Refraction

Spec mapping: OCR H556 Module 4.4 — Waves: refraction at a boundary; refractive index $n = c/v$ ; Snell's law $n_1 \sin\theta_1 = n_2 \sin\theta_2$ ; angles measured from the normal; PAG 3 anchor. (Refer to the official OCR H556 specification document for exact wording.)

When a wave passes from one medium into another, it usually changes speed. If it strikes the boundary at an angle, this change in speed causes the wave to change direction — a phenomenon known as refraction. Refraction is responsible for a host of familiar effects, from the apparent bending of a straw in a glass of water to the focusing of light in a microscope and the twinkling of stars through the restless air above our heads.

This lesson develops the quantitative physics of refraction at A-Level: the refractive index, Snell's law, the geometry of refraction at a boundary between two transparent media, and the wave-theoretic underpinning that links refraction to the constancy of frequency and the wave equation $v = f\lambda$ established in earlier lessons.

What Refraction Is and What It Is Not

Refraction is the change in direction of a wave as it passes from one medium to another, caused by a change in its speed. Note carefully what refraction is not:

It is not reflection (the bouncing back of a wave at a boundary).
It is not diffraction (the bending of a wave around an obstacle or through an aperture, covered later in this module).
It does not change the wave's frequency — only its speed and wavelength change.

The frequency is set by the source and does not change across a boundary (a continuity argument: the number of waves arriving at the boundary per second on the incident side must equal the number leaving per second on the refracted side, otherwise waves would pile up or vanish at the interface). Since $v = f\lambda$ and $f$ is constant, a fall in speed must produce a proportional fall in wavelength.

Refractive Index

The refractive index $n$ of a material is defined as the ratio of the speed of light in vacuum ( $c$ ) to the speed of light in the material ( $v$ ):

$n = \frac{c}{v}$

where $c = 3.00 \times 10^8$ m s $^{-1}$ is the speed of light in vacuum and $v$ is the speed of light in the material (m s $^{-1}$ ).

Because $v \leq c$ in any physical medium, $n \geq 1$ always.

Medium	Refractive index (for visible yellow light)
Vacuum	$1.000$ (exactly)
Air (at $20$ °C, $1$ atm)	$1.0003$ (usually taken as $1.00$ )
Water	$1.33$
Ethanol	$1.36$
Glass (crown)	$1.52$
Perspex	$1.49$
Diamond	$2.42$

A larger refractive index means light travels more slowly in the medium. Diamond's high refractive index ( $2.42$ ) is why it sparkles so brightly — light is slowed dramatically and refracted strongly, and the critical angle for total internal reflection (next lesson) is correspondingly small.

Exam Tip: For air, the refractive index is so close to $1$ that OCR questions treat air as equivalent to vacuum unless explicitly stated otherwise. "Light passes from air into glass" means $n_1 = 1.00$ , $n_2 = 1.52$ .

Snell's Law

When a wave passes from a medium of refractive index $n_1$ into a medium of refractive index $n_2$ , with angles measured from the normal to the boundary (not from the boundary itself), the angles of incidence ( $\theta_1$ ) and refraction ( $\theta_2$ ) are related by Snell's law:

$n_1 \sin \theta_1 = n_2 \sin \theta_2$

This is the fundamental equation of refraction at A-Level.

Direction of Bending

When light passes from a less dense medium (smaller $n$ ) into a more dense medium (larger $n$ ), it slows down and bends towards the normal. $\theta_2 < \theta_1$ .
When light passes from a more dense medium into a less dense medium, it speeds up and bends away from the normal. $\theta_2 > \theta_1$ .

Exam Tip: To remember the direction: slow media (high $n$ ) bend rays towards the normal. A convenient mnemonic: "into slower, towards normal".

The Normal

The normal is the imaginary line perpendicular to the boundary at the point where the wave meets it. Angles of incidence, reflection and refraction are always measured between the ray and the normal, never between the ray and the boundary. OCR mark schemes will deduct marks for angles measured from the surface — this is a recurring source of dropped marks.

Deriving Snell's Law from Wave Speeds

Snell's law can be written in a more physical form. The refractive index ratio equals the speed ratio: substituting $n_1 = c/v_1$ and $n_2 = c/v_2$ into Snell's law, the $c$ 's cancel:

$\frac{\sin \theta_1}{\sin \theta_2} = \frac{v_1}{v_2}$

In words: the ratio of the sines of the angles equals the ratio of the speeds. This reveals the physical cause of refraction — the wave slows down in the denser medium, and the wavefronts "bunch up" against the boundary, tilting the direction of propagation. This is exactly the form that Huygens's construction predicts: each point on a wavefront emits a spherical wavelet, and the envelope of these wavelets on the slower-medium side is closer to the boundary, producing the tilted wavefront.

Since $f$ is constant across the boundary and $v = f\lambda$ :

$\frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2} = \frac{\sin \theta_1}{\sin \theta_2}$

The wavelength therefore also shrinks in the slower medium in the same proportion as the speed.

Worked Example 1 — Light into Glass

Q. A light ray strikes a glass block ( $n = 1.52$ ) from air at an angle of incidence of $30°$ . Calculate the angle of refraction in the glass.

A. Applying Snell's law with $n_1 = 1.00$ (air), $n_2 = 1.52$ (glass):

$1.00 \times \sin 30° = 1.52 \times \sin \theta_2$ $\sin \theta_2 = \frac{0.500}{1.52} = 0.329$ $\theta_2 = \arcsin(0.329) = 19.2°$

The ray bends towards the normal, as expected when entering a denser medium.

Worked Example 2 — Light out of Water

Q. A light ray travelling in water ( $n = 1.33$ ) strikes the water-air boundary from below at an angle of $40°$ to the normal. Calculate the angle of refraction in air.

A. Applying Snell's law with $n_1 = 1.33$ , $n_2 = 1.00$ :

$1.33 \times \sin 40° = 1.00 \times \sin \theta_2$ $\sin \theta_2 = 1.33 \times 0.643 = 0.855$ $\theta_2 = \arcsin(0.855) = 58.8°$

The ray bends away from the normal, as expected when emerging into a less dense medium. If the angle of incidence were larger (above the critical angle $\arcsin(1/1.33) \approx 48.8°$ ), no refraction into air would occur — the ray would be totally internally reflected, as we will see in the next lesson.

Worked Example 3 — Speed in a Medium

Q. In a transparent liquid, light travels at $1.75 \times 10^8$ m s $^{-1}$ . Calculate (a) the refractive index, (b) the wavelength in the liquid of light whose vacuum wavelength is $600$ nm.

(a) Using $n = c/v$ :

$n = \frac{3.00 \times 10^8}{1.75 \times 10^8} = 1.71$

(b) The frequency is unchanged. In vacuum $f = c/\lambda_0 = (3.00 \times 10^8)/(6.00 \times 10^{-7}) = 5.00 \times 10^{14}$ Hz. In the liquid:

$\lambda_{\text{liquid}} = \frac{v}{f} = \frac{1.75 \times 10^8}{5.00 \times 10^{14}} = 3.50 \times 10^{-7} \text{ m} = 350 \text{ nm}$

Equivalently, $\lambda_{\text{liquid}} = \lambda_0 / n = 600/1.71 = 350$ nm.

Refraction at Two Parallel Boundaries

If a ray passes through a parallel-sided slab of glass, it refracts at the first surface and then refracts again at the second surface, in the opposite sense. The two refractions cancel out (because the slab has parallel sides and the angles obey symmetrical Snell's law), so the emerging ray is parallel to the incident ray but displaced sideways.

This is why a glass window does not distort straight-line viewing even though it bends light at each surface — the lateral displacement is usually small enough to be unnoticeable for a thin pane of glass at normal viewing distance, even though it is real.

You should be able to apply Snell's law twice to calculate the angle inside the slab and then confirm that the exit angle equals the entrance angle.

Worked Example 4 — Parallel-Sided Slab

Q. A light ray strikes a parallel-sided glass slab ( $n = 1.50$ ) from air at an angle of incidence of $45°$ . Calculate (a) the angle of refraction inside the slab, (b) the angle of incidence at the second surface, (c) the angle at which the ray emerges into the air on the far side.

(a) At the first surface, $1.00 \sin 45° = 1.50 \sin \theta_2$ , so $\sin \theta_2 = 0.707/1.50 = 0.471$ , giving $\theta_2 = 28.1°$ .

(b) The ray inside the glass travels at $28.1°$ from the normal to the first surface; this normal is parallel to the normal at the second surface (since the slab has parallel sides). The angle of incidence at the second surface is therefore also $28.1°$ .

(c) Applying Snell's law at the second surface in reverse, $1.50 \sin 28.1° = 1.00 \sin \theta_3$ , giving $\sin \theta_3 = 1.50 \times 0.471 = 0.707$ , so $\theta_3 = 45.0°$ . The emerging ray is parallel to the incident ray, as expected.

Huygens's Construction at the Boundary

A useful intuition for why Snell's law has the sine form, rather than (say) a tangent form, comes from Huygens's construction. Consider a plane wavefront approaching a horizontal boundary at angle $\theta_1$ to the normal. The end of the wavefront closer to the boundary reaches it first; the far end reaches it a moment later. During the time it takes the far end to reach the boundary, the near end has already entered the second medium and travelled at the new speed $v_2$ .

Concretely: let the perpendicular distance from the far end to the boundary be $a$ , so the time taken for the far end to reach the boundary is $a / v_1 \sin\theta_1$ (the path along the boundary measured by the geometry). During that time, the near end travels a distance $v_2 \times (\text{same time})$ at speed $v_2$ into the second medium. Setting up the right-angled triangles on both sides of the boundary and applying $\sin = \text{opposite}/\text{hypotenuse}$ gives:

$\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}$

which rearranges to Snell's law. The sine arises because the wavefront makes contact with the boundary along a line that projects onto the propagation direction by a factor of $\sin\theta$ — and the perpendicular distance the wave has to travel scales the same way.

Dispersion (Why Prisms Split White Light)

The refractive index of a material is slightly different for different wavelengths. In glass, violet light (~400 nm) has n ≈ 1.53, while red light (~700 nm) has n ≈ 1.51. Because violet light is refracted more strongly than red light, white light entering a prism is separated into its constituent colours: this is dispersion, and is the origin of the rainbow.

Dispersion is beyond the main OCR A-Level content but may appear in extension contexts. It is an important reminder that refractive index is really a function of wavelength — quoted values are for a standard reference wavelength (often yellow sodium, 589 nm).

Refraction of Sound and Water Waves

Refraction is not unique to light — it happens to all waves.

Water waves refract as they enter shallower regions: the wave speed in shallow water is approximately $v = \sqrt{gh}$ where $g$ is gravitational acceleration and $h$ is depth, so a smaller depth means a slower speed. As ocean swells approach a coastline at an angle, the part of the wavefront over shallower water slows first; the wavefronts rotate and line up parallel to the depth contours. This is why ocean swells reach beaches roughly parallel to them, even if the swells started oblique far offshore — refraction has bent them into alignment.

Sound waves refract in the atmosphere due to temperature (and hence sound-speed) gradients. The speed of sound in air is approximately $v = 331 + 0.6 T$ m s $^{-1}$ where $T$ is the Celsius temperature. On a clear, still night the ground cools below the warmer air above (a temperature inversion), and sound waves rising obliquely from the ground refract back downwards — the listener can sometimes hear distant traffic or train whistles unusually clearly. By day, the ground is warmer than the air above, and sound waves refract upwards away from the listener; distant sounds are then comparatively muted.

For these non-light waves, the "refractive index" concept is usually replaced by explicit wave speeds in each medium, and the form of Snell's law that derives directly from Huygens's construction is the one to use:

$\frac{\sin \theta_1}{\sin \theta_2} = \frac{v_1}{v_2}$

The defining $n = c/v$ form is specific to light because $c$ is a universal constant; for water waves there is no equivalent universal speed to which all speeds can be referred.

Refraction

Refraction

What Refraction Is and What It Is Not

Refractive Index

Snell's Law

Direction of Bending

The Normal

Deriving Snell's Law from Wave Speeds

Worked Example 1 — Light into Glass

Worked Example 2 — Light out of Water

Worked Example 3 — Speed in a Medium

Refraction at Two Parallel Boundaries

Worked Example 4 — Parallel-Sided Slab

Huygens's Construction at the Boundary

Dispersion (Why Prisms Split White Light)

Refraction of Sound and Water Waves

Synoptic Links

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