OxfordAQA International GCSE Physics: Reflection, refraction and total internal reflection (9203)

This guide follows sections 3.3.4 (Reflection) and 3.3.5 (Refraction and total internal reflection) of the OxfordAQA International GCSE Physics (9203) specification, for exams from May/June 2018 onwards. The qualification is untiered, and every outcome is taught with worked examples.

Use it with the revision notes and the practice questions. Course pages: the OxfordAQA International GCSE Physics hub, the printable checklist and the free diagnostics.

What this unit covers

Spec	What you must be able to do
3.3.4 a	State and use the law of reflection
3.3.4 b	Define the normal
3.3.4 c	Describe the image in a plane mirror and construct reflection ray diagrams
3.3.5 a	Explain refraction as a change of wave speed, and when direction changes
3.3.5 b	Say when light bends towards or away from the normal; use wavefront diagrams
3.3.5 c	Describe dispersion and the spectrum from a prism
3.3.5 d	Define refractive index using wave speed
3.3.5 e	Use n = sin i / sin r; required practical on refraction
3.3.5 f	Use n = 1 / sin c
3.3.5 g	Explain total internal reflection
3.3.5 h	Describe optical fibres in medicine and communications

3.3.4 Reflection

The law of reflection: when a wave is reflected, the angle of incidence equals the angle of reflection.

The normal is a construction line drawn at 90° to the reflecting surface at the point where the ray hits. Measure both angles from the normal, not from the mirror.

The image in a plane mirror

A plane mirror is a flat mirror. The image it forms is:

• virtual: the reflected rays only appear to come from the image. They do not pass through it, so it cannot be caught on a screen
• upright: the same way up as the object
• laterally inverted: left and right are swapped, so your right hand looks like the left hand of your reflection

The image is also the same size as the object and as far behind the mirror as the object is in front.

Constructing a ray diagram

1. Mark the object point O in front of the mirror.
2. Draw two rays from O to the mirror, at different angles.
3. At each point of incidence draw a dashed normal.
4. Draw each reflected ray with angle to the normal equal to the incident angle.
5. Extend each reflected ray backwards behind the mirror with dashed lines. They cross at the image, I.
6. Use solid lines with arrows for real rays and dashed lines behind the mirror.

Worked example 1: angles at a mirror

A ray strikes a plane mirror and makes an angle of 32° with the surface of the mirror. Find the angle of incidence and the angle between the incident and reflected rays.

Normal is 90° to the surface, so
angle of incidence = 90° - 32° = 58°
angle of reflection = 58°
angle between rays = 58° + 58° = 116°

Worked example 2: distance to the image

You stand 3.5 m from a plane mirror. The image is 3.5 m behind it, so you are 7.0 m from your image. Walking towards the mirror at 1.2 m/s, you and the image each approach it at 1.2 m/s, so the gap closes at 2.4 m/s.

3.3.5 a-b: refraction as a change of speed

The speed of a wave depends on the medium. A wave crossing into a different medium changes speed. If it arrives at a slant, it also changes direction. This is refraction. A wave travelling along the normal (angle of incidence 0°) still changes speed but keeps its direction.

For light:

• entering an optically denser medium (light travels more slowly), the ray bends towards the normal
• entering a less dense medium (light travels faster), the ray bends away from the normal

Explaining it with wavefronts

A wavefront is a line joining points on a wave that are in step. Picture parallel wavefronts reaching a glass surface at an angle:

 air (fast)      wavefronts closer together in glass
 ----------------------------------------- surface
 glass (slow)    the end that enters first slows first

1. One end of each wavefront reaches the glass first and slows down.
2. The other end is still travelling at the faster speed, so it catches up.
3. The wavefront pivots, so the direction of travel turns towards the normal.
4. The frequency does not change, so a slower wave has a shorter wavelength (v = f × λ). The wavefronts bunch closer together in the glass.

Glass to air does the reverse, and the ray turns away from the normal.

Worked example 3: wavelength change

Light of frequency 5.50 × 10¹⁴ Hz travels in air at 3.00 × 10⁸ m/s. Its speed in a glass is 2.00 × 10⁸ m/s. Find its wavelength in each.

In air:   λ = v / f = 3.00 × 10⁸ / 5.50 × 10¹⁴ = 5.45 × 10⁻⁷ m (545 nm)
In glass: λ = 2.00 × 10⁸ / 5.50 × 10¹⁴ = 3.64 × 10⁻⁷ m (364 nm)

The frequency is the same in both. The wavelength fell to two-thirds because the speed did.

3.3.5 c: prisms and dispersion

White light is a mixture of colours. When it passes through a glass prism, it refracts at both faces, and different colours are refracted by different amounts. This is dispersion. The light fans out into a spectrum.

Red is bent least and violet most, so the order is red, orange, yellow, green, blue, indigo, violet. Each colour travels at a slightly different speed in glass, so each bends by a slightly different amount.

For example, in one glass n is 1.51 for red and 1.53 for violet (data supplied here). Violet travels more slowly, about 1.96 × 10⁸ m/s against 1.99 × 10⁸ m/s for red, so it bends more.

3.3.5 d: refractive index from wave speed

The refractive index of a medium is:

n = (speed of light in vacuum (air)) / (speed of light in the medium) = c / v

Light is slower in any material than in a vacuum, so n is greater than 1. It has no unit. A larger n means slower light and more bending.

Worked example 4: speed in glass

A glass has n = 1.55. Take c = 3.00 × 10⁸ m/s.

v = c / n = 3.00 × 10⁸ / 1.55 = 1.94 × 10⁸ m/s

Worked example 5: finding n

Light travels at 2.05 × 10⁸ m/s in a clear oil.

n = c / v = 3.00 × 10⁸ / 2.05 × 10⁸ = 1.46

3.3.5 e: Snell’s relationship and the required practical

For light going from air into a medium:

n = sin i / sin r

where i is the angle in air and r is the angle inside the medium. Both are measured from the normal.

Worked example 6: finding the refraction angle

Light in air hits a perspex block (n = 1.49) at i = 40°.

sin r = sin i / n = sin 40° / 1.49 = 0.6428 / 1.49 = 0.4314
r = 25.6°

The ray bends towards the normal, so r is smaller than i.

Worked example 7: finding n

A ray in air hits a plastic at i = 62° and refracts at r = 35°.

n = sin 62° / sin 35° = 0.8829 / 0.5736 = 1.54

Required practical: refraction by different substances

Aim: find the refractive index of a transparent substance.

1. Place a rectangular block (glass, perspex) on paper and draw round it. For a liquid, use a rectangular transparent tank.
2. Draw a normal at the middle of one long side.
3. Direct a thin ray from a ray box along a line at a chosen angle of incidence to the normal.
4. Mark the ray’s path with crosses: two before the block and two after the ray leaves it.
5. Remove the block. Join the exit crosses to the point where the ray entered, then draw the ray inside the block.
6. Measure i and r from the normal with a protractor.
7. Repeat for at least five values of i from about 10° to 70°.
8. Calculate sin i and sin r. Plot sin i on the y-axis against sin r on the x-axis. The line should pass through the origin and its gradient is n.
9. Repeat with a different substance and compare the gradients.

Improving accuracy: use a thin ray, mark crosses far apart and draw a sharp normal. A gradient from a best-fit line uses all the results, which is better than one calculation.

Example results for a block: i = 15°, 30°, 45°, 60°, 75° gave r = 10°, 19°, 28°, 35°, 40°. The gradient between the origin and the last point is sin 75° / sin 40° = 0.966 / 0.643 = 1.50, so n ≈ 1.50.

3.3.5 f-g: critical angle and total internal reflection

When light goes from a denser medium to a less dense one, it bends away from the normal. As the angle of incidence increases, the angle of refraction grows faster and reaches 90°, so the ray travels along the boundary. The critical angle, c, is the angle of incidence, measured in the denser medium, at which the refracted ray just skims along the boundary.

For a medium with air outside:

n = 1 / sin c

Total internal reflection (TIR) is a special case of refraction. It happens when light inside the denser medium meets the boundary at an angle of incidence that exceeds the critical angle. No light is refracted out. All of it is reflected back inside, obeying the law of reflection.

Angle of incidence in the denser medium	What happens
less than c	refracts out, with some partial reflection
equal to c	refracted ray travels along the boundary
greater than c	total internal reflection

Worked example 8: critical angle

For glass with n = 1.50:

sin c = 1 / n = 1 / 1.50 = 0.6667
c = 41.8°

A ray inside the glass hits the glass-air boundary at 50°. This is greater than 41.8°, so it is totally internally reflected. At 30° the ray refracts out: sin r = n sin i = 1.50 × sin 30° = 0.75, so r = 48.6°.

Worked example 9: n from a measured critical angle

A liquid has a critical angle of 36.0° with air.

n = 1 / sin 36.0° = 1 / 0.5878 = 1.70

3.3.5 h: optical fibres

An optical fibre is a thin strand of glass or plastic. Light or infrared radiation is shone into one end. It meets the fibre wall at a large angle of incidence, above the critical angle, so it is totally internally reflected again and again. It zigzags along the fibre, even round gentle bends. Visible light and infrared can both be sent this way.

In medicine: a bundle of fibres in an endoscope carries light inside the body and carries an image back out. Doctors can see inside the body and operate through a very small cut or an opening such as the mouth, which avoids large surgery.

In communications: telephone, broadband and television signals are sent as pulses of light or infrared. Benefits include carrying large amounts of information, little loss of signal over long distances, no electrical interference and thin, light cables.

Worked example 10: transit time

A fibre has n = 1.48 and length 2.0 km.

v = c / n = 3.00 × 10⁸ / 1.48 = 2.03 × 10⁸ m/s
t = s / v = 2000 / 2.03 × 10⁸ = 9.9 × 10⁻⁶ s

Common errors

• Measuring angles from the mirror or block surface instead of the normal.
• Saying the image in a plane mirror is “real” or “reversed”. It is virtual and laterally inverted.
• Writing that light “speeds up” when it enters glass.
• Saying the frequency changes at a boundary.
• Using n = 1 / sin c for light going from air into glass. TIR needs light starting in the denser medium.
• Drawing the extended rays behind the mirror as solid lines.

Official syllabus

OxfordAQA International GCSE Physics (9203), for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.3.4 and 3.3.5.