearly normal rays. Calculate the apparent separation they would estimate.

c) Explain why the two observers report different separations even though neither marker moves.

Solution 2

The water’s refractive index is approximately \(1.33\). The underwater observer estimates an apparent separation of \(1.60\,\mathrm m\), compared with the air observer’s \(1.20\,\mathrm m\).

a) Recover the refractive index.

For the observer in air, marker A is in the same medium as the observer, so its height remains \(0.60\,\mathrm m\).

Only marker B’s apparent depth is reduced by the water-air boundary.

Therefore:

\[ 1.20=0.60+\frac{0.80}{n_{\mathrm w}} \]

Rearranging:

\[ \begin{aligned} 0.60&=\frac{0.80}{n_{\mathrm w}}\\ n_{\mathrm w}&=\frac{0.80}{0.60}\\ &=1.33 \end{aligned} \]

The refractive index is dimensionless.

b) Calculate the underwater observer’s apparent separation.

For this observer, marker B lies in the same water and remains \(0.80\,\mathrm m\) below the interface.

Marker A is in air. Because the light travels from air into water, its apparent height increases.

\[ \begin{aligned} h_{\mathrm{app,A}} &=0.60\times\frac{1.33}{1.00}\\ &\approx0.80\,\mathrm m \end{aligned} \]

The apparent separation is:

\[ \begin{aligned} s_{\mathrm{app}} &=0.80+0.80\\ &=1.60\,\mathrm m \end{aligned} \]

c) Explain the difference.

The two observers receive different refracted rays and therefore construct different apparent positions.

The air observer sees the underwater marker shifted towards the surface. The water observer sees the above-water marker shifted further away from the surface.

A tempting mistake is to apply the water-to-air depth formula to both markers, regardless of where the observer is. That would ignore which ray actually crosses the interface.

The real separation remains \(1.40\,\mathrm m\). Neither apparent measurement changes the physical positions.

Question 3

A transparent tank contains two stable, unmixed liquid layers with horizontal boundaries.

From top to bottom:

• An upper liquid X, \(0.30\,\mathrm m\) deep, with refractive index \(1.50\).
• Water, \(0.90\,\mathrm m\) deep, with refractive index \(1.33\).

A coin rests on the tank floor beneath both layers. An observer looks almost vertically down from air, with refractive index \(1.00\).

a) Estimate the coin’s apparent depth below the top surface.

b) The upper liquid is replaced by water, keeping the total liquid depth and coin position unchanged. Calculate the new apparent depth.

c) Does the coin’s apparent position move upwards or downwards? Explain why applying one average refractive index can be misleading.

Solution 3

The coin initially appears approximately \(0.877\,\mathrm m\) below the surface. Replacing the upper liquid with water makes it appear approximately \(0.902\,\mathrm m\) deep, so its apparent position moves downwards by about \(0.026\,\mathrm m\).

a) Calculate the initial apparent depth.

For nearly normal rays crossing parallel boundaries, the apparent-depth contributions of the individual layers add:

\[ d_{\mathrm{app}}\approx \frac{n_{\mathrm a}}{n_{\mathrm X}}d_{\mathrm X} + \frac{n_{\mathrm a}}{n_{\mathrm w}}d_{\mathrm w} \]

Here \(d_{\mathrm X}\) and \(d_{\mathrm w}\) are the actual thicknesses of the upper liquid and water.

Substituting:

\[ \begin{aligned} d_{\mathrm{app}} &\approx\frac{0.30}{1.50} +\frac{0.90}{1.33}\\ &=0.200+0.677\\ &=0.877\,\mathrm m \end{aligned} \]

b) Replace the upper layer with water.

The entire \(1.20\,\mathrm m\) is now water.

\[ \begin{aligned} d_{\mathrm{app,new}} &\approx\frac{1.20}{1.33}\\ &=0.902\,\mathrm m \end{aligned} \]

c) Explain the apparent movement.

The difference is:

\[ \begin{aligned} \Delta d_{\mathrm{app}} &=0.902-0.877\\ &=0.025\,\mathrm m \end{aligned} \]

Using unrounded values gives approximately \(0.0256\,\mathrm m\), or \(2.6\,\mathrm{cm}\).

The coin appears slightly deeper after the upper layer is replaced.

Why? The original upper liquid has a larger refractive index than water. For the same physical thickness, it contributes a smaller amount to the apparent depth seen from air.

The tempting shortcut is to calculate a simple arithmetic average of \(1.50\) and \(1.33\), then treat the entire tank as one medium. But the layers have different thicknesses, and each contributes according to its own refractive index.

The result follows from tracing nearly normal rays across both interfaces, not from assigning the tank an unweighted average refractive index.

Question 4

A coin is fixed to the floor of a rectangular tank. Initially, the water surface is \(1.60\,\mathrm m\) above the coin.

Water is slowly drained until the depth is \(1.00\,\mathrm m\). The coin and tank floor do not move.

An observer in air looks almost vertically down throughout. Assume \(n_{\mathrm w}=1.33\) and \(n_{\mathrm a}=1.00\).

A student argues:

“Because the water is becoming shallower, the apparent image of the coin must rise towards the observer.”

a) Calculate the apparent height of the coin above the fixed tank floor before and after draining.

b) State whether the apparent image actually moves upwards or downwards relative to the floor, and by how much.

c) Predict where the apparent image ends up when the tank is completely empty. Use this limiting case to check your conclusion.

Solution 4

The apparent image moves downwards by approximately \(0.149\,\mathrm m\) relative to the fixed floor. Its height above the floor decreases from about \(0.397\,\mathrm m\) to \(0.248\,\mathrm m\).

a) Calculate the initial apparent height.

At the beginning, the real water depth is \(1.60\,\mathrm m\).

The coin’s apparent depth measured from the water surface is:

\[ \begin{aligned} d_{\mathrm{app,1}} &=\frac{1.60}{1.33}\\ &=1.203\,\mathrm m \end{aligned} \]

The image height above the floor is:

\[ \begin{aligned} h_1&=1.60-1.203\\ &=0.397\,\mathrm m \end{aligned} \]

After draining, the water depth is \(1.00\,\mathrm m\):

\[ \begin{aligned} d_{\mathrm{app,2}} &=\frac{1.00}{1.33}\\ &=0.752\,\mathrm m \end{aligned} \]

The new image height above the fixed floor is:

\[ \begin{aligned} h_2&=1.00-0.752\\ &=0.248\,\mathrm m \end{aligned} \]

b) Determine the direction of movement.

\[ \begin{aligned} \Delta h&=h_1-h_2\\ &=0.397-0.248\\ &=0.149\,\mathrm m \end{aligned} \]

The image moves downwards by approximately \(0.15\,\mathrm m\), approaching the real coin.

The student’s statement feels reasonable because the apparent depth measured from the water surface becomes smaller as water is removed.

But the surface itself has moved down by \(0.60\,\mathrm m\). Apparent depth measured from a moving surface is not the same thing as apparent height measured from a fixed floor.

That changing reference point is the crucial part of the problem.

c) Check the limiting case.

With no water remaining, the coin is viewed directly through air. There is no water-air refraction between it and the observer.

Its apparent height above the floor is therefore \(0\,\mathrm m\), the same as its real position.

This agrees with the calculated downward movement: as the water disappears, the apparent image approaches the real coin.

Question 5

A light-emitting point lies \(1.20\,\mathrm m\) below a perfectly flat water surface. Three rays leave this same point and eventually enter air.

The rays emerge into air at angles of \(0^\circ\), \(30^\circ\), and \(60^\circ\) to their respective surface normals. They travel in the same vertical plane, and the two non-vertical rays emerge on the same side of the point directly above the source.

Take \(n_{\mathrm w}=1.33\) and \(n_{\mathrm a}=1.00\).

A student reasons:

“All three rays come from one physical point. If an observer traces them backwards through the air, they must all lead to one exact virtual image point. That point should be \(1.20/1.33\) metres below the surface.”

a) Use Snell’s law to calculate the water-side angles for the \(30^\circ\) and \(60^\circ\) air rays.

b) For each of those two rays, calculate where its straight backward extension in the air direction crosses the vertical line through the real source. Give the depth below the surface.

c) Explain why the \(0^\circ\) ray does not settle the disagreement between these two depths.

d) Decide whether one exact virtual point can explain all three rays. Explain why the familiar near-normal apparent-depth formula is still useful, and why the source does not necessarily appear as three separate objects.

Assume geometrical optics, a uniform medium on either side, and that no rays are blocked.

Solution 5

The student’s conclusion is not exact. The \(30^\circ\) and \(60^\circ\) air rays, when independently extended backwards to the vertical line through the source, cross it at different depths: approximately \(0.843\,\mathrm m\) and \(0.594\,\mathrm m\). Therefore, all three rays cannot be represented by one exact virtual point.

The familiar result \(d_{\mathrm{app}}\approx d/1.33\) is a near-normal approximation, not an angle-independent image position.

a) Calculate the water-side angles.

Snell’s law gives:

\[ 1.33\sin\theta_{\mathrm w} =1.00\sin\theta_{\mathrm a} \]

For the \(30^\circ\) air ray:

\[ \begin{aligned} \sin\theta_{\mathrm w} &=\frac{\sin30^\circ}{1.33}\\ &=0.376\\ \theta_{\mathrm w}&=22.1^\circ \end{aligned} \]

For the \(60^\circ\) air ray:

\[ \begin{aligned} \sin\theta_{\mathrm w} &=\frac{\sin60^\circ}{1.33}\\ &=0.651\\ \theta_{\mathrm w}&=40.6^\circ \end{aligned} \]

These are physically possible refracted rays. The water-air critical angle is approximately \(48.8^\circ\) in water, and both incident angles are below it.

b) Calculate each backward crossing depth.

Let \(x\) be the horizontal displacement from the vertical line through the source to the ray’s emergence point.

For each ray:

\[ x=d\tan\theta_{\mathrm w} \]

The straight backward extension of its air segment reaches that vertical line at depth:

\[ d_{\mathrm{cross}} =\frac{x}{\tan\theta_{\mathrm a}} \]

For the \(30^\circ\) air ray:

\[ \begin{aligned} x_{30} &=1.20\tan22.1^\circ\\ &=0.487\,\mathrm m \end{aligned} \]

Therefore:

\[ \begin{aligned} d_{30} &=\frac{0.487}{\tan30^\circ}\\ &=0.843\,\mathrm m \end{aligned} \]

For the \(60^\circ\) air ray:

\[ \begin{aligned} x_{60} &=1.20\tan40.6^\circ\\ &\approx1.03\,\mathrm m \end{aligned} \]

Therefore, retaining unrounded intermediate values:

\[ \begin{aligned} d_{60} &=\frac{x_{60}}{\tan60^\circ}\\ &=0.594\,\mathrm m \end{aligned} \]

The two rays produce different backward crossing depths.

c) Explain the vertical ray.

For the \(0^\circ\) ray, the angle in water is also \(0^\circ\), so there is no bending.

Its straight backward continuation lies along the vertical line through the actual source.

But every point on that vertical line is also on the ray’s straight backward path. The ray therefore doesn’t establish one particular depth.

It cannot be used to insist that the virtual image must be at the source’s real position.

d) Test the single-image claim.

If all three backward ray extensions passed through one exact virtual point, that point would have to lie on the vertical \(0^\circ\) ray.

But the other two backward extensions intersect that same vertical at different depths.

So there is no common intersection point for all three rays.

The near-normal formula gives:

\[ d_{\mathrm{app}}\approx \frac{1.20}{1.33} =0.902\,\mathrm m \]

This is the limiting depth approached by the backward crossing construction for rays very close to the normal.

At larger angles, the tangent ratios in the exact ray geometry no longer match the sine ratios used in the approximation.

There is a further detail: two backward extensions always intersect somewhere in a two-dimensional diagram, but their intersection need not lie on the vertical through the actual source. A third ray exposes whether that intersection is truly common to all the rays.

This doesn’t mean an observer necessarily sees three separate glowing dots.

An individual eye receives only a restricted bundle of rays determined by its position and pupil. Different viewing positions can produce different apparent locations, and a wide bundle may not form a perfectly sharp point image. The three rays in this problem travel towards different observing directions.

The deeper lesson is that an apparent image is constructed from ray directions. A fixed object does not guarantee that rays refracted by a flat boundary have backward extensions meeting at one exact point for every angle.

07Where this idea leads next

When you see an object through a flat water surface, the reliable procedure is to identify the object medium and observer medium, draw the normal at the crossing point, determine the direction of bending, and then trace the emerging rays backwards. Use the simple apparent-depth formula only when the near-normal approximation is justified.

At increasingly large angles, another question becomes important: can the underwater ray escape into air at all? The same water-air interface has a critical angle beyond which total internal reflection prevents an ordinary refracted ray from emerging.

That is the next useful boundary test, covered in How to Find the Critical Angle Between Two Named Media.