Refraction of Light & Snell's Law

Dip a straight straw into a glass of water and it snaps at the surface — the lower half slides sideways and looks broken. Stand a coin in an empty bowl until it just vanishes below the rim, then pour in water: the coin floats back into view. A swimming pool always looks shallower than the day you learn, the hard way, that it is not. All three are the same phenomenon, and all three have the same one-line cause: when light crosses from one transparent medium into another, it changes speed, and a wave that changes speed at an angle must change direction. That bending is refraction, and the exact rule that governs it is Snell's law. The speed change is captured by a single number, the refractive index $n = c/v$ : how many times slower light travels in the medium than in vacuum. Glass has $n \approx 1.5$ , so light crawls through it at two-thirds of $c$ ; water is $4/3$ ; diamond a remarkable $2.42$ . Whenever light enters a slower (higher- $n$ , "optically denser") medium it bends toward the normal; entering a faster medium it bends away. Snell's law makes this quantitative: $n_1 \sin\theta_1 = n_2 \sin\theta_2$ , with every angle measured from the normal, not the surface. This article builds Snell's law from the wavefront picture — the part of a wavefront that reaches the slow medium first lags behind and tilts the whole front, exactly like a marching column wheeling when one flank slows — so you see why the ratio of sines is the ratio of indices, not merely that it is. Then we pin down the companions: frequency is unchanged across a boundary but wavelength shrinks as $\lambda/n$ , and rays are perfectly reversible. Refraction is the gateway to the entire second half of ray optics — slabs, prisms, fibres and lenses all begin here.

Key takeaways

Refraction is the bending of light when it crosses a boundary and changes speed; the refractive index $n = c/v \ge 1$ measures the slowing (glass $1.5$ , water $4/3$ , diamond $2.42$ ).
Snell's law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$ , all angles from the normal. Into a denser medium light bends toward the normal; into a rarer one, away.
Speeds: $v_{\text{glass}} = c/1.5 = 2.0\times10^8$ m/s, $v_{\text{water}} = 2.25\times10^8$ , $v_{\text{diamond}} = 1.24\times10^8$ m/s.
Worked anchors: air→glass at $30^\circ$ gives $\theta_2 = 19.47^\circ$ ; air→water at $45^\circ$ gives $\theta_2 = 32.03^\circ$ ; the reverse glass→air ray at $19.47^\circ$ returns to $30^\circ$ (reversibility).
Frequency is unchanged across the boundary (the source sets it); wavelength shrinks, $\lambda_{\text{medium}} = \lambda_{\text{vac}}/n$ (so $600$ nm light becomes $400$ nm in glass).
Key idea: light changes speed at a boundary, so it changes direction, and $n_1 \sin\theta_1 = n_2 \sin\theta_2$ with $n = c/v$ fixes exactly how much — measured from the normal, every time.

The core insight

Crossing into a medium where it travels at $v = c/n$ , light bends because a tilted wavefront reaches the new speed unevenly; Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ quantifies the bend, frequency stays fixed while wavelength scales as $\lambda/n$ , and every angle is measured from the normal.

Where this lives today

Snell's law is the working equation of every device that steers light by bending it. Eyeglasses and camera lenses shape their surfaces so that rays refract to a chosen focus; the index of the glass, plugged into refraction at each surface, is what the designer tunes. Optical fibres carry the internet on light that refracts into a glass core and then stays trapped — the total-internal-reflection story of the next-but-one article, whose critical angle is just Snell's law pushed to its limit. Prisms in binoculars and spectrometers refract (and, because $n$ depends slightly on colour, disperse) light in controlled ways. Even a mirage on a hot road is refraction: light curving through air whose index changes with temperature. And the whole chapter ahead — apparent depth, the sideways shift through a slab, refraction at a curved surface, and finally the lens — is Snell's law applied to progressively cleverer geometries. In fact the coming lens equation, $1/v - 1/u = 1/f$ with $m = +v/u$ , carries a minus where the mirror formula $1/v + 1/u = 1/f$ carried a plus (as the mirror-formula article flags); that sign flip, and the New Cartesian conventions behind it, all trace back to bending governed by $n_1\sin\theta_1 = n_2\sin\theta_2$ .

The physical picture

Light is a wave, and a wave is defined by how fast it travels. In vacuum every colour races at $c = 3.0\times10^8$ m/s (more precisely $2.998\times10^8$ ); inside glass or water the electric field of the wave keeps interacting with the atoms, and the net forward progress slows to some $v < c$ . The refractive index simply names the slowdown: $n = c/v$ . Because nothing outruns light in vacuum, $n \ge 1$ always — glass $\approx 1.5$ , water $\approx 4/3 \approx 1.33$ , diamond $2.42$ , and air a whisker above one at $1.0003$ , which we round to $1$ . Now send that light at a boundary obliquely. The wavefront — the line of "same phase", perpendicular to the ray — meets the surface at an angle, so one edge of the front crosses into the slow medium while the other edge is still in the fast one. The edge that has entered the slow side falls behind, and the whole front pivots, like a line of marchers wheeling when the soldiers on one end take shorter steps. The ray, always perpendicular to its front, swings with it — toward the normal when entering a slower medium, away when entering a faster one. That is refraction, and the geometry of the pivot is exactly Snell's law. It is worth saying what does not change: the wave's frequency. The boundary cannot create or destroy wave crests, so they arrive and leave at the same rate; what shrinks in the slow medium is the spacing between crests, the wavelength.

Figure

A light ray crossing from air into glass at a horizontal boundary — the normal drawn vertical and dashed, the incident ray at 30 degrees from the normal in air of index 1, the refracted ray bent toward the normal at 19.47 degrees in glass of index 1.5, labelled with Snell's law n1 sin theta1 equals n2 sin theta2

Snell's law and the refractive index

Two equations carry the whole chapter. The first defines the index from the speed:

n = \frac{c}{v} \qquad (1)

where: $c$ is the speed of light in vacuum and $v$ its speed in the medium, so $n \ge 1$ and a larger $n$ means slower light. From it, the relative index between two media is $n_{21} = \dfrac{n_2}{n_1} = \dfrac{v_1}{v_2}$ — how much the light slows going specifically from medium 1 into medium 2. The second is Snell's law, the bending rule:

n_1 \sin\theta_1 = n_2 \sin\theta_2 \qquad (2)

where: $\theta_1$ is the angle of incidence and $\theta_2$ the angle of refraction, both measured from the normal, and $n_1, n_2$ are the indices on the incident and refracted sides. Read it as a conserved quantity: the product $n\sin\theta$ has the same value on both sides of any boundary. Its consequences are immediate — if $n_2 > n_1$ (into a denser medium) then $\sin\theta_2 < \sin\theta_1$ , so $\theta_2 < \theta_1$ and the ray bends toward the normal; if $n_2 < n_1$ it bends away. Because the equation is symmetric under swapping $(1)\leftrightarrow(2)$ , running a ray backwards retraces its path exactly — rays are reversible. A third relation ties the wave picture together. Since $v = f\lambda$ and the frequency $f$ is fixed across the boundary while $v$ drops to $c/n$ , the wavelength must shrink in step:

\lambda_{\text{medium}} = \frac{\lambda_{\text{vac}}}{n} \qquad (3)

where: $\lambda_{\text{vac}}$ is the vacuum (or air) wavelength. So yellow light of $600$ nm becomes $400$ nm inside glass, while its frequency — and hence its colour — is unchanged. This is why a submerged object keeps its colour even though the light inside the water is "shorter".

Limiting cases

Case	Result	Meaning
$\theta_1 = 0$ (normal incidence)	$\theta_2 = 0$	a ray along the normal passes straight through, unbent
$n_2 = n_1$	$\theta_2 = \theta_1$	matched media — no bending (index-matching fluids)
$n_2 > n_1$ (into denser)	$\theta_2 < \theta_1$	bends toward the normal (air → glass)
$n_2 < n_1$ (into rarer)	$\theta_2 > \theta_1$	bends away from the normal (glass → air)
$\theta_1 \to 90^\circ$ , air → glass	$\theta_2 \to \arcsin(1/n) = 41.8^\circ$	grazing incidence fills the "cone" — the seed of the critical angle

Figure

The wavefront explanation of refraction — parallel wavefronts in air approach the boundary obliquely, the edge that enters the glass first travels a shorter distance v2 t while the trailing edge covers v1 t in air, so the front tilts and the ray swings toward the normal

Derivation ASnell's law from the wavefront picture

Snell's law is not a decree; it is what geometry forces once you accept that the wavefront moves slower in the denser medium. The goal here is to derive $n_1\sin\theta_1 = n_2\sin\theta_2$ using only the wavefront construction and $n = c/v$ , with no calculus — the least-time principle behind it (Fermat's, the portrait of the total-internal-reflection article) is stated but not minimised. Picture a plane wavefront in medium 1 meeting the flat boundary at incidence angle $\theta_1$ .

Step 1. Track two edges of the wavefront. One edge, call it the "leading" point $A$ , reaches the boundary and starts crossing into medium 2; the other edge $B$ has yet to arrive. Let the time for $B$ to travel across to the boundary be $t$ . In that same time $t$ , the light that already entered medium 2 at $A$ advances into it.

Step 2. Write the two distances. In medium 1, the trailing edge covers $BX = v_1 t$ along its ray to reach the boundary. In medium 2, the entered light covers $AY = v_2 t$ . Both distances are measured over the same time $t$ , and $v_2 < v_1$ , so $AY < BX$ .

Step 3. Read the two angles off the same boundary segment. The wavefront makes angle $\theta_1$ with the boundary in medium 1 and $\theta_2$ in medium 2. Sharing the hypotenuse (the boundary segment $AB'$ of length $L$ ), geometry gives $\sin\theta_1 = \dfrac{v_1 t}{L}$ and $\sin\theta_2 = \dfrac{v_2 t}{L}$ .

Step 4. Divide and convert to indices. $\dfrac{\sin\theta_1}{\sin\theta_2} = \dfrac{v_1 t / L}{v_2 t / L} = \dfrac{v_1}{v_2}$ . Now use $n = c/v$ , so $v_1 = c/n_1$ and $v_2 = c/n_2$ , giving $\dfrac{v_1}{v_2} = \dfrac{n_2}{n_1}$ . Therefore $\dfrac{\sin\theta_1}{\sin\theta_2} = \dfrac{n_2}{n_1}$ , i.e. $n_1\sin\theta_1 = n_2\sin\theta_2$ . Test it: air → glass at $\theta_1 = 30^\circ$ gives $\sin\theta_2 = \dfrac{\sin 30^\circ}{1.5} = \dfrac{0.5}{1.5} = 0.3333$ , so $\theta_2 = 19.47^\circ$ — bent toward the normal, exactly as the front's tilt predicted. Verified: $\theta_2 = 19.47^\circ$ from $30^\circ$ , and the index-from-speed values, are recomputed in verify_hs2-refraction-snell.py.

Go deeper · Derivation B · 1 minFrequency stays, wavelength shrinks by n

Derivation BFrequency stays, wavelength shrinks by n

The wave companions of Snell's law — what happens to frequency and wavelength — are as heavily examined as the bending itself, and students routinely swap them. The goal is to show that frequency is conserved across a boundary while wavelength scales as $\lambda_{\text{vac}}/n$ , and to compute the glass wavelength of $600$ nm light. The anchor is the universal wave relation $v = f\lambda$ .

Step 1. Argue frequency is conserved. Wave crests cannot pile up or disappear at the boundary — every crest that arrives must leave — so the number passing per second is the same on both sides. That count is the frequency, so $f_1 = f_2$ . The source, not the medium, sets the frequency.

Step 2. Apply $v = f\lambda$ on each side. In vacuum, $c = f\lambda_{\text{vac}}$ ; in the medium, $v = f\lambda_{\text{medium}}$ , with the same $f$ . Dividing, $\dfrac{\lambda_{\text{medium}}}{\lambda_{\text{vac}}} = \dfrac{v}{c} = \dfrac{1}{n}$ , so $\lambda_{\text{medium}} = \dfrac{\lambda_{\text{vac}}}{n}$ .

Step 3. Put numbers in. For $600$ nm light entering glass ( $n = 1.5$ ): $\lambda_{\text{glass}} = \dfrac{600}{1.5} = 400$ nm. Its frequency stays $f = \dfrac{c}{\lambda_{\text{vac}}} = \dfrac{3.0\times10^8}{600\times10^{-9}} = 5.0\times10^{14}$ Hz.

Step 4. Cross-check inside the glass. There, $v = c/n = 2.0\times10^8$ m/s and $\lambda_{\text{glass}} = 400$ nm, so $f = \dfrac{v}{\lambda} = \dfrac{2.0\times10^8}{400\times10^{-9}} = 5.0\times10^{14}$ Hz — the same frequency, confirming it is what carries across the boundary while the wavelength is what compresses. Verified: $\lambda_{\text{glass}} = 400$ nm and $f = 5.0\times10^{14}$ Hz computed both ways are recomputed in verify_hs2-refraction-snell.py.

Common trapMeasure angles from the normal, and never confuse n equals c over v with its reciprocal

Two habits quietly drain marks. First, the angles in Snell's law are measured from the normal — the line perpendicular to the surface — not from the surface itself. A ray at " $60^\circ$ to the surface" is at $30^\circ$ to the normal, and using $60^\circ$ in the equation gives nonsense. Draw the normal first, always, and measure to it. Second, the index is $n = c/v$ , vacuum speed over medium speed — the medium is slower, so $n > 1$ ; writing $v/c$ inverts everything and predicts $n < 1$ , which never happens for a real medium. Two more slips to name: "optically denser" means higher index (slower light), which is not the same as physically denser or heavier — many oils are less dense than water yet optically denser; and it is the frequency that is conserved across a boundary, not the wavelength — the wavelength always shrinks by $n$ going into the medium. Get the normal, the ratio, the meaning of "denser", and the frequency/wavelength split right, and Snell problems are almost free.

Worked examples

Board level — bending, speed and wavelength into glass (CBSE and IGCSE pattern)

Light of wavelength $600$ nm in air strikes a glass block ( $n = 1.5$ ) at $30^\circ$ to the normal. Find (a) the angle of refraction, (b) the speed and (c) the wavelength of the light inside the glass.

Step 1. (a) Snell's law with $n_1 = 1$ , $n_2 = 1.5$ , $\theta_1 = 30^\circ$ : $\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} = \dfrac{1 \times 0.5}{1.5} = 0.3333$ , so $\theta_2 = \arcsin(0.3333) = \boxed{19.47^\circ}$ — bent toward the normal, since glass is denser.

Step 2. (b) Speed inside the glass from $n = c/v$ : $v = \dfrac{c}{n} = \dfrac{3.0\times10^8}{1.5} = \boxed{2.0\times10^8\ \mathrm{m/s}}$ , two-thirds of the vacuum speed.

Step 3. (c) Wavelength inside from $\lambda_{\text{medium}} = \lambda_{\text{vac}}/n$ : $\lambda_{\text{glass}} = \dfrac{600}{1.5} = \boxed{400\ \mathrm{nm}}$ , while the frequency $5.0\times10^{14}$ Hz and the colour are unchanged. IGCSE note: your syllabus writes Snell's law for an air-to-medium boundary as $n = \dfrac{\sin i}{\sin r}$ , which is exactly equation (2) with $n_1 = 1$ ; here $1.5 = \dfrac{\sin 30^\circ}{\sin 19.47^\circ}$ .

Competitive level — water, reversibility and a relative index (JEE Main and NEET pattern)

(a) Light in air meets a water surface ( $n = 4/3$ ) at $45^\circ$ ; find the angle of refraction. (b) A ray inside the glass of the previous example travels back out to air at $19.47^\circ$ — what angle does it leave at? (c) What is the refractive index of glass relative to water?

Step 1. (a) $\sin\theta_2 = \dfrac{\sin 45^\circ}{4/3} = \dfrac{0.7071}{1.3333} = 0.5303$ , so $\theta_2 = \arcsin(0.5303) = \boxed{32.03^\circ}$ .

Step 2. (b) By reversibility, running the air → glass ray backwards: $n_2\sin\theta_2 = n_1\sin\theta_1$ becomes $1.5 \times \sin 19.47^\circ = 1 \times \sin\theta_{\text{out}}$ , so $\sin\theta_{\text{out}} = 1.5 \times 0.3333 = 0.5000$ and $\theta_{\text{out}} = \boxed{30^\circ}$ — the ray retraces its incoming direction exactly.

Step 3. (c) The relative index of glass with respect to water is $n_{\text{water}\to\text{glass}} = \dfrac{n_{\text{glass}}}{n_{\text{water}}} = \dfrac{1.5}{4/3} = \boxed{1.125}$ . Because it exceeds $1$ , glass is optically denser than water, so a ray going from water into glass still bends toward the normal. Timing note: about 60 seconds for all three. The reversibility part (b) needs no fresh calculation — recognise that a reversed ray gives back the original angle, and just state $30^\circ$ .

Exam traps

Angles from the surface, not the normal. Snell's law uses angles measured from the normal. " $60^\circ$ to the surface" means $30^\circ$ to the normal. Draw the normal and measure to it.

Writing $n = v/c$ . The index is $c/v$ (vacuum over medium), always $\ge 1$ . The reciprocal predicts impossible $n < 1$ and inverts the bending.

Swapping frequency and wavelength. Frequency is conserved across the boundary; wavelength shrinks by $n$ . Saying "the frequency changes in glass" is a classic error.

"Optically denser" = "heavier". It means higher index (slower light), independent of mass density. Some light oils are optically denser than water.

Bending direction. Into a denser medium (higher $n$ ): toward the normal. Into a rarer medium: away. Getting this backward flips the whole diagram.

Board scope. Refraction and Snell's law sit on all ten boards this lane serves. IGCSE 0625 uses the air-to-medium form $n = \sin i / \sin r$ ; AP Physics 2 and IB (both SL and HL, as wave behaviour) use $n_1\sin\theta_1 = n_2\sin\theta_2$ . Where this lane tags "A-Level" it means Edexcel International A-Level — note that CIE 9702 A-Level contains no geometric optics at all, so a 9702 candidate does not study refraction for that paper.

Origins · 1 minWillebrord Snellius and how this was discovered

Willebrord Snellius

1580–1626 · Leiden

Snellius — the Latinised name of the Dutch astronomer and mathematician Willebrord Snel van Royen — is the eponym of the law this article is built on, and his story is a lesson in the difference between discovering and publishing. Around 1621, working at Leiden, he found empirically that the ratio of the cosecants (equivalently, the ratio of the sines) of the angles is constant for a given pair of media — the sine law of refraction. But he never published it; the result sat in his manuscripts, and Christiaan Huygens later reported having seen it there. The honest scope: Snellius established the law as an experimental fact but did not put it in print, so the first printed statement came from René Descartes in La Dioptrique of 1637 (the portrait of the glass-slab article). Historians credit Snellius with the empirical discovery and Descartes with publication and the geometric treatment; naming the law for Snellius honours the prior, if private, finding. Earlier still, the Baghdad scholar Ibn Sahl had written an equivalent geometric construction around 984.

History

984

— Ibn Sahl, in On Burning Instruments, states a geometric construction equivalent to the sine law of refraction.

1621

— Willebrord Snellius finds the sine law of refraction empirically at Leiden, but leaves it unpublished.

1637

— René Descartes prints the sine law and a coordinate-geometry treatment of refraction in La Dioptrique.

1662

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

1678

— Christiaan Huygens, presenting his wavefront principle, explains refraction as a change of wave speed.

Going further

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), refraction and Snell's law · NCERT Science Class 10, Ch. 10, refraction and the refractive index · H. C. Verma, Concepts of Physics Vol. 1, "Light Waves" and "Geometrical Optics" · Cambridge IGCSE Physics 0625, refraction and refractive index.

Provenance. Standard wave optics; every number recomputed in verify_hs2-refraction-snell.py. Recompute: $v = c/n$ gives $2.0\times10^8$ (glass), $2.25\times10^8$ (water $4/3$ ), $1.24\times10^8$ (diamond $2.42$ ); air → glass $30^\circ \to \theta_2 = 19.47^\circ$ ; air → water $45^\circ \to 32.03^\circ$ ; reverse glass → air $19.47^\circ \to 30^\circ$ ; relative index $1.5/(4/3) = 1.125$ ; $\lambda_{\text{glass}} = 600/1.5 = 400$ nm with $f = 5.0\times10^{14}$ Hz both sides.

Prerequisites. The Ray Model and Reflection — the ray picture, the normal, and the law of reflection that refraction sits beside. Leads to. Refraction Through a Glass Slab and Apparent Depth (why the pool looks shallow), Total Internal Reflection and Critical Angle (Snell pushed to its limit), and the UG seam Snell's Law of Refraction.

Go deeper (UG). Why the index takes the value it does — the atomic re-radiation that slows the wave, and its dependence on colour — is developed in Snell's Law of Refraction in the university optics track. Enrichment only; every result here follows from the wavefront picture and $n = c/v$ , with no need for the electromagnetic derivation.

Why does light bend at all when it enters glass — what pushes it?

Nothing pushes it sideways; the bend is a consequence of the wave slowing down unevenly. Think of the wavefront as a wide line crossing the boundary at an angle: the part that reaches the glass first immediately slows, while the part still in air keeps its faster pace for a moment longer. That mismatch pivots the whole front, and since the ray always points perpendicular to its front, the ray direction swings too — toward the normal on entering the slower medium. It is exactly like a car driving off tarmac onto sand at an angle: the wheel that hits the sand first slows, and the car turns. No force acts along the ray; the change of speed at an angle is the entire cause.

Does the frequency or the wavelength change when light enters a medium?

The wavelength changes; the frequency does not. Frequency is fixed because wave crests cannot accumulate or vanish at the boundary — whatever rate they arrive at, they must leave at, so $f$ is the same on both sides. The speed drops to $v = c/n$ , and since $v = f\lambda$ with $f$ fixed, the wavelength must shrink to $\lambda/n$ . So $600$ nm light in air becomes $400$ nm in glass, but its frequency stays $5.0\times10^{14}$ Hz. This is why the colour of a submerged object does not change — colour is tied to frequency, which the medium leaves alone. Confusing the two is one of the most common exam mistakes.

What is the difference between absolute and relative refractive index?

The absolute index of a medium is measured against vacuum: $n = c/v$ , always at least $1$ (glass $1.5$ , water $4/3$ ). The relative index compares two media directly: $n_{21} = n_2/n_1 = v_1/v_2$ , the factor by which light slows going specifically from medium 1 into medium 2. For example, glass relative to water is $1.5 / (4/3) = 1.125$ — so light slows by a further $12.5\%$ crossing from water into glass, and bends toward the normal. When a boundary is between two non-vacuum media, it is the relative index that appears in the ratio $\sin\theta_1/\sin\theta_2$ . You can always rebuild a relative index from two absolute ones by dividing.

Why is the refractive index always greater than one?

Because light travels fastest in vacuum, so its speed in any material medium is smaller, and $n = c/v$ is therefore a number bigger than one. Physically, inside matter the light wave keeps driving the atoms' electrons, which re-radiate slightly out of step, and the sum of the original and re-radiated waves advances more slowly than $c$ . Air is barely slower than vacuum ( $n = 1.0003$ ), so we round it to $1$ ; water slows light to three-quarters of $c$ ( $n = 4/3$ ); diamond to about two-fifths ( $n = 2.42$ ). You may hear of "metamaterials" engineered to behave oddly, but for every ordinary transparent substance the index exceeds one, which is why the equation is written to give $n \ge 1$ .

If I reverse a light ray, does it retrace its path exactly?

Yes — this is the principle of reversibility, and it falls straight out of Snell's law. The equation $n_1\sin\theta_1 = n_2\sin\theta_2$ is symmetric: swap which side you call incident and which refracted, and it still holds with the same angles. So a ray that goes air → glass bending $30^\circ \to 19.47^\circ$ will, sent backwards from inside the glass at $19.47^\circ$ , emerge into air at exactly $30^\circ$ . Reversibility is enormously useful: it lets you solve "what angle does the light leave at?" without any new computation, and it underlies why the same lens can form an image of an object or, run backwards, project that image back to the object. It holds for reflection too.

Is refraction on my board, and does the A-Level part apply to me?

Refraction and Snell's law are on every board this article serves — CBSE, ICSE, JEE Main, JEE Advanced, NEET, AP Physics 2, IB (both SL and HL, under wave behaviour), IGCSE and A-Level. But "A-Level" needs a caveat: Cambridge International A-Level (CIE 9702) contains no geometric optics at all — its waves topics are about superposition, diffraction and interference, not refraction or lenses. So when this lane says A-Level, it means Edexcel International A-Level, which does cover refraction, reflection and total internal reflection. If you sit CIE 9702, you will not be examined on Snell's law for that paper, though it remains excellent background. IGCSE students use the equivalent form $n = \sin i / \sin r$ .