A Pond Is Always Deeper Than It Looks, and by an Exact Amount

Put a coin at the bottom of a bucket of water and it looks as though it is floating well above the base. Reach for it and your fingers meet the bottom before they meet the coin. The water has not moved the coin. It has bent the light coming from it.

And the bending is not vague. A bucket of water $4$ m deep makes the bottom appear to be at $3$ m — exactly three quarters of the true depth, because the refractive index of water is $\frac{4}{3}$. The same rule makes a stick dipped obliquely in water look bent at the surface and makes the bottom of a swimming pool look shallower than it is.

That bending of light on passing from one transparent medium into another is refraction, and it happens because light travels at different speeds in different media.

- Light travels fastest in vacuum, at about $3\times 10^8$ m s${}^{-1}$
- In water it slows to about three quarters of that speed
- In glass to about two thirds
- In diamond to less than half

The ratio of the two speeds is the refractive index, and it is the single number that decides how sharply a medium bends light.

Four things follow from it, and they make up this part of the chapter.

- The laws of refraction, which say where the refracted ray goes, and which of the wave's properties change on the way
- The refractive index, defined from the speeds and used in every numerical
- The path of a ray through a rectangular glass block, which explains lateral displacement and the faint extra images you see in a thick mirror
- Real and apparent depth, which is the coin in the bucket turned into a formula

**Take the speed of light in vacuum as $3\times 10^8$ m s${}^{-1}$** throughout.

This page covers the first part of the ICSE Class 10 Physics chapter on light: the laws of refraction, refractive index, refraction through a glass block, and real and apparent depth.

Two laws fix where the refracted ray goes, and of the wave's three properties only the frequency survives unchanged.

The first law. The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.

The second law, called Snell's law. For light of a given colour passing from one medium into another, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant:

$$\frac{\sin i}{\sin r}=\text{constant}=\mu$$

and that constant is the refractive index of the second medium with respect to the first.

Which way the ray bends.

- Going from a rarer to a denser medium — air into glass — the ray bends toward the normal, so $r<i$
- Going from a denser to a rarer medium — glass into air — the ray bends away from the normal, so $r>i$

Now the three properties of the light wave. On crossing into a denser medium:

- The speed decreases. This is the cause of the bending, not a consequence of it
- The wavelength decreases, in exactly the same proportion as the speed
- The frequency does not change at all

Why the frequency cannot change. The frequency is set by the source that emitted the light, and the number of waves arriving at the boundary each second must equal the number leaving it — otherwise waves would pile up at the surface. So the frequency is carried across unaltered, and since $V=\nu \lambda$, a smaller speed forces a smaller wavelength.

Worked example — the wavelength in glass. Light of wavelength $6000$ angstrom in air enters glass of refractive index $1.5$. Find its wavelength and frequency in glass, given that $1$ angstrom $=10^{-10}$ m.

The speed in glass:

$$V=\frac{c}{\mu }=\frac{3\times 10^8}{1.5}=2\times 10^8{\text{ m s}}^{-1}$$

The wavelength in glass, reduced in the same ratio:

$${\lambda }_{\text{glass}}=\frac{6000}{1.5}=4000\text{ angstrom}$$

The frequency in air:

$$\nu =\frac{c}{\lambda }=\frac{3\times 10^8}{6\times 10^{-7}}=5\times 10^{14}\text{ Hz}$$

And in glass, as a check:

$$\nu =\frac{V}{{\lambda }_{\text{glass}}}=\frac{2\times 10^8}{4\times 10^{-7}}=5\times 10^{14}\text{ Hz}$$

The same frequency, which confirms that only the speed and the wavelength changed.

Worked example 2 — Snell's law both ways. Light travels from air into a glass of refractive index $\sqrt{2}$ at an angle of incidence of $45^{\circ }$. Find the angle of refraction. Then find the angle of refraction when light leaves the same glass into air at $30^{\circ }$.

Air into glass:

$$\sin r=\frac{\sin 45^{\circ }}{\sqrt{2}}=\frac{0.7071}{1.4142}=0.5⟹r=30^{\circ }$$

Glass into air, where the roles reverse so $\frac{\sin i}{\sin r}=\frac{1}{\sqrt{2}}$:

$$\sin r=\sqrt{2}\times \sin 30^{\circ }=1.4142\times 0.5=0.7071⟹r=45^{\circ }$$

The path is exactly reversed, which illustrates the principle of reversibility of light: a ray that goes from A to B by some path will travel from B to A by the same path if its direction is reversed.

The two cases in which a ray passes undeviated, and both are examined:

- When the ray strikes the surface normally, so $i=0^{\circ }$. Then $\sin i=0$, so $\sin r=0$ and $r=0^{\circ }$. The ray goes straight through, though its speed still changes
- When the two media have the same refractive index. Then $\mu =1$ and $r=i$, so there is no bending at all. A glass rod immersed in a liquid of the same refractive index becomes practically invisible, because no light is bent or reflected at its surface

The second case is worth noting carefully. Refraction is not caused by the boundary itself but by the change in speed across it. No change in speed means no refraction, however sharp the boundary looks.

It bends toward the normal on entering, bends away from the normal on leaving, and emerges parallel to its original direction but shifted sideways.

Tracing the path, step by step.

- At the first face the ray passes from air into glass, so it bends toward the normal and $r_1<i_1$
- Inside the glass it travels in a straight line to the second face
- At the second face the two faces are parallel, so the angle of incidence inside the glass equals $r_1$. The ray now passes from glass into air, bending away from the normal
- On emerging, the angle of emergence equals the original angle of incidence, $e=i_1$

So the emergent ray is parallel to the incident ray. The two bendings are equal and opposite, because the two faces are parallel — which is why a glass block shifts an image sideways but does not tilt it.

Lateral displacement is the perpendicular distance between the direction of the incident ray produced forward and the emergent ray.

Three things increase it, and a question on this asks for all three:

- A thicker block — the ray travels further inside before emerging
- A larger angle of incidence — the ray is bent more steeply and travels a longer slanting path
- A larger refractive index — the ray is bent more sharply on entering

And it is zero when the ray enters normally, since then there is no bending at all.

Partial reflection at each face. At every boundary between two transparent media, part of the light is reflected and part is refracted. So at the first face of the block a faint reflected ray comes back into the air, and at the second face a faint ray is reflected back into the glass. The refracted beam is therefore always a little weaker than the incident beam, and that partial reflection is the cause of the next effect.

Multiple images in a thick glass mirror. An ordinary mirror is a sheet of glass silvered on the back, so light from an object meets two surfaces: the front glass surface and the silvered back surface.

- A faint first image is formed by partial reflection at the front glass surface, before the light has entered the glass at all
- A bright second image is formed by reflection at the silvered back surface. This is the image we normally see, and it is the brightest
- Further images, progressively fainter, are formed when the light reflected from the silvering is partly reflected again at the front surface, travels back to the silvering, and returns

So a thick mirror shows a series of images, of which the second is much the brightest. The faint ones are easiest to see when looking at a bright object at a large angle, and they are the reason a good optical mirror is silvered on the front surface rather than the back.

The same effect appears in a thick glass plate without any silvering, where several faint images of a bright object can be seen because of repeated partial reflections between the two faces.

Worked example — the emergent ray. A ray strikes one face of a rectangular glass block of refractive index $1.5$ at an angle of incidence of $45^{\circ }$. Find the angle of refraction inside the glass and the angle of emergence.

$$\sin r_1=\frac{\sin 45^{\circ }}{1.5}=\frac{0.7071}{1.5}=0.4714⟹r_1\approx 28^{\circ }$$

At the second face, the angle of incidence inside the glass is $28^{\circ }$ and the ray leaves into air, so

$$\sin e=1.5\times \sin 28^{\circ }=1.5\times 0.4714=0.7071⟹e=45^{\circ }$$

The angle of emergence equals the angle of incidence, so the emergent ray is parallel to the incident ray. That result holds for any angle and any refractive index, provided the two faces are parallel — and it is the reason a glass window does not distort the view, while a prism does.

Divide the real depth by the refractive index.

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

so

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

and the amount by which the object appears raised is

$$\text{apparent shift}=\text{real depth}-\text{apparent depth}=\text{real depth}\left(1-\frac{1}{\mu }\right)$$

Why the object appears raised. Rays from the coin leave the water at the surface and bend away from the normal, since they are going from a denser to a rarer medium. To an eye above, those rays appear to come from a point higher up, where their backward extensions meet — and that point is the apparent position of the coin. The image is virtual, formed by the brain producing the rays backwards, which is why you cannot catch it.

Worked example 1 — a tank of water. A coin lies at the bottom of a tank of water $4$ m deep. Taking the refractive index of water as $\frac{4}{3}$, find the apparent depth and how much the coin appears raised.

$$\text{apparent depth}=\frac{4}{4/3}=4\times \frac{3}{4}=3\text{ m}$$
$$\text{apparent shift}=4-3=1\text{ m}$$

So the bottom looks three quarters of its true depth, and that fraction is $\frac{1}{\mu }$ — the same for any depth of water.

Worked example 2 — a mark under a glass slab. A mark on a table is viewed through a glass slab $6$ cm thick of refractive index $1.5$. Find the apparent depth of the mark and how much it appears raised.

$$\text{apparent depth}=\frac{6}{1.5}=4\text{ cm}$$
$$\text{apparent shift}=6-4=2\text{ cm}$$

Check with the shift formula:

$$6\left(1-\frac{1}{1.5}\right)=6\left(1-\frac{2}{3}\right)=6\times \frac{1}{3}=2\text{ cm}$$

The same value, and the shift formula is the quicker route when only the shift is asked for.

Worked example 3 — finding the refractive index. A coin at the bottom of a vessel containing a liquid $12$ cm deep appears to be at a depth of $9$ cm. Find the refractive index of the liquid.

$$\mu =\frac{12}{9}=\frac{4}{3}=1.33$$

So the liquid is water, or something with the same refractive index — and this is the standard laboratory method of measuring a refractive index without any angles at all.

Worked example 4 — a pool. A swimming pool is $3$ m deep. How deep does it appear to a person standing at the edge and looking straight down? Take $\mu =1.33$.

$$\text{apparent depth}=\frac{3}{1.33}=2.26\text{ m}$$

The pool appears about three quarters of a metre shallower than it is, which is a genuine hazard and the reason depths are painted on the side of a pool rather than judged by eye.

The bent stick, explained. A straight stick held obliquely in water appears bent at the surface.

- The part above the water is seen directly, by light travelling in a straight line
- The part below the water is seen by refracted light, which bends away from the normal on leaving the water
- So every submerged point appears raised, and the submerged part of the stick appears lifted toward the surface
- The result is a sharp apparent bend exactly at the water line, with the submerged part appearing to slope upward

And the stick appears to bend upward, not downward, which is the detail to state. If the stick is held vertically it appears shorter but not bent, since every point is raised along the same vertical line — a useful boundary case that shows the bending is caused by the sideways view.

One condition attached to the formula. The relation $\mu =\frac{\text{real depth}}{\text{apparent depth}}$ holds only when the object is viewed normally, that is from almost directly above. At a large viewing angle the apparent depth is smaller still, and the simple formula no longer applies — which is why the derivation assumes the eye is nearly overhead.