Refraction Through a Glass Slab & Apparent Depth

A coin at the bottom of a bucket of water sits lower than it looks. A fish hangs higher than it swims. A pencil standing in a glass jumps sideways where it pierces the surface, and a thick glass paperweight slides the print beneath it visibly off to one side. These are not separate curiosities — they are the two everyday faces of the same physics: what refraction does when light passes through a transparent block or across a flat surface, rather than merely bending once and moving on. There are two results to own, and both come straight from Snell's law. The first is apparent depth: look straight down into a medium and everything appears raised, because the rays leaving an underwater object bend away from the normal as they exit and, traced back by your eye, seem to come from a shallower point. The rule is beautifully simple — apparent depth $=$ real depth $/\,n$ — so a pool looks only three-quarters as deep as it is ( $n_{\text{water}} = 4/3$ ), and every underwater distance is compressed by the index. The second is the lateral shift through a parallel-sided slab: a ray entering obliquely bends at the first surface, bends back by an equal and opposite amount at the second, and emerges parallel to its original path but displaced sideways by $x = t\,\sin(i - r)/\cos r$ . The slab does not change the ray's direction — only its position. This article derives both from Snell's law using nothing beyond trigonometry and small-angle geometry, then works the standard exam numbers: a $4$ cm pool that looks $3$ cm deep, a $6$ cm glass slab whose $2$ cm normal shift and $4$ cm apparent thickness follow instantly, and a $10$ cm slab that shifts a $60^\circ$ ray sideways by $5.125$ cm. Master these, and the flat surface is done — the next step is to curve it, which is where lenses begin.

Key takeaways

Apparent depth $= \dfrac{\text{real depth}}{n}$ : looking straight down, everything in a medium appears raised — a $4$ cm pool ( $n = 4/3$ ) looks $3$ cm deep; a coin under $8$ cm of water ( $n = 1.33$ ) appears at $6.02$ cm.
A glass slab ( $t = 6$ cm, $n = 1.5$ ) has apparent thickness $t/n = 4.0$ cm and raises objects by a normal shift $t(1 - 1/n) = 2.0$ cm.
Lateral shift through a parallel slab: the emergent ray is parallel to the incident ray but displaced by $x = t\,\dfrac{\sin(i - r)}{\cos r}$ ; for $t = 10$ cm, $i = 60^\circ$ , $n = 1.5$ (so $r = 35.26^\circ$ ), $x = 5.125$ cm.
A slab offsets a ray without deviating it — direction unchanged, position shifted.
Both effects follow from Snell's law — apparent depth from near-normal incidence, lateral shift from the slab's equal-and-opposite bends.
Key idea: a flat surface raises what you see by the factor $n$ , and a parallel slab slides a ray sideways without turning it — both are Snell's law in disguise.

The core insight

Through a flat boundary an object viewed near the normal appears raised to depth (real depth) $/n$ , and through a parallel-sided slab a ray emerges parallel to itself but shifted sideways by $t\sin(i-r)/\cos r$ — the slab offsets without deviating — and both results are Snell's law applied to simple geometry.

Where this lives today

The flat-surface and slab effects run through everyday sight and precision optics alike. Spearfishers and herons must aim below an apparent fish, correcting instinctively for the raising by $n$ ; divers judging depth, and anyone reaching for something underwater, are fooled the same way. Glass and acrylic blocks in a school lab demonstrate the lateral shift directly, and the same offset must be accounted for when a laser beam passes through a window or a protective flat in an instrument. Plane-parallel plates are used deliberately in interferometers and beam-splitters precisely because they shift a beam without turning it. The apparent-thickness rule $t/n$ matters whenever you look through a thick optical window or measure a specimen under a cover slip in a microscope — the true depth is $n$ times what the focus says. And conceptually, the apparent-depth result is the flat-surface limit ( $R \to \infty$ ) of refraction at a curved surface, $n_2/v - n_1/u = (n_2 - n_1)/R$ , which is the direct ancestor of the lens. When that curved-surface physics is assembled into a lens, its imaging equation reads $1/v - 1/u = 1/f$ with $m = +v/u$ — a minus where the mirror formula $1/v + 1/u = 1/f$ has a plus, as the mirror-formula article flags. The shallow pool, in other words, is the first step toward the camera lens.

The physical picture

Everything here is one refraction result — Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$ — used twice over, once for looking into a medium and once for passing through a slab. When you look down at a submerged coin, the light that reaches your eye left the coin, travelled up through the water, and bent away from the normal as it crossed into the less dense air (rarer medium, so the ray swings away). Your eye and brain, which assume light travels in straight lines, trace those bent rays back along their final straight-line directions to a point that is higher than the coin actually is. The coin looks raised; the water looks shallow. Nothing about the coin moved — only the direction of the light did. A parallel-sided slab is the same law at two surfaces. A ray strikes the top face and refracts toward the normal (into the denser glass), crosses the slab as a straight line, then strikes the bottom face and refracts away from the normal by exactly the same angle it gained at the top — because the two faces are parallel and Snell's law is symmetric. So it comes out travelling in precisely its original direction, undeviated. But during its slanted passage through the glass it drifted sideways, and that drift is frozen in as a permanent sideways offset between where the ray would have been and where it actually emerges. Direction preserved, position shifted: that is the signature of a slab.

Figure

A ray entering a parallel-sided glass slab from the top-left, refracting toward the normal inside the glass, then refracting back at the lower face to emerge parallel to its original direction but displaced sideways, with the perpendicular lateral shift x marked between the dashed original path and the emergent ray

Apparent depth and the shifts

Three formulas cover every flat-surface and slab question, and all three descend from Snell's law. First, apparent depth for near-normal viewing:

d_{\text{apparent}} = \frac{d_{\text{real}}}{n} \qquad (1)

where: $d_{\text{real}}$ is the true depth of the object below the surface, $n$ the index of the medium it sits in (viewed from air), so the object always looks raised because $n > 1$ . A pool of water ( $n = 4/3$ ) shows only $3/4$ of its depth; the raising is more dramatic in denser media. For a slab of thickness $t$ viewed along the normal, two companion quantities follow at once — the apparent thickness and the amount an object seen through it appears to shift up:

t_{\text{apparent}} = \frac{t}{n}, \qquad \text{normal shift} = t\left(1 - \frac{1}{n}\right) \qquad (2)

where: the apparent thickness is the slab's depth compressed by $n$ , and the normal shift — the distance an object under the slab appears to rise — is the difference between the real and apparent thicknesses, $t - t/n$ . For a $6$ cm glass slab this is a clean $6 - 4 = 2$ cm. For an oblique ray through a parallel slab, the emergent ray is parallel to the incident one but laterally displaced by

x = t\,\frac{\sin(i - r)}{\cos r} \qquad (3)

where: $i$ is the angle of incidence, $r$ the angle of refraction inside the slab (from Snell's law, $\sin r = \sin i / n$ ), and $t$ the slab thickness. The shift grows with thickness and with the obliqueness $i - r$ ; it vanishes at normal incidence ( $i = 0$ , so $r = 0$ and $x = 0$ ) and at $t = 0$ . Note that only the oblique ray shifts sideways — looking straight down through a slab gives the depth shift of equation (2), not a lateral $x$ .

Limiting cases

Case	Result	Meaning
view along the normal	$d_{\text{app}} = d_{\text{real}}/n$	maximum raising; the pool looks $1/n$ as deep
$n \to 1$ (matched medium)	$d_{\text{app}} \to d_{\text{real}}$	no raising, no shift — the block "disappears" optically
$i = 0$ (normal incidence on slab)	$x = 0$	a straight-down ray is offset in depth only, not sideways
$t \to 0$ (thin slab)	$x \to 0$	nothing to drift through; no lateral shift
$i \to 90^\circ$ (grazing)	$x \to t\tan(\text{large})$ , largest	the most oblique rays shift the most
flat surface as $R \to \infty$	single-surface refraction reduces to $d_{\text{app}} = d_{\text{real}}/n$	the seed of the lens story

Figure

The apparent-depth construction — a coin at real depth 4 cm under water sends two rays up to an eye, the rays refract away from the normal at the surface, and traced back they meet at a shallower apparent depth of 3 cm, so the coin looks raised

Derivation AApparent depth equals real depth over n

The "shallow pool" result is Snell's law read at small angles, and doing it once shows exactly why the raising factor is the index and nothing else. The goal is to derive $d_{\text{apparent}} = d_{\text{real}}/n$ for an object at depth $d$ in a medium of index $n$ , viewed nearly straight down from air, using only Snell's law and the small-angle approximation $\sin\theta \approx \tan\theta$ . Picture the object $O$ at depth $d$ , and two rays leaving it: one straight up (undeviated) and one at a small angle that strikes the surface a little to the side.

Step 1. Refract the slanted ray. Inside the medium it makes a small angle $\theta_1$ with the normal; on crossing into air it bends away to $\theta_2$ , with $n\sin\theta_1 = 1\cdot\sin\theta_2$ (object in the denser medium, emerging into air).

Step 2. Locate the two crossing points on the surface. If the slanted ray meets the surface a horizontal distance $x$ from the vertical ray, then from the geometry $\tan\theta_1 = \dfrac{x}{d_{\text{real}}}$ (real triangle, from the true object) and $\tan\theta_2 = \dfrac{x}{d_{\text{apparent}}}$ (apparent triangle, from where the eye traces the ray back).

Step 3. Take the ratio. $\dfrac{d_{\text{apparent}}}{d_{\text{real}}} = \dfrac{\tan\theta_1}{\tan\theta_2}$ . For the small angles of near-normal viewing, $\tan\theta \approx \sin\theta$ , so $\dfrac{\tan\theta_1}{\tan\theta_2} \approx \dfrac{\sin\theta_1}{\sin\theta_2}$ .

Step 4. Substitute Snell's law. From $n\sin\theta_1 = \sin\theta_2$ we get $\dfrac{\sin\theta_1}{\sin\theta_2} = \dfrac{1}{n}$ , hence $d_{\text{apparent}} = \dfrac{d_{\text{real}}}{n}$ . For a pool $4$ cm deep in water, $d_{\text{apparent}} = \dfrac{4}{4/3} = 3.0$ cm — raised by $1$ cm. The same law gives the slab's normal shift $t - t/n = t(1 - 1/n)$ : for $t = 6$ cm and $n = 1.5$ , a shift of $6(1 - 2/3) = 2.0$ cm and apparent thickness $6/1.5 = 4.0$ cm. Verified: $4/(4/3) = 3.0$ cm, $8/1.33 = 6.02$ cm, and the slab's $2.0$ cm shift and $4.0$ cm apparent thickness are recomputed in verify_hs2-refraction-glass-slab.py.

Go deeper · Derivation B · 1 minThe lateral shift through a parallel slab

Derivation BThe lateral shift through a parallel slab

The sideways offset a slab gives an oblique ray is the article's signature result, and it is pure slab geometry once Snell's law fixes the internal angle. The goal is to derive $x = t\,\sin(i - r)/\cos r$ for a ray crossing a parallel-sided slab of thickness $t$ , and to compute it for the standard $t = 10$ cm, $i = 60^\circ$ , $n = 1.5$ case. Set up the ray: it strikes the top face at incidence $i$ , refracts to angle $r$ inside, travels to the bottom face, and emerges parallel to its entry direction.

Step 1. Find the internal angle. Snell's law at the top face: $\sin i = n\sin r$ , so $\sin r = \dfrac{\sin i}{n} = \dfrac{\sin 60^\circ}{1.5} = \dfrac{0.8660}{1.5} = 0.5774$ , giving $r = 35.26^\circ$ and $\cos r = 0.8165$ .

Step 2. Find the path length inside the slab. The ray crosses the perpendicular thickness $t$ while travelling at angle $r$ to the normal, so its actual path length in the glass is $\ell = \dfrac{t}{\cos r}$ (the hypotenuse of a right triangle with the thickness as one leg).

Step 3. Project onto the perpendicular to the incident direction. The lateral shift $x$ is the component of that internal path measured perpendicular to the original ray direction. The internal ray is turned from the original direction by the angle $(i - r)$ , so $x = \ell\,\sin(i - r) = \dfrac{t}{\cos r}\sin(i - r) = t\,\dfrac{\sin(i - r)}{\cos r}$ .

Step 4. Put the numbers in. $i - r = 60^\circ - 35.26^\circ = 24.74^\circ$ , so $x = 10\cdot\dfrac{\sin 24.74^\circ}{\cos 35.26^\circ} = 10\cdot\dfrac{0.4184}{0.8165} = 5.125$ cm. The emergent ray runs parallel to the incident ray — the slab did not turn it — but it is displaced sideways by just over $5$ cm. At normal incidence ( $i = 0$ ) the formula gives $x = 0$ , as it must. Verified: $r = 35.26^\circ$ and $x = 5.125$ cm for $t = 10$ cm, $i = 60^\circ$ , $n = 1.5$ are recomputed in verify_hs2-refraction-glass-slab.py.

Common trapObjects look shallower, not deeper — and a slab shifts a ray without bending it

Two sign-of-the-effect errors dominate here. First, apparent depth is real depth divided by $n$ , not multiplied — since $n > 1$ , an object always looks shallower (raised), never deeper. Writing "apparent depth $= n \times$ real depth" makes the pool look $5.3$ cm deep instead of $3$ cm and is simply backwards; the water always flatters the depth. Second, a parallel-sided slab does not deviate the emergent ray — it comes out exactly parallel to how it went in. The slab's whole effect is a sideways offset $x$ , not a change of direction; a solution that has the ray leaving the slab at a different angle has broken the parallel-faces symmetry. Two finer traps: the lateral-shift formula uses the refraction angle $r$ in $\cos r$ and the difference $(i - r)$ — putting $i$ where $r$ belongs is a frequent slip; and the sideways shift $x$ needs an oblique ray, so a "straight-down" viewing gives only the depth shift of $t(1 - 1/n)$ , with no lateral $x$ at all. Keep "shallower, by dividing" and "offset, not bent" in mind and the whole topic stays straight.

Worked examples

Board level — the shallow pool and the raised coin (IGCSE and CBSE pattern)

(a) A swimming pool is $4$ cm of clear water deep at its edge ( $n = 4/3$ ); how deep does it look from directly above? (b) A coin lies under $8$ cm of water ( $n = 1.33$ ); at what depth does it appear?

Step 1. Use apparent depth $= \dfrac{\text{real depth}}{n}$ for straight-down viewing.

Step 2. (a) $d_{\text{apparent}} = \dfrac{4}{4/3} = 4 \times \dfrac{3}{4} = \boxed{3.0\ \mathrm{cm}}$ — the pool looks exactly three-quarters as deep, raised by $1$ cm.

Step 3. (b) $d_{\text{apparent}} = \dfrac{8}{1.33} = \boxed{6.02\ \mathrm{cm}}$ — the coin appears nearly $2$ cm higher than it is.

Step 4. In both cases the object looks shallower, because dividing by $n > 1$ shrinks the depth. This is exactly why a river or pool is deceptively deeper than it looks — and why you should never judge water depth by eye. IGCSE note: your syllabus states this as $n = \dfrac{\text{real depth}}{\text{apparent depth}}$ , which rearranges to equation (1); here $\dfrac{4}{3} = \dfrac{4}{3.0}$ .

Competitive level — the glass slab, its shifts, and the lateral offset (JEE Main and NEET pattern)

A glass slab has $n = 1.5$ . (a) For a slab $6$ cm thick, find its apparent thickness and the normal shift of an object seen straight through it. (b) For a slab $10$ cm thick, a ray strikes at $i = 60^\circ$ ; find the angle of refraction inside and the lateral shift of the emergent ray.

Step 1. (a) Apparent thickness $= \dfrac{t}{n} = \dfrac{6}{1.5} = \boxed{4.0\ \mathrm{cm}}$ . Normal shift $= t\left(1 - \dfrac{1}{n}\right) = 6\left(1 - \dfrac{1}{1.5}\right) = 6 \times \dfrac{1}{3} = \boxed{2.0\ \mathrm{cm}}$ .

Step 2. (b) Internal angle from Snell's law: $\sin r = \dfrac{\sin 60^\circ}{1.5} = \dfrac{0.8660}{1.5} = 0.5774$ , so $r = \boxed{35.26^\circ}$ and $\cos r = 0.8165$ .

Step 3. Lateral shift: $x = t\,\dfrac{\sin(i - r)}{\cos r} = 10 \times \dfrac{\sin(60^\circ - 35.26^\circ)}{\cos 35.26^\circ} = 10 \times \dfrac{\sin 24.74^\circ}{0.8165} = 10 \times \dfrac{0.4184}{0.8165} = \boxed{5.125\ \mathrm{cm}}$ .

Step 4. The emergent ray leaves parallel to the incident ray — undeviated — but shifted sideways by $5.125$ cm. That is why writing viewed through a thick glass paperweight appears displaced but not tilted. Timing note: about 80 seconds. The efficiency trick is to compute $\cos r$ once from $\sin r = 0.5774$ (so $\cos r = \sqrt{1 - 0.5774^2} = 0.8165$ ) and reuse it; and remember it is $(i - r)$ in the sine and $r$ in the cosine — never $i$ in the cosine.

Exam traps

Multiplying instead of dividing. Apparent depth $=$ real depth $/\,n$ . Since $n > 1$ , the object looks shallower. Multiplying by $n$ is backwards.

Thinking the slab bends the ray. A parallel-sided slab leaves the emergent ray parallel to the incident ray — it offsets, it does not deviate. Only a change of position, never of direction.

Using $i$ where $r$ belongs. The lateral-shift formula is $x = t\sin(i - r)/\cos r$ : the difference $(i - r)$ in the sine, the refraction angle $r$ in the cosine. Swapping them is a standard slip.

Expecting a lateral shift from straight-down viewing. The sideways $x$ needs an oblique ray. Looking along the normal gives only the depth/normal shift, no lateral offset.

Board scope. Apparent depth and lateral shift are examined by CBSE, ICSE, JEE Main, JEE Advanced, NEET, IGCSE and A-Level (Edexcel International A-Level). IGCSE 0625 owns "real vs apparent depth" and $n = \text{real}/\text{apparent}$ qualitatively; the lateral-shift formula is the CBSE/JEE/NEET emphasis. This topic is not part of AP Physics 2 or IB. (As with all refraction here, "A-Level" means Edexcel IAL — CIE 9702 has no geometric optics.)

Origins · 1 minRené Descartes and how this was discovered

René Descartes

1596–1650 · Netherlands & Paris

Descartes gave refraction its first printed law and its first coordinate-geometry treatment — exactly the geometry this article computes. In La Dioptrique (1637), one of the essays appended to his Discourse on Method, he set out the sine law of refraction in print for the first time and used it to analyse how light crosses media, deriving results for lenses and, in the companion essay Les Météores, explaining the rainbow through refraction and one internal reflection in each droplet. The honest scope: the sine law itself was found earlier and privately by Willebrord Snellius (around 1621, unpublished — the portrait of the previous article), so Descartes is credited with publication and geometric method, not first discovery, and the two dates must be kept distinct: Snell $\approx 1621$ (empirical, unpublished), Descartes 1637 (first in print). His coordinate approach — attaching numbers and axes to geometry — is the very habit behind the apparent-depth and lateral-shift calculations here, and behind the "Cartesian" sign convention the chapter uses.

History

1621

— Willebrord Snellius finds the sine law of refraction empirically, but does not publish it.

1637

— René Descartes prints the sine law in La Dioptrique and explains the rainbow in Les Météores.

1662

— Pierre de Fermat derives the sine law from the principle of least time.

1678

— Christiaan Huygens accounts for refraction and apparent depth with his wavefront construction.

1690

— Huygens publishes the Traité de la Lumière, systematising the wave view of refraction.

Going further

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics"), refraction at plane surfaces and apparent depth · NCERT Science Class 10, Ch. 10, refraction through a glass slab · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics", slab and apparent-depth problems · Cambridge IGCSE 0625, real and apparent depth.

Provenance. Standard geometric optics; every number recomputed in verify_hs2-refraction-glass-slab.py. Recompute: pool $4/(4/3) = 3.0$ cm; coin $8/1.33 = 6.02$ cm; slab $t = 6$ , $n = 1.5$ : apparent thickness $6/1.5 = 4.0$ cm, normal shift $6(1 - 1/1.5) = 2.0$ cm; lateral shift $t = 10$ , $i = 60^\circ$ : $r = \arcsin(\sin 60^\circ/1.5) = 35.26^\circ$ , $x = 10\sin(24.74^\circ)/\cos(35.26^\circ) = 5.125$ cm.

Prerequisites. Refraction of Light and Snell's Law — the bending rule and $n = c/v$ that every result here rests on. Leads to. Total Internal Reflection and Critical Angle (what happens when a ray cannot get out), Refraction at a Spherical Surface (the curved-boundary generalisation), and the UG seam Refractive Index and Optical Path Length.

Go deeper (UG). The idea that a slab adds "optical path length" $n t$ , and how that governs interference and the design of plane-parallel plates, is developed in Refractive Index and Optical Path Length in the university optics track. Enrichment only; every formula here follows from Snell's law and trigonometry.

Why does a swimming pool always look shallower than it is?

Because the light from the pool floor bends away from the normal as it leaves the water and enters the air (water is optically denser, so exiting into the rarer air the ray swings away from the normal). Your eye assumes light came in a straight line, so it traces those exit rays back to a point higher than the true floor — the floor looks raised, and the water looks shallow. The exact factor is the refractive index: apparent depth $=$ real depth $/\,n$ , so with $n_{\text{water}} = 4/3$ the pool shows only three-quarters of its depth. A $4$ cm depth looks $3$ cm; a $2$ m deep end looks about $1.5$ m. It is a genuine safety issue — natural water is deeper than it appears.

Does the glass slab bend the light or not?

It bends the light inside the slab, but the ray comes out travelling in exactly its original direction, so the net effect is no deviation — only a sideways shift. Here is why: at the first face the ray refracts toward the normal by some amount; at the second face, which is parallel to the first, it refracts away from the normal by precisely the same amount (Snell's law is symmetric and the normals are parallel). The two bends cancel in direction, leaving the emergent ray parallel to the incident one. What does not cancel is the sideways drift it accumulated crossing the tilted path through the glass — that survives as the lateral shift $x = t\sin(i - r)/\cos r$ . So: bent within, parallel without, shifted sideways.

What is the difference between the normal shift and the lateral shift?

They answer different questions. The normal shift, $t(1 - 1/n)$ , is how much an object seen straight through a slab appears to move toward you (raised along the line of sight) — it is a depth effect and needs no oblique ray. The lateral shift, $x = t\sin(i - r)/\cos r$ , is how far an oblique ray is displaced sideways as it passes through — a position effect perpendicular to the ray. For a $6$ cm slab of index $1.5$ the normal shift is $2$ cm; a $60^\circ$ ray through a $10$ cm slab has a lateral shift of $5.125$ cm. One is about looking through along the normal; the other is about a slanted ray's parallel offset. Keep them apart by asking: am I looking straight down (normal shift) or is the ray oblique (lateral shift)?

Why do we use the small-angle approximation for apparent depth?

Because the simple result apparent depth $=$ real depth $/\,n$ holds exactly only for rays that reach the eye close to the vertical (near-normal viewing), which is how you usually look into water. For those small angles, $\tan\theta \approx \sin\theta$ , and the ratio of tangents that sets the depth ratio becomes the ratio of sines, which Snell's law fixes as $1/n$ . If you look at a steep, oblique angle, the apparent depth changes and the object also appears distorted, because different rays now bend by different amounts — the image blurs and shifts. So the clean $/n$ rule is a near-normal result; boards always intend straight-down (or nearly straight-down) viewing when they quote it, which is why the approximation is legitimate.

Does the colour of the light change how much the coin is raised?

Slightly, yes — because the refractive index depends a little on colour (this is dispersion, developed in a later chapter). Blue light has a marginally higher index than red in glass and water, so blue is raised a touch more than red, and the apparent depth differs by a small amount across the spectrum. For exam purposes you use a single quoted index for the medium (like $n = 4/3$ for water) and ignore the tiny colour dependence, so the coin has one apparent depth. But the effect is real and is why fine optical instruments and rainbows care about colour: the same surface bends different colours by slightly different amounts. Here, treat $n$ as one number and the apparent depth as a single value.

Is this topic on my exam board?

Apparent depth and lateral shift are examined by CBSE, ICSE, JEE Main, JEE Advanced, NEET, IGCSE and A-Level — but with a split in emphasis. IGCSE 0625 owns the qualitative "real versus apparent depth" and the relation $n = \text{real}/\text{apparent}$ ; it does not require the lateral-shift formula. The lateral-shift calculation $x = t\sin(i - r)/\cos r$ is a CBSE/JEE/NEET (and Edexcel IAL) topic. This material is not part of AP Physics 2 or IB Physics, which cover refraction and Snell's law but not the slab shifts. And "A-Level" here means Edexcel International A-Level; CIE 9702 has no geometric optics, so a 9702 candidate does not study slabs or apparent depth for that paper.