OxfordAQA International AS and A-level Physics: Refraction, total internal reflection and optical fibres (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow section 3.5.8 (Refraction at a plane surface) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, which is shared AS and A-level content. Use c = 3.00 × 10⁸ m s⁻¹ and treat air as having refractive index 1. Use the study guide and revision notes first, and see the course hub, the printable checklist and the free diagnostics.

Questions

1. Light travels at 1.95 × 10⁸ m s⁻¹ in a plastic. Calculate its refractive index. [2]

2. A ray in air meets the surface of a liquid of refractive index 1.36 at 62° to the normal. Calculate the angle of refraction. [3]

3. A ray travels in glass of refractive index 1.62 towards a boundary with a liquid of refractive index 1.45.

(a) Calculate the critical angle at this boundary. [2] (b) The angle of incidence is 38°. Calculate the angle of refraction in the liquid. [2]

4. A transparent block has a critical angle of 44.0° when surrounded by air.

(a) Show that its refractive index is 1.44, to 3 significant figures. [2] (b) Calculate its critical angle when it is immersed in water of refractive index 1.33. [2]

5. A right-angled glass prism of refractive index 1.55 reflects a ray that meets its long face at 45°.

(a) Show that the ray is totally internally reflected in air. [2] (b) Explain why the ray is not totally internally reflected if the prism is immersed in water of refractive index 1.33. [2]

6. A step index optical fibre has a core and a cladding. Describe the structure of the fibre and explain the functions of the cladding. [5]

7. A step index fibre is 800 m long. The core has refractive index 1.52 and the cladding has refractive index 1.48.

(a) Calculate the critical angle at the core-cladding boundary. [2] (b) Calculate the largest angle to the fibre axis at which a ray can travel in the core and still be guided. [1] (c) Calculate the difference in transit time between a ray along the axis and a ray travelling at the critical angle. [3]

8. Explain what is meant by material dispersion in an optical fibre, and why a laser gives less material dispersion than a light-emitting diode. [4]

9. A fibre loses 18% of the power entering each kilometre.

(a) Calculate the fraction of the input power that remains after 6.0 km. [2] (b) A receiver needs at least 10% of the input power. Calculate the greatest length of fibre that can be used without an amplifier. [3] (c) Name the process in the fibre that causes this loss. [1]

10. A step index fibre of length 5.0 km has core refractive index 1.50 and cladding refractive index 1.47.

(a) Calculate the critical angle at the core-cladding boundary. [2] (b) Calculate the difference in transit time between the axial ray and the ray at the critical angle. [3] (c) Assume that pulses can be sent no closer together than the time difference in (b). Calculate the greatest number of pulses per second. [2] (d) Suggest two changes that would allow a higher pulse rate. [2]

11. A student measures angles of incidence i in air and angles of refraction r in a rectangular block.

i / degrees	20	30	40	50	60	70
r / degrees	13	20	26	31	35	39

(a) The student plots sin i against sin r. Use the first and last points to find the gradient of the line, and state what the gradient represents. [3] (b) Calculate the speed of light in the block. [1] (c) Explain why plotting a graph is better than using one pair of angles, and state one way to reduce the uncertainty in each angle. [2]

Answers

1. n = c / cs [1] = 3.00 × 10⁸ / 1.95 × 10⁸ = 1.54 [1] [2] Examiner insight: Write the ratio the right way round: a refractive index below 1 for a solid shows that c and cs have been swapped.

2. 1 × sin 62° = 1.36 × sin θ [1]. sin θ = 0.8829 / 1.36 = 0.6492 [1]. θ = 40.5° [1] [3] Examiner insight: Show the substitution into Snell’s law before the answer, and check that the refracted angle is smaller than 62°, as it must be for a ray entering a denser medium.

3. (a) sin θc = 1.45 / 1.62 [1] = 0.8951, so θc = 63.5° [1] (b) 38° is less than θc, so the ray refracts. 1.62 × sin 38° = 1.45 × sin θ [1]. sin θ = 0.9973 / 1.45 = 0.6878, so θ = 43.5° [1] [4] Examiner insight: Comparing 38° with the critical angle first shows that refraction, not total internal reflection, occurs; the angle then means something.

4. (a) n = 1 / sin 44.0° [1] = 1 / 0.6947 = 1.44 [1] (b) sin θc = 1.33 / 1.44 = 0.9236 [1], so θc = 67.5° [1] [4] Examiner insight: On a “show that” question write the unrounded value (1.4396) before the given 1.44, so that the last step is visible.

5. (a) sin θc = 1 / 1.55 = 0.645, so θc = 40.2° [1]. The angle of incidence, 45°, is greater than 40.2°, and light is going from glass to air, so there is total internal reflection [1] (b) In water, sin θc = 1.33 / 1.55, so θc = 59.1° [1]. The ray’s 45° is now less than the critical angle, so most of the light is refracted out [1] [4] Examiner insight: A comparison of the incident angle with the critical angle, in words, is needed as well as the number. The critical angle is larger because the two refractive indices are closer.

6. The fibre has a thin central core of glass or plastic [1] surrounded by cladding of lower refractive index [1]. The lower index in the cladding means light meeting the boundary above the critical angle is totally internally reflected, keeping it in the core [1]. The cladding protects the core surface from scratches that would let light escape [1]. It stops light passing between neighbouring fibres in contact, so signals do not cross [1] [5] Examiner insight: “Lower refractive index” is the key phrase. A function such as protection earns credit only when it is tied to light escaping or leaking.

7. (a) sin θc = 1.48 / 1.52 [1] = 0.9737, so θc = 76.8° [1] (b) 90° − 76.8° = 13.2° [1] (c) Axial time = 800 × 1.52 / 3.00 × 10⁸ = 4.053 × 10⁻⁶ s [1]. Critical-angle path is 1.52 / 1.48 times longer, so time = 4.163 × 10⁻⁶ s [1]. Difference = 1.10 × 10⁻⁷ s [1] [6] Examiner insight: The path length at the critical angle is L / sin θc, so the ratio of times is 1 / sin θc. Using the angle in degrees as the ratio is a common slip.

8. The refractive index of the glass depends on wavelength, so different wavelengths travel at different speeds [1]. A pulse from a source with a range of wavelengths therefore spreads out in time as the wavelengths arrive at different times [1]. An LED emits a wide range of wavelengths [1]. A laser emits a very narrow range, so there is little difference in speed within the pulse and less broadening [1] [4] Examiner insight: Explain the cause (index depends on wavelength) before the cure. A reference to a single wavelength alone does not explain why the pulse spreads.

9. (a) Fraction per km = 0.82 [1]. 0.82⁶ = 0.304 [1] (b) 0.82ˣ = 0.10 [1]. x = ln 0.10 / ln 0.82 [1] = 11.6 km [1] (c) Absorption (and scattering) of the light by the glass [1] [6] Examiner insight: The loss is multiplicative. Subtracting 18% for each kilometre (6 × 18 = 108%) is a clear sign the power law has been missed.

10. (a) sin θc = 1.47 / 1.50 [1] = 0.98, so θc = 78.5° [1] (b) Axial time = 5000 × 1.50 / 3.00 × 10⁸ = 2.50 × 10⁻⁵ s. Critical-angle time = 2.50 × 10⁻⁵ × 1.50 / 1.47 = 2.551 × 10⁻⁵ s [2]. Difference = 5.1 × 10⁻⁷ s [1] (c) Rate = 1 / 5.1 × 10⁻⁷ [1] = 1.96 × 10⁶ pulses per second [1] (d) Any two of: use a narrower (single-mode) core so only the axial path is possible; reduce the difference between core and cladding refractive indices; shorten the length between regenerators [2] [9] Examiner insight: In (d) each suggestion needs an effect on the path or time difference, not simply “better fibre”. Give two distinct changes rather than one change reworded.

11. (a) sin 70° = 0.9397, sin 20° = 0.3420, sin 39° = 0.6293, sin 13° = 0.2250 [1]. Gradient = (0.9397 − 0.3420) / (0.6293 − 0.2250) = 1.48 [1]. The gradient is the refractive index of the block [1] (b) cs = 3.00 × 10⁸ / 1.48 = 2.03 × 10⁸ m s⁻¹ [1] (c) A graph uses all the readings, so one poor measurement shows up as a point off the line and its effect is reduced [1]. Draw a thin ray and mark it with distant points, or use a protractor with a fine scale and a normal drawn with a set square [1] [6] Examiner insight: A gradient from two widely spaced points on the line is better than two adjacent readings, and the answer should carry no unit because n is a ratio.

Where marks are usually lost

• Taking angles from the surface of the boundary instead of the normal.
• Using n2/n1 the wrong way round and getting a sine greater than 1.
• Calculating a critical angle for light moving into a denser medium, where there is none.
• Stopping at “total internal reflection occurs” with no comparison of the angles.
• Describing the function of the cladding without the words “lower refractive index”.
• Saying that dispersion is caused by absorption, or that absorption lengthens pulses.
• Subtracting a percentage loss per kilometre instead of multiplying.
• Giving a refractive index with a unit.
• Rounding intermediate values and carrying the error into a time difference that is small.

Next steps

• Review the revision notes and the study guide.
• Tick off the topic on the printable checklist and visit the course hub.
• Take one of the free 10-minute diagnostics to check what to revise next.
• Book a free trial class if you want help with any of these questions.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Section 3.5.8 Refraction at a plane surface.