OxfordAQA International GCSE Physics: Reflection, refraction and total internal reflection (9203) – 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 cover sections 3.3.4 and 3.3.5 of the OxfordAQA International GCSE Physics (9203) specification, for exams from May/June 2018 onwards. The qualification is untiered. Where a question needs the speed of light, it is given as 3.00 × 10⁸ m/s. The study guide and revision notes help if you get stuck.

Course pages: the OxfordAQA International GCSE Physics hub, the printable checklist and the free diagnostics.

Questions

1. State three properties of the image formed by a plane mirror. [3]

2. A ray of light hits a plane mirror and makes an angle of 24° with the surface of the mirror.

(a) Calculate the angle of incidence. [1] (b) Calculate the angle between the incident ray and the reflected ray. [2] (c) State what is meant by the normal. [1]

3. A small lamp O stands in front of a plane mirror. Describe how you would construct a ray diagram to find the position of its image and show that the image is virtual. [4]

4. Light passes from air into a block of glass at an angle.

(a) State what happens to the speed, the frequency and the wavelength of the light. [3] (b) Use wavefronts to explain why the ray bends towards the normal. [3]

5. Light travels at 1.85 × 10⁸ m/s in a transparent plastic. The speed of light in a vacuum is 3.00 × 10⁸ m/s.

(a) Calculate the refractive index of the plastic. [2] (b) A ray in air hits the plastic at an angle of incidence of 48°. Calculate the angle of refraction. [3]

6. A narrow beam of white light passes through a glass prism and forms a spectrum on a screen. Describe what is seen and explain why it forms. [4]

7. A student investigates refraction in a clear plastic block. She shines a ray from air into the block at different angles of incidence, i, and measures the angle of refraction, r.

i (degrees)	20	35	50	65	80
r (degrees)	13	22	30	36	40

(a) Calculate sin i and sin r when i = 65° and r = 36°. [2] (b) The student plots sin i against sin r. Describe the line she should expect and how to use it to find the refractive index. [2] (c) Use the results for i = 80° to calculate the refractive index. [2] (d) Suggest one way to make the readings more accurate and explain why it helps. [2]

8. (a) Define the critical angle. [2] (b) A liquid has a critical angle of 39.0° with air. Calculate its refractive index. [2] (c) State two conditions needed for total internal reflection. [2]

9. A dense glass has a refractive index of 1.65. A ray inside the glass meets the glass-air boundary.

(a) Calculate the critical angle. [2] (b) Describe and calculate what happens when the angle of incidence is 30°. [2] (c) Show what happens when the angle of incidence is 45°. [1]

10. (a) Explain how light travels along an optical fibre. [3] (b) Describe one use of optical fibres in medicine and one benefit. [2] (c) State one benefit of using optical fibres rather than copper cables in communications. [1]

11. A lamp is fixed under the water in a tank. The refractive index of water is 1.33. Light from the lamp reaches the surface.

(a) Calculate the critical angle at the water-air surface. [2] (b) A ray reaches the surface at an angle of incidence of 40°. Calculate the angle at which it leaves the water. [3] (c) Describe what happens to a ray that reaches the surface at 55°. [2] (d) Explain why the ray in part (b) bends the way it does, in terms of the speed of light. [1]

12. A rectangular block of glass with parallel sides has refractive index 1.56. A ray in air hits one face at an angle of incidence of 35°.

(a) Calculate the speed of light in the glass. [2] (b) Calculate the angle of refraction at the first face. [3] (c) State the angle of incidence at the second face and the angle at which the ray emerges into air. [3] (d) Show that total internal reflection does not occur at the second face. [2]

Answers

1. Virtual [1]; upright [1]; laterally inverted [1] (same size and same distance behind the mirror are also true of the image). [3] Examiner insight: Use the exact terms. “Reversed” and “real” are not the terms the specification uses.

2. (a) The normal is at 90° to the surface, so i = 90° - 24° = 66° [1] (b) Angle of reflection = 66° [1]; angle between rays = 66° + 66° = 132° [1] (c) A construction line at 90° to the mirror surface at the point where the ray hits [1] Examiner insight: A common slip is to give 24° as the angle of incidence. Check whether an angle is to the surface or to the normal.

3. Draw the mirror and point O in front [1]. Draw two rays from O to the mirror, with normals, and reflect them so that angle of incidence equals angle of reflection [1]. Extend the reflected rays backwards behind the mirror as dashed lines [1]. They cross at the image I, which is behind the mirror; the rays only appear to come from it and do not pass through it, so it is virtual [1]. [4] Examiner insight: Use solid lines for real rays and dashed lines behind the mirror. Solid lines there suggest light is really present.

4. (a) Speed decreases [1]; frequency stays the same [1]; wavelength decreases [1]. [3] (b) One end of each wavefront reaches the glass first and slows down [1]. The other end is still travelling faster in the air, so it catches up [1]. This turns the wavefront and the ray bends towards the normal [1]. [3] Examiner insight: An explanation of bending needs the idea that different parts of the wavefront change speed at different times. “Light is slower so it bends” is incomplete.

5. (a) n = c / v [1] = 3.00 × 10⁸ / 1.85 × 10⁸ = 1.62 [1] (b) sin r = sin i / n = sin 48° / 1.62 [1] = 0.7431 / 1.62 = 0.4587 [1]; r = 27.3° [1] Examiner insight: Show the rearrangement before substituting, then each step of the working, not just a final angle.

6. A band of colours (a spectrum) is seen [1], with red at one end and violet at the other [1]. White light is a mixture of colours that travel at slightly different speeds in glass [1], so each is refracted by a different amount, with red bent least and violet most [1]. [4] Examiner insight: Naming the order of colours and saying which end is bent most are separate points. State both and give the reason.

7. (a) sin 65° = 0.906 [1]; sin 36° = 0.588 [1] (b) A straight line through the origin [1]. The gradient (sin i divided by sin r) is the refractive index [1]. (c) n = sin 80° / sin 40° [1] = 0.9848 / 0.6428 = 1.53 [1] (d) For example: use a narrower ray [1], because a thinner ray allows the points on its path to be marked more precisely [1]. (Other improvements: repeat the readings and average; mark crosses far apart.) [2] Examiner insight: Link an improvement to its reason. “Be more careful” names nothing specific to change.

8. (a) The angle of incidence, measured in the denser medium [1], at which the refracted ray travels along the boundary (angle of refraction 90°) [1]. (b) n = 1 / sin c [1] = 1 / sin 39.0° = 1 / 0.6293 = 1.59 [1] (c) The light travels from a denser medium towards a less dense one [1], and the angle of incidence exceeds the critical angle [1]. Examiner insight: Give both conditions. “Angle greater than critical” alone misses that the light must start in the denser medium.

9. (a) sin c = 1 / 1.65 = 0.6061 [1]; c = 37.3° [1] (b) 30° is less than 37.3°, so the ray refracts out [1]. sin r = 1.65 × sin 30° = 0.825, so r = 55.6° [1] (c) At 45°, 1.65 × sin 45° = 1.167, which is greater than 1, so no refracted ray is possible; the ray is totally internally reflected [1]. Examiner insight: The calculation gives a sine larger than 1 when TIR occurs. Comparing the angle with c is cleaner, but state the conclusion in words.

10. (a) Light enters one end of the thin fibre and meets the fibre wall [1]. The angle of incidence exceeds the critical angle [1], so it is totally internally reflected, again and again, along the fibre [1]. [3] (b) An endoscope uses fibres to see inside the body [1]; surgery or examination needs only a small cut or opening, so recovery is faster [1]. [2] (c) Carries large amounts of information, or less signal loss, or no electrical interference [1] Examiner insight: A use and a benefit are separate points. Name the device (endoscope) and the benefit to the patient.

11. (a) sin c = 1 / 1.33 [1] = 0.7519; c = 48.8° [1] (b) 40° is less than 48.8°, so the light refracts out [1]. sin r = 1.33 × sin 40° = 0.8549 [1]; r = 58.7° [1] (c) 55° exceeds the critical angle [1], so the ray is totally internally reflected back into the water [1]. (d) The light speeds up as it passes from water into air, so it bends away from the normal [1]. Examiner insight: A direction claim such as “away from the normal” is strongest when tied to the change in speed.

12. (a) v = c / n [1] = 3.00 × 10⁸ / 1.56 = 1.92 × 10⁸ m/s [1] (b) sin r = sin 35° / 1.56 [1] = 0.5736 / 1.56 = 0.3677 [1]; r = 21.6° [1] (c) The faces are parallel, so the normals are parallel and the angle of incidence at the second face is 21.6° [1]. The ray leaves the glass at 35° [1], parallel to the incident ray but shifted sideways [1]. (d) sin c = 1 / 1.56 = 0.6410, so c = 39.9° [1]; 21.6° is less than 39.9°, so there is no TIR [1]. Examiner insight: In a multi-part question each part uses the result of the previous one. Carry unrounded values from one part to the next and set out the working clearly, so the reasoning in each part can be followed.

Where marks are usually lost

• Measuring from the mirror or block surface rather than the normal.
• Using the words “real” or “reversed” for a plane mirror image.
• Drawing refracted rays that bend the wrong way at a boundary.
• Explaining bending without mentioning a change in speed.
• Applying n = 1 / sin c to light starting in air.
• Rounding too early, which shifts a later answer.
• Leaving out units on a speed.
• Describing the required practical without repeat readings or a graph of sin i against sin r.

Next steps

• Re-read the revision notes for any answers you missed.
• Go back to the study guide for full explanations.
• Open the OxfordAQA International GCSE Physics hub and the printable checklist.
• Take one of the free 10-minute diagnostics to find your weakest topic.
• Book a free trial class if you want help with the unit.

Official syllabus

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