Refraction at a plane surface — AQA A-Level Physics
Teaching note checked against the official specification.
Refraction at a plane surface is the change in direction of a wave when it crosses a boundary between two media with different refractive indices, governed by Snell's law: n1 sin(theta1) = n2 sin(theta2). For AQA A-Level Physics, you must be able to calculate refractive index, critical angle, and analyse ray diagrams for light passing through parallel-sided blocks and optical fibres.
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Refraction at a plane surface explained
Refraction at a plane surface covers 3 specification points, including Optical fibres will be limited to step index… and Material and modal dispersion..
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Open a point below for its explanation and revision guidance.
Optical fibres will be limited to step index only.
A step-index optical fibre has a core of higher refractive index surrounded by a cladding of lower refractive index, producing a sharp jump in index at the boundary. Light entering within the acceptance angle strikes the core–cladding boundary above the critical angle and undergoes total internal reflection, so it zig-zags along the core with little loss. The specification limits fibre study to this step-index type, so you should not analyse graded-index fibres, where the index changes gradually and rays curve. You should be able to sketch the path, identify core, cladding and buffer, and explain how the critical angle and acceptance angle follow from refractive indices.
Material and modal dispersion.
Dispersion in an optical fibre broadens a pulse of light so that pulses can overlap and limit the bit rate. Material dispersion arises because the refractive index of the core depends on wavelength, so different wavelengths in a source travel at different speeds and arrive at different times. Modal dispersion arises because rays taking different paths through a step-index core have different path lengths, so they arrive at different times even at the same wavelength. You should be able to distinguish the two causes, explain how each broadens a pulse, and suggest how using a monochromatic source or a narrower core reduces the effect.
Students are expected to understand the principles and consequences of pulse broadening and absorption.
When a light pulse travels along an optical fibre, it spreads in time: this is pulse broadening. Two causes are modal dispersion (different rays take different path lengths) and material dispersion (different wavelengths travel at different speeds because refractive index varies with wavelength). Broadening limits the bit rate because pulses overlap, causing errors. Absorption removes energy from the signal as it is converted to other forms, reducing amplitude and requiring amplification or limiting repeater spacing. In a step-index multimode fibre, a ray along the axis arrives before a ray that zig-zags, so a short input pulse becomes a longer output pulse. Absorption depends on wavelength and material purity, so operating at a low-absorption wavelength window reduces loss.
Your focus
- Describe the structure of a step-index optical fibre, naming the core, cladding and buffer and stating their roles.
- Explain how total internal reflection at the core–cladding boundary guides light along the fibre.
- Apply sin C = n₂ / n₁ to calculate a critical angle and relate the acceptance angle to whether a ray is guided.
Show all 9 objectives
- Distinguish material dispersion from modal dispersion by their causes in an optical fibre.
- Explain how each type of dispersion broadens a pulse and limits the bit rate.
- Suggest and justify practical measures that reduce material or modal dispersion.
- Describe how pulse broadening occurs in an optical fibre and state its effect on signal transmission.
- Distinguish between modal dispersion and material dispersion in terms of their causes.
- Explain how absorption reduces signal amplitude and describe one method used to reduce its impact.
Which command word is this?
You must use the correct formula, substitute the given values, and show your working. The answer must include the correct unit and be given to an appropriate number of significant figures. No marks are awarded for the correct answer without working unless stated otherwise.
Refraction at a plane surface exam tips
Topic Overview
Refraction at a plane surface is a fundamental topic in AQA A-Level Physics that explains how light changes direction when it passes from one medium to another. It introduces the refractive index, Snell's law, and the conditions for total internal reflection, which are essential for understanding lenses, optical fibres, and many real-world applications. You will learn to calculate angles of refraction, critical angles, and speeds of light in different materials.
This topic builds directly on your knowledge of waves and ray diagrams from GCSE and provides the foundation for more advanced topics such as lenses, diffraction, and polarisation. It is also heavily examined in both AS and A-Level papers, often combined with practical skills like drawing ray diagrams and using Snell's law in experiments. Mastering refraction is crucial for achieving high marks in waves and optics questions.
Key Concepts
- →Refractive index n of a medium is defined as n = c / v, where c is the speed of light in a vacuum and v is the speed of light in the medium. It is always greater than or equal to 1.
- →Snell's law relates the angles of incidence and refraction to the refractive indices of the two media: n1 sin(theta1) = n2 sin(theta2), where angles are measured from the normal.
- →When light travels from a denser to a less dense medium, there is a critical angle C given by sin C = 1/n (for the denser medium relative to the less dense one). At angles greater than C, total internal reflection occurs.
- →The frequency of light remains constant during refraction, but its speed and wavelength change. The colour (frequency) does not change, but the wavelength changes according to v = f lambda.
- →Ray diagrams for refraction at a plane surface must show the normal, incident ray, refracted ray, and angles measured from the normal. For a parallel-sided block, the emergent ray is parallel to the incident ray but laterally displaced.
Marking Points
- A step-index fibre consists of a core of higher refractive index surrounded by cladding of lower refractive index, with an abrupt change of index at the core–cladding boundary.
- Total internal reflection occurs at the core–cladding boundary when light meets it at an angle of incidence greater than the critical angle, so light is repeatedly reflected along the core.
- The critical angle is related to the refractive indices by sin C = n₂ / n₁, where n₁ is the core index and n₂ is the cladding index, and n₁ must exceed n₂ for guidance to be possible.
- The acceptance angle at the entrance face determines whether an incoming ray can be guided; rays entering too steeply are refracted out through the cladding rather than trapped.
- Cladding protects the core surface, reduces scattering losses and prevents light crossing between adjacent fibres, while the buffer provides mechanical protection.
- Only step-index behaviour is required: the index changes sharply at one boundary, unlike graded-index fibres where the index varies continuously and rays follow curved paths.
- Material dispersion occurs because the refractive index of the core varies with wavelength, so different wavelengths travel at different speeds and spread out in time.
- Modal dispersion occurs because rays follow different zig-zag paths of different lengths in a step-index core, so they take different times to reach the far end.
- Both effects broaden an input pulse, causing adjacent pulses to overlap and limiting the maximum reliable bit rate over a given length of fibre.
- Using a monochromatic source such as a laser reduces material dispersion because the range of wavelengths present is very small.
- Reducing the core diameter or using a single-mode fibre reduces modal dispersion because fewer ray paths are available.
- Material and modal dispersion are separate mechanisms: one depends on wavelength spread, the other on path-length spread, and both can occur together in a step-index fibre.
- Pulse broadening is the temporal spreading of a signal pulse as it propagates, so the output pulse has a longer duration than the input pulse.
- Modal dispersion arises because rays travelling different paths in a multimode fibre have different path lengths and therefore different transit times.
- Material dispersion arises because the refractive index of the fibre depends on wavelength, so different spectral components travel at different speeds.
- Consequences of broadening include overlap of adjacent pulses, which limits the maximum bit rate and can cause detection errors.
- Absorption is the loss of optical power as light is absorbed by the fibre material or impurities, converting energy to other forms such as thermal energy.
- Absorption reduces signal amplitude, so the signal must be amplified or regenerated, and it limits the maximum transmission distance between repeaters.
- Using a monomode fibre reduces modal dispersion because only one mode propagates, and choosing a wavelength with low absorption reduces loss.
Examiner Tips
- 💡Draw a labelled longitudinal section showing core, cladding, buffer, the normal at the boundary and the zig-zag ray path, marking angles of incidence and reflection.
- 💡When calculating the critical angle, identify clearly which refractive index is n₁ and which is n₂ before substituting into sin C = n₂ / n₁.
- 💡Link the refractive index difference to the size of the critical angle: a larger difference gives a smaller critical angle and traps a wider range of rays.
- 💡Use a sketch of two rays with different path lengths to show how modal dispersion delays one ray relative to another.
- 💡When comparing the two effects, state the cause, the quantity that varies and the consequence for pulse arrival time in each case.
- 💡Link pulse broadening to bit rate by explaining that overlapping pulses make it impossible to distinguish successive bits reliably.
- 💡When explaining pulse broadening, refer explicitly to the effect on the output pulse duration and how that affects the ability to distinguish adjacent bits.
- 💡Use a labelled diagram or ray paths to show a short axial path and a longer zig-zag path in a multimode fibre, then state which arrives first.
- 💡For absorption, distinguish between the cause (energy absorbed by the material) and the consequence (reduced amplitude and need for amplification or regeneration).
- 💡Always draw the normal as a dashed line perpendicular to the surface and measure angles from it. Label angles clearly with the correct symbols (i, r, C).
- 💡When using Snell's law, write the full equation n1 sin(theta1) = n2 sin(theta2) before substituting numbers. This shows your method and can earn method marks even if the final answer is wrong.
- 💡For total internal reflection questions, always state the direction of travel (e.g., 'from glass to air') and compare the angle of incidence to the critical angle. Use the phrase 'angle of incidence is greater than the critical angle' to secure the mark.
Common Mistakes
- Thinking total internal reflection happens at the outer surface of the cladding rather than at the core–cladding boundary; correct this by stating that reflection occurs where the refractive index changes, at the core–cladding interface.
- Believing any ray entering the fibre will be trapped; correct this by explaining that the ray must strike the boundary above the critical angle, which requires entry within the acceptance angle.
- Confusing step-index with graded-index behaviour by describing a gradual change of refractive index; correct this by stating that in a step-index fibre the core index is uniform and changes abruptly at the cladding.
- Treating material and modal dispersion as the same effect; correct this by stating that material dispersion depends on wavelength-dependent refractive index while modal dispersion depends on different path lengths.
- Claiming that modal dispersion is reduced by using a brighter source; correct this by explaining that reducing the number of available paths, for example with a narrower core, is what reduces modal dispersion.
- Assuming a monochromatic source removes all dispersion; correct this by stating that it removes material dispersion but modal dispersion can still occur in a step-index fibre.
- Confusing pulse broadening with absorption: broadening changes pulse duration, whereas absorption reduces pulse amplitude. Correction: state that broadening is temporal spreading and absorption is energy loss.
- Thinking that all rays in a fibre travel at the same speed regardless of path: in a multimode fibre, rays with longer paths take longer, causing modal dispersion. Correction: link path length differences to arrival time differences.
- Assuming absorption is independent of wavelength: absorption varies with wavelength and material. Correction: refer to low-absorption wavelength windows when explaining how loss is minimised.
- Students often think that light bends because it hits the surface at an angle, but the real reason is the change in speed as it enters a different medium. Correction: Refraction occurs because the speed of light changes, causing the wavefront to change direction.
- Many believe that total internal reflection can happen when light travels from air into glass. Correction: Total internal reflection only occurs when light travels from a denser medium to a less dense medium (e.g., glass to air) and the angle of incidence exceeds the critical angle.
- Students sometimes think that the frequency of light changes during refraction. Correction: The frequency remains constant; only the speed and wavelength change. This is why the colour of light does not change upon refraction.
Revision Plan
- 1Day 1-2: Revise the definitions of refractive index and Snell's law. Practice calculating refractive index and angles using past paper questions. Ensure you can rearrange the formula confidently.
- 2Day 3-4: Learn the conditions for total internal reflection and the critical angle formula. Practice drawing ray diagrams for semicircular blocks and optical fibres. Complete at least five calculations for critical angle.
- 3Day 5-6: Work through exam-style questions on refraction at plane surfaces, including those with parallel-sided blocks and real-life applications. Focus on explaining why light bends and the effects on speed and wavelength.
- 4Day 7: Create a summary sheet with key formulas, definitions, and common pitfalls. Test yourself with active recall prompts and past paper questions under timed conditions.
- 5Day 8-10: Review any weak areas by re-reading textbook sections and watching video tutorials. Attempt a full past paper question on waves and optics to integrate your knowledge.
Exam Question Types
- 📋Calculation of refractive index or angle of refraction using Snell's law. Advice: Always write the formula, substitute values, and give your answer to an appropriate number of significant figures (usually 2 or 3).
- 📋Determination of critical angle and explanation of total internal reflection. Advice: State the direction of travel, use sin C = 1/n, and compare the angle of incidence to the critical angle in your explanation.
- 📋Ray diagram drawing for a light ray passing through a parallel-sided glass block or optical fibre. Advice: Use a ruler, draw the normal at each boundary, and ensure the emergent ray is parallel to the incident ray for a parallel-sided block.
- 📋Practical-based question on measuring refractive index using a glass block and ray tracing. Advice: Describe the procedure clearly, including how to measure angles and reduce errors, and include a suitable graph (e.g., sin i vs sin r) to determine n.
Command Word Expectations (AQA)
You must use the correct formula, substitute the given values, and show your working. The answer must include the correct unit and be given to an appropriate number of significant figures. No marks are awarded for the correct answer without working unless stated otherwise.
You must provide a detailed reasoning that links cause and effect. For example, 'explain why total internal reflection occurs' requires stating that the angle of incidence exceeds the critical angle and that light travels from a denser to a less dense medium. Use correct terminology and refer to the physics principles involved.
You must give a detailed account of the steps or observations without necessarily giving reasons. For example, 'describe how to measure the refractive index of a glass block' requires a clear, sequential method including apparatus, measurements, and calculations. Marks are awarded for each correct point.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A ray of light travelling in air strikes a flat surface of a transparent plastic block at an angle of incidence of 55.0 degrees. The refractive index of the plastic is 1.45. Calculate the angle of refraction inside the plastic and the speed of light in the plastic. (Speed of light in air = 3.00 x 10^8 m/s).
- 1.Step 1: Identify given facts: angle of incidence i = 55.0 degrees, refractive index of plastic n = 1.45, speed of light in air c = 3.00 x 10^8 m/s. Light travels from air (n=1.00) into plastic.
- 2.Step 2: Apply Snell's law: n1 sin i = n2 sin r. Here n1 = 1.00, n2 = 1.45. So sin r = (1.00 / 1.45) x sin 55.0 = 0.6897 x 0.8192 = 0.5650. Therefore r = sin^-1(0.5650) = 34.4 degrees.
- 3.Step 3: Calculate speed in plastic using n = c / v, so v = c / n = (3.00 x 10^8) / 1.45 = 2.07 x 10^8 m/s.
- 4.Step 4: State final answers with units and correct significant figures: angle of refraction = 34.4 degrees, speed = 2.07 x 10^8 m/s.
Question: A student shines a laser into a semicircular glass block of refractive index 1.52. The light enters the curved surface normally and strikes the flat surface from inside the glass. Calculate the critical angle at the glass-air boundary and describe what happens when the angle of incidence at the flat surface is 40 degrees and when it is 45 degrees.
- 1.Step 1: Identify given facts: refractive index of glass n = 1.52, light travels from glass to air. Critical angle C is given by sin C = 1/n.
- 2.Step 2: Calculate sin C = 1 / 1.52 = 0.6579. Therefore C = sin^-1(0.6579) = 41.1 degrees.
- 3.Step 3: Compare given angles to critical angle. At 40 degrees, angle of incidence < critical angle, so light refracts out into air with an angle of refraction greater than 40 degrees, and some light is partially reflected internally.
- 4.Step 4: At 45 degrees, angle of incidence > critical angle, so total internal reflection occurs: all light is reflected back into the glass, obeying the law of reflection (angle of reflection = 45 degrees).
- 5.Step 5: State final conclusion: critical angle = 41.1 degrees; at 40 degrees refraction occurs; at 45 degrees total internal reflection occurs.