Light
Key idea: Reflection, refraction, refractive index and the image characteristics of converging lenses.
Continue where you stopped
The core idea
Syllabus and review details
For this course, focus on reflection, refraction, refractive index and describing converging-lens images. You do not need critical angle, total internal reflection or a memorised table of images at different object distances.
- K326 / K327 Science Physics componentK326 / K327 · 2027Checked against the syllabus · complete topic coverageK326/K327 2027 syllabus, Light topic 11
Reflection and angles from the normal
Measure every reflection angle from the normal
- Normal
- A line at 90° to the surface at the point where the ray meets it.
- Angle of incidence, i
- The angle between the incident ray and the normal.
- Angle of reflection, r
- The angle between the reflected ray and the normal.
i = r
The incident ray, reflected ray and normal lie in one plane. Draw arrowheads to show direction and place the normal exactly at the point of incidence.
Worked reflection construction
A ray makes 32° with a plane mirror. Its angle to the normal is90° − 32° = 58°. Therefore i = r = 58°. Draw the reflected ray on the opposite side of the normal at 58°.
Try first: an incoming ray makes 25° with the mirror surface. Find its angle to the normal and the reflected angle. Explain why simply copying 25° would be wrong.
The normal and surface are perpendicular, so i = 90° − 25° = 65°. The reflected ray makes r = 65° with the normal. Copying the surface angle labels the wrong angle.
Independent construction: a horizontal mirror is struck from above-left by a ray at 40° to its upward normal. Draw the reflected ray with an arrow. A learner draws it below the mirror instead; explain the first error.
The ray leaves the point of incidence upwards-right at 40° to the normal, remaining on the incident side of the mirror. A path below the mirror would represent transmission through it, not the reflected ray. Equal angles alone do not fix that wrong-side construction.
Refraction and refractive index
A speed change can change a ray's direction at a boundary
The angle of refraction, r, is measured between the refracted ray and the normal. At oblique incidence, a ray crossing into a medium where light travels more slowly bends towards the normal; a ray crossing into a faster medium bends away from it. At normal incidence its speed and wavelength change, but its direction does not.
Use the sine ratio for a fixed pair of media
sin i / sin r = constant
Both angles are measured from the normal. For the same light and a fixed direction through a fixed pair of media, this ratio is constant. It compares the two media: sin i / sin r = v1 / v2 = n2 / n1, where the n values are absolute refractive indices. Reversing the direction gives the reciprocal ratio. Do not replace the sines with the angles themselves.
Refractive index compares light speeds
n = c / v
Here c is the speed of light in vacuum and v is its speed in the medium. A larger n means a smaller speed in that medium.
The measured sine ratio equals the second medium's absolute index only when the first medium has n ≈ 1, as in the usual air-to-material approximation. Between two other media, it is a relative index. For example, supplied indices n1 = 1.20 and n2 = 1.50 give a sine ratio of 1.50/1.20 = 1.25, not 1.50.
| Trial | Incidence i / ° | Refraction r / ° |
|---|---|---|
| A | 20.0 | 13.2 |
| B | 40.0 | 25.4 |
| C | 60.0 | 35.3 |
Worked interpretation: in trial A, sin 20.0° / sin 13.2° ≈ 1.50. The ray has a smaller angle to the normal inside the block, consistent with a smaller speed there. The angles themselves need not stay in a fixed ratio: refraction relates their sines.
Try first: calculate the sine ratios for B and C in degree mode. Do the rounded observations support the same constant? Compare this with using i/r.
Both sine ratios round to 1.50. The angle ratios are about 1.57 and 1.70, so i/r is not the constant. Small departures in the sine ratios are expected because the supplied angles are rounded.
Independent check: a medium has absolute index 1.60. Using c = 3.00 × 10⁸ m/s, find its light speed. Would a measured sine ratio of 1.60 always identify that absolute index?
v = c/n = 1.875 × 10⁸ m/s, or 1.88 × 10⁸ m/s to three significant figures. The sine ratio compares the second medium with the first, so it equals the second absolute index only if the first index is approximately 1.
Converging lenses and image properties
A thin converging lens brings a parallel beam to a focus
Rays parallel to the principal axis refract through the principal focus on the far side of the lens. The focal length is the distance from the optical centre of the lens to that principal focus.
Try first: on an optical bench, a lens centre is at the 25.0 cm mark and a parallel beam comes to a sharp focus at 41.0 cm. Find the focal length and correct the claim that it is 41.0 cm.
The focal length is the separation, 41.0 − 25.0 = 16.0 cm. The 41.0 cm reading is a position measured from the bench zero, not a distance from the lens centre. The parallel incident beam is what identifies this meeting point as the principal focus.
Describe an image using three independent pairs
Real or virtual
A real image is formed where rays meet and can be caught on a screen. A virtual image is located where backward extensions appear to meet.
Magnified or diminished
Compare image height with object height; do not infer size from whether the image is real.
Upright or inverted
Compare the image orientation with the object orientation.
Independent interpretation: a sharp image forms on a screen. It points downwards while its object points upwards, and its height is 2.4 cm compared with the object's 1.2 cm. Give three characteristics and repair the claim “real means diminished”.
The image is real because it forms on the screen, inverted because its orientation is reversed, and magnified because it is twice the object's height. Screen formation establishes where rays meet; it does not determine the image's size.
Common mistakes
“Angles are measured from the surface.”
Remember: measure i and r from the normal.
“Light always bends towards the normal.”
For an oblique transmitted ray, slowing bends it towards the normal and speeding up bends it away. A ray along the normal keeps its direction even when its speed changes.
“Higher n means higher speed.”
Remember: n = c/v, so higher n means lower v.
“Every converging-lens image is real and inverted.”
Remember: classify real/virtual, upright/inverted and magnified/diminished separately from the stated observation.
Worked examples
Link a calculation to the ray direction
Light enters glass from air, taking the air index as approximately 1, withi = 45° and r = 28°. Find the sine ratio and the speed of light if the glass has n = 1.51.
1. Use the two measured angles: sin 45° / sin 28° = 1.51 to three significant figures.
2. Rearrange n = c/v: v = c/n.
3. Substitute: v = (3.00 × 108 m s−1) / 1.51 = 1.99 × 108 m s−1.
Check: v is below c, as it must be. The ray bends towards the normal because it enters the slower medium.
Build a complete image description from evidence
An image can be caught sharply on a screen, appears upside down and is shorter than its object. A complete answer is: “The image is real because it can be formed on a screen, inverted because its orientation is reversed, and diminished because its height is smaller than the object's.”
Each observation supports one characteristic; none of the three words is guessed from another.
Spot and correct an angle error
A student writes “the incidence angle is 20°” beside a ray drawn 20° from the glass surface. The value is wrong because incidence is measured from the normal. The correct incidence angle is 90° − 20° = 70°.
Check your understanding
Guided practice
A ray is 25° to a mirror. Find i and r.
Both are 65° because angles are measured from the normal.
What changes at normal incidence from air into glass?
Speed and wavelength decrease; frequency and direction stay unchanged.
A medium has n = 1.50. Find v using c = 3.00 × 10⁸ m s⁻¹.
v = c/n = 2.00 × 10⁸ m s⁻¹.
Independent self-check
- Define the normal, angle of incidence and angle of reflection. Then state the law of reflection.
- An oblique ray crosses from glass into air and its speed increases. State and explain its bend direction.
- For one fixed pair of media, i = 50° and r = 30°. Calculate sin i / sin r.
- A medium has refractive index 1.60. Calculate the speed of light in it using c = 3.00 × 108 m s−1.
- A converging lens forms an image that cannot be caught on a screen, is the same way up as the object and is twice its height. Describe the image, then define focal length.
Check the independent answers
- The normal is a line at 90° to the surface at the point of incidence. The angle of incidence is between the incident ray and the normal; the angle of reflection is between the reflected ray and the normal. The law is i = r.
- The ray bends away from the normal because it speeds up as it enters air.
- sin 50° / sin 30° = 0.766/0.500 = 1.53 to three significant figures.
- v = c/n = (3.00 × 108)/1.60 = 1.88 × 108 m s−1 to three significant figures.
- The image is virtual, upright and magnified. Focal length is the distance from the lens's optical centre to its principal focus.
Try this next
Close the answers and retry with new angles and a new refractive index. Review only the relationship that was difficult before moving on.
Explore reflection, refraction and converging lenses using only the ideas taught above.
Practise this topic
Use the course-specific Light check after the lesson. It covers reflection and refraction as well as lenses; if one part is difficult, return to the matching explanation and worked example before trying again later.
Continue with the next resource in this course.
Course and syllabus information
- Course
- SEC G3 Combined Science Physics component
- Edition
- SEC G3 Combined Science Physics component 2027