Put a one-rupee coin at the bottom of an empty steel bucket. Step back slowly until the rim of the bucket just hides the coin. Now stand still while a friend pours water gently into the bucket. The coin comes back into view, although neither you nor the coin has moved.
Water looks shallower than it is
The last lesson ended with a straw that seemed to break at the surface of a drink. The part under the drink looked raised. The same effect lifts the coin in the bucket into view. Light from the coin bends away from the normal as it leaves the water. The eye traces each ray straight back, and the rays seem to come from a point above the coin.
How far above? Take a coin at depth \(h\). A ray from it meets the surface at an angle \(\theta_w\) in the water and leaves at \(\theta_a\) in the air, with \(\mu\sin\theta_w = \sin\theta_a\). The eye sees the coin at the apparent depth \(h'\), where the ray traced back meets the vertical through the coin.
\[ x = h\tan\theta_w \]
Why. the ray rises through the depth \(h\) at the angle \(\theta_w\) to the vertical. It reaches the surface a distance \(x\) from the point straight above the coin.
\[ x = h'\tan\theta_a \]
Why. the eye traces the ray in the air straight back at the angle \(\theta_a\). It meets the vertical through the coin at the depth \(h'\), the same distance \(x\) away.
\[ h' = h\,\dfrac{\tan\theta_w}{\tan\theta_a} \]
Why. equate the two expressions for \(x\).
\[ h' \approx h\,\dfrac{\sin\theta_w}{\sin\theta_a} = \dfrac{h}{\mu} \]
Read it. for a view close to the normal the angles are small, and \(\tan\theta \approx \sin\theta\). Snell's law gives \(\sin\theta_a = \mu\sin\theta_w\).
A coin under \(12\,\text{cm}\) of water appears at \(12 \div \tfrac43 = 9\,\text{cm}\). It seems raised by \(3\,\text{cm}\), the normal shift. A pond \(4\,\text{m}\) deep looks \(3\,\text{m}\) deep from its bank. Many people have misjudged a pond or a river this way, so never judge the depth of water by eye.
Looking straight down, the readout above gives \(9.0\,\text{cm}\), the value of \(h/\mu\). Tilt the line of sight, and the image rises: \(7.3\,\text{cm}\) at \(30°\) from the vertical, only \(2.5\,\text{cm}\) at \(60°\). It also moves towards you. So the formula \(h/\mu\) holds only for viewing near the normal. Seen from the side, a pool looks shallower still.
The formula \(h' = h/\mu\) needs a view close to the normal, because it replaces \(\tan\theta\) by \(\sin\theta\). A problem that says "viewed from above" or "viewed normally" means exactly this.
Looking out from the water
Now reverse the situation. A fish in the water looks up at a bird in the air. Rays from the bird bend towards the normal as they enter the water. Traced back, they seem to come from a point higher than the bird. The same derivation, with the media swapped, gives an apparent height of \(\mu\) times the real height.
A single rule covers both cases. The apparent distance from the surface is the real distance multiplied by \(\mu_{\text{observer}}/\mu_{\text{object}}\). From air, water shrinks depths by \(\tfrac34\). From water, air stretches heights by \(\tfrac43\).
A fish is \(2\,\text{m}\) below the surface of a lake. A bird is \(3\,\text{m}\) above the surface, on the same vertical line. Water has \(\mu = \tfrac43\). How far away does each one see the other?
Each sees the other near the normal. The fish looks from water into air; the bird looks from air into water.
Fish: apparent height of the bird \(= \tfrac43 \times 3\). Bird: apparent depth of the fish \(= 2 \div \tfrac43\).
The bird appears \(4\,\text{m}\) above the surface, so \(2 + 4 = 6\,\text{m}\) from the fish. The fish appears \(1.5\,\text{m}\) below the surface, so \(3 + 1.5 = 4.5\,\text{m}\) from the bird.
The fish sees the bird \(6\,\text{m}\) away; the bird sees the fish \(4.5\,\text{m}\) away. The real distance, \(5\,\text{m}\), is between the two.
A swimming pool is filled to a depth of \(2.0\,\text{m}\) with water of \(\mu = \tfrac43\). How deep does it look to a person looking straight down into it, in metres?
Layers add
A coin lies in a beaker under \(8\,\text{cm}\) of water, with \(6\,\text{cm}\) of oil of \(\mu = 1.5\) floating on the water. Seen from above, each layer shrinks only its own thickness. The water layer looks \(8 \div \tfrac43 = 6\,\text{cm}\) thick, and the oil layer \(6 \div 1.5 = 4\,\text{cm}\). So the coin appears \(6 + 4 = 10\,\text{cm}\) below the top of the oil, raised by \(14 - 10 = 4\,\text{cm}\).
Why may the layers be taken one at a time? The water forms an image of the coin. For the oil, that image is the object, and the oil forms an image of it in turn. At each flat boundary only the relative index acts, and the relative indices multiply out so that each layer contributes its own \(t/\mu\).
A tank holds \(20\,\text{cm}\) of water (\(\mu = \tfrac43\)) with \(3\,\text{cm}\) of oil (\(\mu = 1.5\)) on top. How far below the top of the oil does a mark on the bottom appear, seen from above, in cm?
A moving object, a slower image
The apparent depth is \(h/\mu\) at every instant. So if the real depth changes, the apparent depth changes \(\mu\) times more slowly. A fish rising at \(4\,\text{m/s}\) appears, from above, to rise at \(4 \div \tfrac43 = 3\,\text{m/s}\). Seen from the water, an object in air moves \(\mu\) times faster than it really does.
The fish and the bird of the example above move towards each other at the same instant. The fish rises at \(4\,\text{m/s}\); the bird dives at \(3\,\text{m/s}\). At what speed does each see the other approach?
The bird sees the fish's image rise at \(4 \div \tfrac43 = 3\,\text{m/s}\). The bird itself dives at \(3\,\text{m/s}\). So it sees the fish approach at \(3 + 3 = 6\,\text{m/s}\).
The fish sees the bird's image fall at \(\tfrac43 \times 3 = 4\,\text{m/s}\), and itself rises at \(4\,\text{m/s}\). So it sees the bird approach at \(4 + 4 = 8\,\text{m/s}\).
Each observer's own motion is real; only the motion across the surface is scaled.
A slab on a printed page
Lay a glass slab \(6\,\text{cm}\) thick, with \(\mu = 1.5\), on a page of your physics book. Looking down, the letters appear \(6 \div 1.5 = 4\,\text{cm}\) below the top face. They seem raised by \(6 - 4 = 2\,\text{cm}\). This raising is the normal shift of the slab, \(t(1 - 1/\mu)\).
Now lift the slab a few centimetres above the page. The letters stay raised by exactly \(2\,\text{cm}\). The shift does not depend on where the slab sits between the object and the eye. The derivation below shows why.
\[ \text{in air: sideways drift} = x\tan\theta, \qquad \text{in glass: } t\tan\theta_g \]
Why. a ray leaves the object at a small angle \(\theta\) to the normal. It crosses air gaps of total thickness \(x\) at \(\theta\), and the slab at \(\theta_g\), where \(\sin\theta = \mu\sin\theta_g\).
\[ \text{traced back: } (x + t)\tan\theta - (\text{true drift}) = t\tan\theta - t\tan\theta_g \]
Why. the eye traces the final ray back at the angle \(\theta\) as if there were air all the way. The drift it expects falls short of the true drift only inside the slab.
\[ \text{shift} = t\left(1 - \dfrac{\tan\theta_g}{\tan\theta}\right) \approx t\left(1 - \dfrac{1}{\mu}\right) \]
Why. the missing drift, divided by \(\tan\theta\), is how much nearer the image appears. Near the normal \(\tan\theta_g/\tan\theta \approx \sin\theta_g/\sin\theta = 1/\mu\).
\[ \text{the gap } x \text{ has cancelled} \]
Read it. the air paths bend no ray, so they add the same drift in the real path and in the traced-back path. Only the slab counts.
Under water the slab does less, because only the relative index \(\mu/\mu_m\) acts. A glass slab \(9\,\text{cm}\) thick shifts an object by \(9(1 - 1/1.5) = 3\,\text{cm}\) in air. Under water the relative index is \(1.5 \div \tfrac43 = \tfrac98\), and the shift is only \(9(1 - \tfrac89) = 1\,\text{cm}\).
A microscope is focused on a dot on a table. A glass slab \(3\,\text{cm}\) thick, with \(\mu = 1.5\), is then placed over the dot. What must be done to see the dot sharply again?
A beaker is filled with a liquid to a depth of \(12\,\text{cm}\). A mark on its bottom appears raised by \(3\,\text{cm}\). What is the refractive index of the liquid?
A glass slab \(6\,\text{cm}\) thick (\(\mu = 1.5\)) lies on a plane mirror. A point object O is \(20\,\text{cm}\) above the mirror. An eye above O looks down at the image of O in the mirror. How far below O is that image?
Light from O goes down through the slab, reflects, and comes back up through the slab. Seen through the slab, the mirror itself appears raised.
The mirror appears raised by the normal shift \(6(1 - 1/1.5) = 2\,\text{cm}\). The effective mirror is \(20 - 2 = 18\,\text{cm}\) below O.
The image is \(18\,\text{cm}\) behind the effective mirror, that is \(18 + 18 = 36\,\text{cm}\) below O.
The image of O is \(36\,\text{cm}\) below O. The slab has brought it \(4\,\text{cm}\) nearer, two shifts of \(2\,\text{cm}\), one for each crossing.
The \(6\,\text{cm}\) slab lies on a printed page, and the letters appear raised by \(2\,\text{cm}\). The slab is now lifted slowly at \(2\,\text{cm/s}\), staying between the page and the eye. How fast does the image of the letters move?
The ray comes out parallel
A ray meets a glass slab at an angle \(i\). At the first face \(\sin i = \mu\sin r\). The two faces are parallel, so their normals are parallel, and the ray meets the second face at the same angle \(r\). There \(\mu\sin r = \sin e\). So \(\sin e = \sin i\), and the emergent ray is parallel to the incident ray.
The same chain works for any number of parallel slabs: \[ \sin i = \mu_1\sin r_1 = \mu_2\sin r_2 = \mu_3\sin r_3 = \sin e. \] The angle inside any one slab depends only on the angle in air and that slab's own \(\mu\). A ray entering at \(45°\) runs at \(30°\) in a slab of \(\mu = \sqrt2\), whatever slabs lie above it.
The emergent ray is parallel, but not on the same line. It is displaced sideways by the lateral shift \(d\), the perpendicular distance between the incident ray, continued, and the emergent ray.
\[ AB = \dfrac{t}{\cos r} \]
Why. the ray enters at A and leaves at B. Inside, it runs at the angle \(r\) to the normal across the thickness \(t\).
\[ \angle(AB, \text{incident line}) = i - r \]
Why. the incident line makes the angle \(i\) with the normal at A. AB makes the angle \(r\) with the same normal, on the same side.
\[ d = AB\sin(i - r) \]
Why. \(d\) is the perpendicular from B to the incident line continued through A. In a right triangle with hypotenuse AB, it is the side opposite the angle \(i - r\).
\[ d = \dfrac{t\sin(i - r)}{\cos r} \]
Read it. substitute AB from step 1.
Drive the slider from \(0\) to \(85°\) in the figure above. At normal incidence \(d = 0\): the ray goes straight through. The shift grows with the angle and approaches the whole thickness at grazing incidence. At \(i = 90°\) the formula gives \(d = t\sin(90° - r)/\cos r = t\) for any glass.
A ray falls at \(45°\) on a slab \(10\,\text{cm}\) thick with \(\mu = \sqrt2\). Find the angle of refraction and the lateral shift. Take \(\sin 15° = 0.26\) and \(\cos 30° = 0.87\).
\(\sin r = \sin 45°/\sqrt2 = \tfrac12\), so \(r = 30°\).
\(d = 10\sin(45° - 30°)/\cos 30° = 10 \times 0.26 / 0.87 \approx 3\,\text{cm}\).
For small angles instead, \(\sin(i - r) \approx i - r\), \(\cos r \approx 1\) and \(r \approx i/\mu\), so \(d \approx t\,i(1 - 1/\mu)\). A slab \(6\,\text{cm}\) thick of \(\mu = 1.5\) at \(i = 0.1\,\text{rad}\) gives \(d \approx 6 \times 0.1 \times \tfrac13 = 0.2\,\text{cm}\).
A ray from air passes through three parallel slabs of refractive indices \(\mu_1\), \(\mu_2 = \sqrt2\) and \(\mu_3\), in that order, and comes out into air. It enters at \(45°\). Which statements are true?
- Using \(h/\mu\) for an object seen at a large slant.\(h/\mu\) holds only near the normal. At a slant the image is higher still, as the coin figure shows.
- Dividing by \(\mu\) when the observer is in the denser medium.Multiply the real distance by \(\mu_{\text{observer}}/\mu_{\text{object}}\). From water, air distances look longer.
- Dividing the total depth of layers by one average \(\mu\).Divide each layer's thickness by its own \(\mu\), then add.
- Thinking the normal shift depends on how far the slab is from the object.The air gaps bend no ray and cancel. Only \(t\) and \(\mu\) set the shift.
- Confusing the normal shift with the lateral shift.The normal shift moves an image along the line of sight; the lateral shift moves a slanting ray sideways. At normal incidence the first is largest and the second is zero.