OCR A Level Physics A: Electromagnetic spectrum, refraction and total internal reflection (H556)

This study guide is aligned to the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), section 4.4.2 Electromagnetic waves, statements (a) to (e). It sits inside Module 4, Electrons, waves and photons. It teaches the electromagnetic spectrum, polarisation of EM waves, refraction and refractive index, the refraction and total internal reflection practical, the critical angle and total internal reflection.

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Scope of this unit

Reference	Outcome you need
4.4.2(a)	Describe the electromagnetic spectrum and the properties of EM waves
4.4.2(b)	Give the orders of magnitude of wavelength from radio waves to gamma rays
4.4.2(c)	Explain plane polarised waves and polarisation of EM waves; know polarising filters for light and metal grilles for microwaves
4.4.2(d)(i)	Use n = c / v and n₁ sin i₁ = n₂ sin i₂; know that n sin i is constant at a boundary
4.4.2(d)(ii)	Describe techniques for investigating refraction and total internal reflection with ray boxes and rectangular and semi-circular blocks (PAG6)
4.4.2(e)	Use sin C = 1/n; explain total internal reflection for light

4.4.2(a) The electromagnetic spectrum and its properties

In order of increasing frequency (decreasing wavelength) the principal radiations are: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. There are no sharp edges. Neighbouring regions overlap, and X-rays and gamma rays are often separated by how they are produced rather than by wavelength alone.

Properties shared by all EM waves:

• They are transverse waves. Their electric and magnetic fields oscillate perpendicular to one another, and both are perpendicular to the direction the wave moves.
• They travel through a vacuum without needing a medium.
• In a vacuum their speed is identical, c = 3.00 × 10⁸ m s⁻¹.
• They obey v = fλ, so in a vacuum c = fλ.
• They transfer energy. They can be reflected, refracted, diffracted, and they can interfere.
• They can be polarised, because they are transverse.

What differs along the spectrum is frequency, wavelength and how the radiation interacts with matter. Gamma rays do not travel faster than radio waves in a vacuum.

4.4.2(b) Orders of magnitude of wavelength

You need the order of magnitude, not exact limits. Use these typical values.

Radiation	Typical wavelength (order of magnitude)
Radio waves	10⁻¹ m up to 10³ m or more
Microwaves	10⁻³ m to 10⁻¹ m
Infrared	10⁻⁶ m to 10⁻³ m
Visible light	about 4 × 10⁻⁷ m (violet) to 7 × 10⁻⁷ m (red)
Ultraviolet	10⁻⁹ m to 4 × 10⁻⁷ m
X-rays	10⁻¹¹ m to 10⁻⁹ m
Gamma rays	below about 10⁻¹¹ m

Worked example 1

An FM radio station broadcasts at 98 MHz. A kitchen microwave oven works at 2.45 GHz. Find the wavelength of each and say which regions they belong to.

• Radio: λ = c / f = (3.00 × 10⁸) / (98 × 10⁶) = 3.06 m, which is a radio wave (order 10⁰ m).
• Oven: λ = (3.00 × 10⁸) / (2.45 × 10⁹) = 0.122 m, which is a microwave (order 10⁻¹ m).

Another check: an X-ray of frequency 3.0 × 10¹⁸ Hz has λ = (3.00 × 10⁸) / (3.0 × 10¹⁸) = 1.0 × 10⁻¹⁰ m, in the X-ray range.

4.4.2(c) Plane polarised waves and polarisation of EM waves

A transverse wave oscillates at right angles to its direction of travel, and there are many planes at right angles to that direction. In a plane polarised wave the oscillations are confined to one plane only, the plane containing the direction of travel. Longitudinal waves cannot be polarised, so being able to polarise a wave shows that it is transverse.

For an EM wave, the plane of polarisation is normally taken as the plane of the electric field oscillation. Light from a lamp is unpolarised: the electric field oscillates in all possible planes.

Polarising filter for light. A polarising filter (Polaroid) allows through only the component of oscillation along its transmission direction.

• Unpolarised light passing through one filter emerges plane polarised, with reduced intensity.
• Look through two filters and rotate one. The transmitted light changes from bright to dark and back, with two bright and two dark positions in a full turn.
• Darkness occurs when the transmission directions are at 90° to each other (crossed). Light polarised by the first filter has no component along the second.

Metal grille for microwaves. Many microwave transmitters give plane polarised waves. Place a grille of parallel metal rods between transmitter and receiver.

• Rods parallel to the electric field: the field drives electrons along the rods, so energy is absorbed or reflected and the receiver signal falls to a minimum.
• Rods at 90° to the electric field: the wave passes through and the signal is a maximum.
• Rotating the grille through 90° switches between the two. This shows the microwaves are plane polarised and transverse.

4.4.2(d)(i) Refraction and refractive index

Refraction is the change in direction of a wave when it passes between media in which its speed differs, because the wavefronts reach the boundary at different times.

The refractive index of a material is n = c / v, where c is the speed of light in a vacuum and v the speed in the material. It has no unit. Since v < c, n is greater than 1. The refractive index of air is very close to 1, so we use n = 1 for air unless a question says otherwise.

At a boundary between medium 1 and medium 2, with angles i₁ and i₂ measured from the normal (not from the surface):

n₁ sin i₁ = n₂ sin i₂

So n sin i is the same on both sides of the boundary. Going into a higher n, the ray bends towards the normal. Going into a lower n, it bends away from the normal. Frequency stays the same on crossing a boundary. The speed and the wavelength both change, so n = λ(vacuum) / λ(material).

Worked example 2

A ray of light in air meets a sheet of plastic of refractive index 1.47 at an angle of incidence of 38.0°. Find the angle of refraction and the speed of light in the plastic.

• 1.00 × sin 38.0° = 1.47 × sin r, so sin r = 0.6157 / 1.47 = 0.4188.
• r = 24.8° (3 s.f.). It is smaller than 38.0°, as expected for a ray moving into a higher n.
• v = c / n = (3.00 × 10⁸) / 1.47 = 2.04 × 10⁸ m s⁻¹.

Worked example 3

A ray in glass (n = 1.50) meets a glass-water boundary at 35.0° to the normal. Water has n = 1.33. Find the angle in the water.

• 1.50 × sin 35.0° = 1.33 × sin r.
• sin r = (1.50 × 0.5736) / 1.33 = 0.6469, so r = 40.3°.
• The ray bends away from the normal, as water has the lower n.

4.4.2(d)(ii) Investigating refraction and total internal reflection (PAG6)

The specification names ray boxes and transparent rectangular and semi-circular blocks.

Refraction in a rectangular block.

1. Place the block on white paper and draw round it.
2. Use a ray box with a single slit to send a narrow ray at the block. Mark two points on the incident ray, and two on the emergent ray.
3. Remove the block. Join the points, and join the entry and exit points to give the ray inside.
4. Draw the normal at the entry point. Measure i and r with a protractor.
5. Repeat for at least six values of i from about 10° to 70°.
6. Plot sin i (y) against sin r (x). The line is straight and passes through the origin, and its gradient is n. Plotting sin i against sin r gives the straight line; plotting i against r does not.

Critical angle with a semi-circular block. Aim the ray at the centre of the flat face, entering through the curved face along a radius. It meets the curved surface along the normal, so it is not deviated there. Inside the glass it travels to the flat face, where it meets the boundary at an angle you can measure. Increase the angle of incidence at the flat face slowly. The refracted ray bends away from the normal until it travels along the flat surface. That angle of incidence is C. Beyond it the ray is totally internally reflected.

Sources of uncertainty: the width of the ray (mark the centre of the beam), protractor resolution, and judging where the refracted ray vanishes. Use a thin slit and a sharp pencil, and repeat readings.

Worked example 4

A student records the following for a glass block.

i / °	20	70
r / °	13	38.5
sin i	0.342	0.940
sin r	0.225	0.623

Taking two well-separated points: gradient = (0.940 − 0.342) / (0.623 − 0.225) = 0.598 / 0.398 = 1.50. So n = 1.50 (3 s.f.). A best-fit line through all points is better, as it averages out random error.

4.4.2(e) Critical angle and total internal reflection

When light travels from a higher n to a lower n, the refracted ray bends away from the normal. As the angle of incidence increases, the refracted angle reaches 90° and the ray grazes along the boundary. The angle of incidence at this point is the critical angle, C.

Apply Snell’s law with n₁ = n, n₂ = 1 (air) and r = 90°: n sin C = 1 × sin 90°, so

sin C = 1/n

For a boundary between two materials, sin C = n₂ / n₁, where n₁ is the higher index.

Total internal reflection (TIR) occurs when both conditions hold:

1. The light is travelling from a medium of higher refractive index towards one of lower refractive index.
2. The angle of incidence is greater than the critical angle.

All of the light is then reflected back into the denser medium, following the law of reflection. Below C there is a refracted ray and a weak reflected ray.

Worked example 5

A glass prism (n = 1.52) has a ray travelling inside it that meets a glass-air face at 45°. Does TIR occur? Would it occur if the prism were made of material with n = 1.30?

• n = 1.52: sin C = 1/1.52 = 0.658, so C = 41.1°. Since 45° > 41.1°, TIR occurs.
• n = 1.30: sin C = 1/1.30 = 0.769, so C = 50.3°. Since 45° < 50.3°, no TIR. Most of the light is refracted out.

Worked example 6

A light source lies at the bottom of a pool of water (n = 1.33), 0.80 m deep. Light can leave the surface only from a circle directly above the source. Find its radius.

• sin C = 1/1.33 = 0.752, so C = 48.8°.
• The edge of the circle is where the ray reaches the surface at C: radius = depth × tan C = 0.80 × tan 48.8° = 0.91 m.

Common errors

• Measuring angles from the surface instead of the normal.
• Using sin C = 1/n for a glass-water boundary. Use n₂/n₁ there.
• Saying TIR can happen when light enters a denser medium. It cannot.
• Saying the frequency changes on refraction. It does not.
• Plotting i against r and calling the gradient n.

Next steps

Attempt the practice questions, then tick off each statement on the checklist.

Official syllabus

OCR, Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), section 4.4.2 Electromagnetic waves, for first assessment in 2017.