Refraction

Dip a straight pencil into a glass of water and look from the side: the pencil appears to snap sharply at the surface, as if two different pencils had been joined at an angle. Nothing has bent except the light. A swimming pool that is really $2$ m deep looks barely $1.5$ m deep from the edge. A diamond sparkles because light entering it changes direction so violently that it can barely find its way back out. All three are the same phenomenon wearing different costumes: refraction, the bending of a light ray as it crosses the boundary between two transparent media. Refraction happens because light does not travel at the same speed in every medium — it slows down in glass, water, and diamond compared to its top speed in vacuum, and that change of speed is what bends the ray. The rule connecting the bend to the speeds is one of the oldest precise laws in physics, Snell's law, and a single number called the refractive index captures how strongly a given medium bends light. This article builds that idea from the ground up: what refractive index actually measures, how Snell's law predicts the exact angle a ray takes at a boundary, and why the direction of bending — towards the normal or away from it — is completely predictable once you know which medium is denser. These are not just facts to memorise for one question type; refractive index and Snell's law are the load-bearing wall for everything that follows in this chapter — refraction at curved (spherical) surfaces, the lens formula, the lens maker's formula, and the working of prisms and optical instruments. Get this foundation solid, and the rest of ray optics becomes a matter of applying the same one law again and again at different-shaped boundaries.

Key takeaways

Refraction is the change in direction of a light ray when it crosses the boundary between two transparent media of different optical density, caused by a change in the speed of light.
The absolute refractive index of a medium is $n = c/v$ , where $c$ is the speed of light in vacuum and $v$ is its speed in that medium; since $v \le c$ always, $n \ge 1$ for every real medium.
Snell's law relates the angle of incidence $\theta_1$ and angle of refraction $\theta_2$ , both measured from the normal, at a boundary between media of refractive indices $n_1$ and $n_2$ : $n_1\sin\theta_1 = n_2\sin\theta_2$ .
Light bends towards the normal on entering an optically denser medium (higher $n$ , lower $v$ ), and away from the normal on entering a rarer medium (lower $n$ , higher $v$ ); along the normal itself ( $\theta_1 = 0$ ) there is no bending at all.
The relative refractive index of medium 2 with respect to medium 1 is $n_{21} = n_2/n_1 = v_1/v_2$ — "denser" and "rarer" are always comparisons between two named media, never an absolute label.
Refractive index is technically wavelength-dependent (this is why a prism splits white light into colours), but a single chapter problem always quotes $n$ for one fixed colour of light, so Snell's law can be applied with one clean number.
Key idea: refraction is light bending at a boundary because its speed changes, and Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$ , turns that speed change into a precisely calculable angle.

The core insight

Because light travels at different speeds in different transparent media (captured by the refractive index $n = c/v$ ), a ray bends at the boundary between two media according to Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$ , bending towards the normal when entering a denser medium and away from it when entering a rarer one.

Where it fits in the syllabus

In CBSE Class 12 Unit IX, "Ray Optics and Optical Instruments," refraction and refractive index sit at the hinge of the chapter: everything before this point (plane mirrors, spherical mirrors, the mirror formula) is about light bouncing off a surface, and everything after this point (refraction at a single spherical surface, the thin lens formula, the lens maker's formula, prisms, and optical instruments) is about light passing through a surface into a new medium. Snell's law and the definition of refractive index, both examined here, are the single tool reused at every one of those later boundaries, so board questions test this section both directly (angle and index calculations) and indirectly (as a hidden first step inside a lens or prism problem).

The physical picture

Light travels fastest in vacuum, at $c \approx 3\times10^8$ m/s, and more slowly in every real transparent medium — air, water, glass, diamond. The absolute refractive index of a medium measures exactly how much slower:

n = \frac{c}{v}

where: $n$ is the (dimensionless) absolute refractive index of the medium, $c \approx 3\times10^8$ m/s is the speed of light in vacuum, and $v$ is the speed of light inside that medium. Because $v$ can never exceed $c$ , this ratio always satisfies $n \ge 1$ . A larger $n$ means a smaller $v$ — the medium is called optically denser, and light crawls through it more slowly. Air has $n \approx 1.0003$ (close enough to $1$ to be treated as vacuum in most problems), water has $n \approx 1.33$ , and ordinary crown glass has $n \approx 1.5$ . When a ray crosses the boundary from one medium into another, it changes direction — this bending is refraction, and the precise rule governing it is Snell's law. If a ray in a medium of index $n_1$ strikes the boundary with a medium of index $n_2$ , making an angle of incidence $\theta_1$ with the normal (the line perpendicular to the surface at the point of incidence), and refracts into the second medium at an angle $\theta_2$ from the same normal, then:

n_1\sin\theta_1 = n_2\sin\theta_2

where: $\theta_1$ and $\theta_2$ are always measured from the normal, never from the surface itself — this single detail is the most common source of a wrong answer in this topic. The physical reading of the law is that the product "index times sine of angle" is conserved as the ray crosses the boundary; whichever side has the larger index must compensate with a smaller sine, and hence a smaller angle, to keep the product balanced. This is exactly why a ray bends towards the normal when it enters a denser medium ( $n_2 > n_1 \Rightarrow \sin\theta_2 < \sin\theta_1 \Rightarrow \theta_2 < \theta_1$ ) and away from the normal when it enters a rarer one. A ray travelling exactly along the normal has $\theta_1 = 0$ , which forces $\theta_2 = 0$ too — it passes straight through, completely undeviated, regardless of how different the two media are. The relative refractive index of medium 2 with respect to medium 1 follows directly from Snell's law by rearranging it as $n_{21} = n_2/n_1 = \sin\theta_1/\sin\theta_2$ , and this equals the ratio of speeds the other way round, $n_{21} = v_1/v_2$ . This number tells you, for that specific pair of media, which one is optically denser — "denser" and "rarer" are not fixed properties of a single medium, only comparisons between two named media (a glass that is denser than water can still be rarer than diamond).

Figure

A ray of light bending towards the normal as it passes from air into water, with the angle of incidence larger than the angle of refraction, both measured from a dashed normal line at the boundary

Derivation ASnell's law from the constancy of frequency and change of speed

The goal of this derivation is to obtain Snell's law itself from a single physical fact — that a wave's frequency cannot change as it crosses into a new medium, only its speed and wavelength can — using nothing more than plane-geometry relations for the two right triangles formed at the boundary, exactly the algebra level of this chapter and no more.

Step 1. Set up the geometry. A plane wavefront in medium 1 (speed $v_1$ ) approaches the flat boundary with medium 2 (speed $v_2$ ) at an angle of incidence $\theta_1$ from the normal. Consider two rays of the same wavefront that strike the boundary at two nearby points, a small distance $AB$ apart along the surface. While the point of the wavefront at $B$ still has to travel a short extra distance in medium 1 to reach the surface, the part of the wavefront that already crossed at $A$ has begun advancing into medium 2 as a new wavefront bending at angle $\theta_2$ .

Step 2. Write the two triangle relations. In the right triangle formed in medium 1, the extra path the wavefront covers before point $B$ reaches the surface is $v_1\,\Delta t = AB\sin\theta_1$ , where $\Delta t$ is the time this takes. In the right triangle formed in medium 2 during the same interval $\Delta t$ , the wavefront launched from $A$ advances a distance $v_2\,\Delta t = AB\sin\theta_2$ . Both triangles share the same base $AB$ and the same time $\Delta t$ , because the wavefront is continuous across the boundary — this is the key physical input, that the two rays are locked together by a common wavefront.

Step 3. Eliminate $AB$ and $\Delta t$ . Dividing the two relations, $\dfrac{v_1\,\Delta t}{v_2\,\Delta t} = \dfrac{AB\sin\theta_1}{AB\sin\theta_2}$ , so $\dfrac{v_1}{v_2} = \dfrac{\sin\theta_1}{\sin\theta_2}$ . This already says the ratio of speeds fixes the ratio of sines — the geometric core of refraction.

Step 4. Convert speeds to refractive indices. Since $n_1 = c/v_1$ and $n_2 = c/v_2$ , we have $v_1 = c/n_1$ and $v_2 = c/n_2$ , so $\dfrac{v_1}{v_2} = \dfrac{n_2}{n_1}$ . Substituting into Step 3 gives $\dfrac{n_2}{n_1} = \dfrac{\sin\theta_1}{\sin\theta_2}$ , which rearranges to the familiar form:

n_1\sin\theta_1 = n_2\sin\theta_2

Verified: with $v_1 > v_2$ (medium 2 denser), Step 3 gives $\sin\theta_2 < \sin\theta_1$ , so $\theta_2 < \theta_1$ — the ray bends toward the normal on entering the denser medium, matching the physical picture above.

Worked examples

Board level - angle of refraction from air into water

Question: A ray of light travelling in air strikes the surface of water (refractive index $1.33$ ) at an angle of incidence of $45°$ . Find the angle of refraction.

Step 1. List what is given: $n_1 = 1$ (air), $\theta_1 = 45°$ , $n_2 = 1.33$ (water), and $\theta_2$ is to be found.

Step 2. Apply Snell's law directly. $n_1\sin\theta_1 = n_2\sin\theta_2$ gives $1\times\sin45° = 1.33\times\sin\theta_2$ .

Step 3. Solve for $\sin\theta_2$ . $\sin\theta_2 = \dfrac{0.707}{1.33} = 0.532$ , so $\theta_2 = \sin^{-1}(0.532)$ .

Step 4. State the answer and check its direction. $\boldsymbol{\theta_2 \approx 32.1°}$ . Since water is optically denser than air ( $n_2 > n_1$ ), the refracted ray must bend towards the normal, so we expect $\theta_2 < \theta_1$ — and indeed $32.1° < 45°$ , confirming the answer is physically consistent.

Competitive level - relative refractive index and speed in glass

Question: The speed of light in a certain glass is $2\times10^8$ m/s. Find the refractive index of the glass, and determine the relative refractive index of water ( $n = 1.33$ ) with respect to this glass. Hence state, without further calculation, which way a ray bends when it travels from this glass into water.

Step 1. Find the refractive index of the glass from its definition. $n_{\text{glass}} = \dfrac{c}{v_{\text{glass}}} = \dfrac{3\times10^8}{2\times10^8} = 1.5$ .

Step 2. Find the relative refractive index of water with respect to glass using $n_{21} = n_2/n_1$ with medium 1 as glass and medium 2 as water. $n_{\text{water, glass}} = \dfrac{n_{\text{water}}}{n_{\text{glass}}} = \dfrac{1.33}{1.5} \approx 0.887$ .

Step 3. Interpret the ratio. Since $n_{\text{water, glass}} < 1$ , water is optically rarer than this particular glass — light travels faster in water ( $v = c/1.33 \approx 2.26\times10^8$ m/s) than in this glass ( $2\times10^8$ m/s).

Step 4. State the answer and the direction of bending. $\boldsymbol{n_{\text{glass}} = 1.5}$ and $\boldsymbol{n_{\text{water, glass}} \approx 0.887}$ ; a ray travelling from this glass into water crosses into a rarer medium, so it bends away from the normal. This confirms that "denser" and "rarer" are relative labels between two named media, not fixed properties of water or glass alone — the same water is denser than air but rarer than this glass. Timing note: in a board or JEE-style paper this whole pair of sub-questions should take about 3 minutes — the arithmetic is short, so the real time cost is remembering to state the qualitative direction-of-bending conclusion, which is often worth separate marks even when no further numbers are asked for.

Common traps

Thinking a higher refractive index means light travels faster in that medium — it is the exact opposite: $n = c/v$ , so a higher $n$ means a lower speed $v$ .
Measuring the angle in Snell's law from the surface instead of from the normal — this single mix-up inverts sine and cosine and produces a wrong answer even with perfectly correct arithmetic otherwise.
Assuming refraction always bends light "towards" the normal — this only happens going from a rarer to a denser medium; going the other way, from denser to rarer, light bends away from the normal, and along the normal itself there is no bending at all.
Confusing absolute refractive index (always measured relative to vacuum or air) with relative refractive index (measured between two specific named media) — a question asking for $n_{21}$ wants a ratio, not either medium's absolute index alone.
Treating refractive index as one universal constant for a medium regardless of colour — it is technically wavelength-dependent (violet bends more than red in the same glass), though a single chapter problem always fixes one colour so a single value of $n$ can be used throughout that problem.

You should be able to…

Define absolute refractive index as $n = c/v$ and explain why $n \ge 1$ for every real transparent medium.
State Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$ , and use it to calculate an unknown angle of incidence or angle of refraction given the other three quantities.
Predict, without calculation, whether a ray bends towards or away from the normal, given only the relative optical densities of the two media.
Calculate the relative refractive index between two named media from their absolute refractive indices or their speeds of light.
Explain, using refraction, why a submerged object appears displaced from its true position and why a pool looks shallower than it is.
Recognise refractive index as (very slightly) wavelength-dependent, while correctly using a single fixed value of $n$ within any one numerical problem.

Ask the tutor: Want to walk through a few more Snell's law problems with different medium pairs, or see how this same law reappears when we get to refraction at a curved (spherical) surface in the next topic?

Why does a pencil in a glass of water look bent at the surface, when the pencil itself is perfectly straight?

The pencil is straight; what changes is the path the light takes on its way from the submerged part of the pencil to your eye. Light leaving the part of the pencil under water refracts as it exits the water into air, bending away from the normal because it is moving from a denser medium to a rarer one. Your eye and brain assume light always travels in a straight line, so they trace the refracted ray backwards in a straight line and place the underwater part of the pencil at a shifted apparent position — higher up and offset from where it actually is. The break you see at the surface is exactly this jump between the real position and the apparent one.

Is refractive index the same for every colour of light in the same medium?

Not quite, and this is exactly what makes prisms interesting later in the chapter. Refractive index depends slightly on wavelength — in most transparent materials, violet light (shorter wavelength) has a slightly higher refractive index than red light (longer wavelength), so it bends a little more at a boundary. This effect is called dispersion. For the numerical problems in this chapter, though, a question will always specify or imply a single colour of light, so you use one fixed value of $n$ throughout that problem — the wavelength-dependence only becomes the star of the show when a prism is explicitly splitting white light into a spectrum.

If n is greater than or equal to 1 always, can refractive index ever be exactly 1?

Yes, but only for vacuum itself, by definition: $n = c/v$ , and in vacuum $v = c$ , so $n = 1$ exactly. Air is so close to vacuum that its refractive index, about $1.0003$ , is usually rounded to $1$ in this chapter's problems, and Snell's law calculations at an air boundary almost always use $n_{\text{air}} = 1$ without meaningfully changing the answer. No real transparent medium has $n$ below $1$ , because that would require light to travel faster than $c$ inside it, which is not physically possible.

How is this different from total internal reflection, which I've heard is also in this chapter?

Refraction, covered here, is what happens at every boundary between two transparent media when the angle of incidence is small enough — the ray bends and continues into the second medium. Total internal reflection is a special, more dramatic case that can only occur when light tries to go from a denser medium into a rarer one (say, glass into air) at an angle of incidence larger than a specific critical angle; beyond that angle, Snell's law would demand a refraction angle greater than $90°$ , which is impossible, so instead the entire ray reflects back into the denser medium with no refraction at all. You need Snell's law and the idea of denser-versus-rarer media from this article before that topic makes sense, which is exactly why it comes right after this one.