The questions in this review are original Marlbridge practice questions, not past exam questions, and each combines two or more topics. The rubrics are a suggested Marlbridge rubric, not official scoring. Data: c = 3.00 × 10⁸ m/s; n for air = 1.00; n for water = 1.33. Angles are measured from the normal. Sign conventions: f > 0 for a concave mirror or converging lens; sᵢ > 0 for a real image. All data are invented.
Big ideas of the unit
- A ray is a model. It is a line perpendicular to the wavefronts, in the direction light travels, and works when wave effects can be ignored (Topic 13.1).
- The normal is the reference line. Reflection and Snell’s law both measure angles from it. On a curved surface the normal changes from point to point, which is why curved surfaces focus light.
- Smooth or rough decides the beam. Specular reflection keeps parallel rays parallel; diffuse reflection scatters them because the normal varies. The law of reflection holds in both.
- Curved mirrors focus by reflection. Concave mirrors bring parallel rays to a real focal point; convex mirrors spread them from a virtual one; f ≈ R/2 (Topic 13.2).
- Refraction comes from a change of speed. n = c/v, and n₁ sin θ₁ = n₂ sin θ₂. Into a lower n, beyond the critical angle, all the light reflects (Topic 13.3).
- Lenses focus by refraction at both surfaces. How much depends on the shape and on n of the lens compared with its surroundings (Topic 13.4).
- One equation, one convention. 1/sᵢ + 1/sₒ = 1/f and |M| = |hᵢ/hₒ| = |sᵢ/sₒ| work for mirrors and thin lenses. For a single element, real means inverted and virtual means upright.
- Systems chain together. The image from one mirror or lens is the object for the next.
Key relationships and methods
| Idea | Relationship or method | Watch out for |
|---|---|---|
| Law of reflection (13.1) | θᵢ = θᵣ, both from the normal | Using the angle with the surface |
| Specular and diffuse (13.1) | Smooth: one normal direction. Rough: normal varies | “Rough surfaces break the law” |
| Focal length of a mirror (13.2) | f ≈ R/2; concave f > 0, convex f < 0 | Using R as f |
| Image equation (13.2, 13.4) | 1/sᵢ + 1/sₒ = 1/f | Dropping the sign of f |
| Magnification (13.2, 13.4) | |M| = |hᵢ/hₒ| = |sᵢ/sₒ| | Ratio upside down |
| Real and real-side (13.2, 13.4) | Mirror: real image in front. Lens: real image on the far side | Putting a lens’s real image on the object side |
| Index of refraction (13.3) | n = c/v | Larger n means slower, not faster |
| Snell’s law (13.3) | n₁ sin θ₁ = n₂ sin θ₂ | Dividing the angle, not its sine, by n |
| Total internal reflection (13.3) | Only into a lower n; sin θc = n₂/n₁ | Using 1/n when the outside is not air |
| Two-element systems (13.2, 13.4) | Image 1 becomes object 2; total |M| = |M₁| × |M₂| | Measuring the second distance from the wrong element |
Question 1 (multiple choice · mixed)
Topics 13.1, 13.3. A ray in air strikes a flat water surface with an angle of incidence of 50°. Part of the light reflects and part refracts into the water. What is the angle between the reflected ray and the refracted ray?
- (A) 80.0°
- (B) 85.2°
- (C) 94.8°
- (D) 14.8°
Answer and explanation
Answer: (C). The reflected ray is 50° from the normal, above the surface. Snell’s law: sin θ₂ = sin 50° ÷ 1.33 = 0.576, so θ₂ = 35.2° below. The angle between the rays is 180° − 50° − 35.2° = 94.8°.
- (A) is 180° − 2 × 50°, how far reflection turns the ray.
- (B) adds the two angles from the normal, as if both rays were on the same side of the surface.
- (D) is the difference between the angles of incidence and refraction.
Question 2 (multiple choice · mixed)
Topics 13.2, 13.4. A concave mirror and a converging lens have the same focal length f. An object is placed 1.5f in front of each. Which statement is correct?
- (A) Both images are real and 3f away, and both are on the far side of the device from the object.
- (B) Both images are real, inverted and twice the object’s size, 3f away. The mirror’s image is on the object’s side; the lens’s image is on the far side.
- (C) The mirror’s image is virtual and upright; the lens’s image is real and inverted.
- (D) Both images are real, 3f away and on the object’s side, because sᵢ is positive in both cases.
Answer and explanation
Answer: (B). 1/sᵢ = 1/f − 1/(1.5f) = 1/(3f), so sᵢ = +3f and |M| = 2: real and inverted for both. A mirror sends light back, so its real image is in front; a lens lets light through, so its real image is on the far side.
- (A) forgets that a mirror reflects light back to the object’s side.
- (C) The object is outside f, so the mirror’s image is real too.
- (D) Positive sᵢ means the side where the light actually goes.
Question 3 (multiple choice · mixed)
Topics 13.3, 13.4. A thin glass converging lens (n = 1.50) is used in air. It is then placed in water (n = 1.33). How does its focal length change?
- (A) The lens becomes diverging, because water has a higher index than air.
- (B) It decreases, because water slows light down, so the lens bends it more.
- (C) It stays the same, because the shape of the lens has not changed.
- (D) It increases, because the light bends less at each surface when the difference in n is smaller.
Answer and explanation
Answer: (D). Bending at a surface depends on n₂/n₁. Glass to water (1.50/1.33) is much closer to 1 than glass to air, so rays bend less and meet farther away: f increases. Glass still has the higher n, so the lens still converges.
- (A) Water’s n is still lower than the glass’s, so the lens still converges.
- (B) Slower light outside means a smaller speed change at the surface, not a larger one.
- (C) Shape matters, but so does the surrounding medium.
Question 4 (constructed response · mixed)
Topics 13.1, 13.2. A torch uses a small bulb and a concave mirror of focal length 8.0 cm.
(a) State the radius of curvature of the mirror. Explain, using the reversibility of light paths, where the bulb must be placed to give a parallel beam. (b) The bulb is moved to 10.0 cm from the mirror. Calculate the image distance and describe what the beam does. (c) The bulb is instead moved to 6.0 cm from the mirror. Calculate the image distance and describe the beam. (d) A cheaper torch has a reflector with a dull, slightly rough coating. Explain why it gives a spread-out glow rather than a beam.
Worked solution and suggested Marlbridge rubric
(a) R ≈ 2f = 16 cm. Parallel rays reflect through F; reversing the paths, rays from a bulb at F reflect parallel.
(b) 1/sᵢ = 1/8.0 − 1/10.0 = 1/40, so sᵢ = +40 cm. The rays converge to a real image 40 cm in front, then spread out beyond it.
(c) 1/sᵢ = 1/8.0 − 1/6.0 = −1/24, so sᵢ = −24 cm. Virtual: the rays diverge as if from 24 cm behind the mirror, so the beam spreads from the start.
(d) On a rough surface the normal changes from patch to patch, so rays meet patches at different angles of incidence and reflect in many directions (diffuse reflection). θᵢ = θᵣ still holds at each patch.
| Point | What earns it |
|---|---|
| 1 | R = 16 cm |
| 1 | Bulb at F, justified by reversing the parallel-ray rule |
| 1 | sᵢ = +40 cm, with a converging then spreading beam |
| 1 | sᵢ = −24 cm, virtual, with rays diverging from behind the mirror |
| 1 | (d) The normal varies across the rough surface, so rays leave in many directions |
| 1 | (d) States the law of reflection still holds at each patch |
Total: 6 points.
Question 5 (constructed response · mixed)
Topics 13.1, 13.3. A glass prism (n = 1.50) has a cross-section that is a right-angled isosceles triangle. A ray enters one short face along the normal, meets the long face, and leaves through the other short face.
(a) Calculate the speed of light in the glass. Explain why the ray does not bend as it enters. (b) Find the angle of incidence at the long face. Show that, in air, all the light reflects there, and find the total angle the ray turns through. (c) The prism is placed in a coolant liquid (n = 1.25, invented). Show that some light now escapes through the long face, and find its angle of refraction. (d) A silver coating on the long face would make the prism work in the coolant. Explain why, and find the smallest n of glass that would make it work in the coolant without the coating.
Worked solution and suggested Marlbridge rubric
(a) v = c/n = (3.00 × 10⁸) ÷ 1.50 = 2.00 × 10⁸ m/s. Along the normal θ₁ = 0, so θ₂ = 0: no bending, only a change of speed.
(b) The long face is at 45° to the short face, so the ray meets it at 45° to its normal. In air, sin θc = 1.00/1.50, so θc = 41.8°. 45° > 41.8°, so there is total internal reflection. It turns through 180° − 2 × 45° = 90° and leaves the other short face along its normal.
(c) In the coolant, sin θc = 1.25/1.50, so θc = 56.4°. Now 45° < 56.4°, so light refracts out: 1.50 sin 45° = 1.25 sin θ₂ gives sin θ₂ = 0.849, so θ₂ = 58.1°.
(d) A coating reflects by the law of reflection, whatever the outside medium: θᵣ = θᵢ = 45° still gives a 90° turn. Without a coating, TIR at 45° in the coolant needs n sin 45° ≥ 1.25, so n ≥ 1.77.
| Point | What earns it |
|---|---|
| 1 | v = 2.00 × 10⁸ m/s and no bending along the normal |
| 1 | Angle of incidence 45° at the long face |
| 1 | θc = 41.8° in air, so TIR, and a 90° turn |
| 1 | θc = 56.4° in the coolant, so light escapes |
| 1 | Refraction angle 58.1° |
| 1 | Coating: law of reflection holds whatever the outside medium |
| 1 | Minimum n = 1.77 |
Total: 7 points. Using sin θc = 1/n in the coolant earns no point for (c).
Question 6 (constructed response · mixed)
Topics 13.3, 13.4. A projector lens is a converging lens of focal length 5.0 cm. It forms a sharp image of a slide 2.4 cm tall on a wall 3.00 m from the lens.
(a) Using refraction, explain why a lens that is thicker in the middle brings rays parallel to the axis together. (b) Calculate the slide-to-lens distance, the magnification and the height of the image. Explain why the slide is put in upside down. (c) The projector is moved so the wall is 1.50 m from the lens. Should the lens move toward or away from the slide, and by how much? Find the new image height. (d) Explain why a diverging lens could not project an image of the slide onto the wall.
Worked solution and suggested Marlbridge rubric
(a) Entering glass, a ray bends toward the normal; leaving, away from it. Away from the centre the faces tilt toward each other, so both bends turn the ray toward the axis. Rays farther out bend more, so all meet near F.
(b) sᵢ = +300 cm. 1/sₒ = 1/5.0 − 1/300, so sₒ = 5.08 cm. |M| = 300/5.085 = 59, so hᵢ = 59 × 2.4 cm = 142 cm. The image is real, so it is inverted; inserting the slide upside down makes the picture upright.
(c) 1/sₒ = 1/5.0 − 1/150, so sₒ = 5.17 cm. The lens moves away from the slide by 0.088 cm (about 0.9 mm). |M| = 150/5.172 = 29, so hᵢ = 69.6 cm.
(d) A diverging lens spreads the rays from each point, so they never meet. Its image is always virtual and cannot be caught on a screen.
| Point | What earns it |
|---|---|
| 1 | Bending toward the normal on entry and away on exit, both toward the axis |
| 1 | sₒ = 5.08 cm |
| 1 | |M| = 59 and hᵢ = 142 cm |
| 1 | Real image is inverted, so the slide goes in upside down |
| 1 | Lens moves away from the slide by about 0.09 cm |
| 1 | New height 69.6 cm |
| 1 | Diverging lens: rays never meet, image always virtual |
Total: 7 points. Carry forward an error in sₒ once.
Question 7 (constructed response · mixed)
Topics 13.1, 13.2, 13.4. An object 1.2 cm tall is 30 cm to the left of a converging lens of focal length 20 cm. A plane mirror stands 80 cm to the right of the lens, perpendicular to the axis.
(a) Find the position, size and orientation of the image formed by the lens alone. (b) Light passes through this image and reaches the mirror. Find where the mirror’s image of it is, and explain why the lens now sees this as its object. (c) The reflected light passes back through the lens. Find the final image position and height, and describe it. (d) A student says: “Two inversions and a mirror, so the final image must be inverted.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) 1/sᵢ = 1/20 − 1/30 = 1/60, so sᵢ = +60 cm, right of the lens. |M₁| = 60/30 = 2.0, so the image is 2.4 cm, real and inverted.
(b) The image is 80 − 60 = 20 cm in front of the mirror, so the mirror’s image is 20 cm behind it, 100 cm from the lens, same size and orientation. Reflected rays reach the lens as if from that point, so it is the lens’s object, 100 cm away.
(c) 1/sᵢ = 1/20 − 1/100 = 4/100, so sᵢ = +25 cm. The light now travels left, so the real image is 25 cm left of the lens, 5 cm from the original object. |M₂| = 25/100 = 0.25. Total |M| = 2.0 × 1 × 0.25 = 0.50, so the height is 0.60 cm.
(d) Wrong. A plane mirror does not invert. The lens inverts twice, so the final image is upright.
| Point | What earns it |
|---|---|
| 1 | sᵢ = +60 cm and 2.4 cm, inverted |
| 1 | Mirror image 20 cm behind the mirror, 100 cm from the lens |
| 1 | Explains that rays appear to diverge from the mirror image, so it is the lens’s object |
| 1 | Final sᵢ = 25 cm, on the left, real |
| 1 | Total |M| = 0.50 and height 0.60 cm |
| 1 | Plane mirror keeps orientation; two lens inversions give an upright final image |
Total: 6 points. Carry forward an error from (a) once.
How did you do?
Questions 1–3 are worth 1 point each; Questions 4–7 are worth 26. There is no pass mark: use the checklist that matches the points you lost.
- Angles from the normal, specular and diffuse, reversibility (Q1, 4(a), 4(d), 5(d)): Topic 13.1 checklist
- Mirror images, signs and plane mirrors (Q2, 4(b)–(c), 7(b)): Topic 13.2 checklist
- Snell’s law, n = c/v and the critical angle (Q1, 3, 5, 6(a)): Topic 13.3 checklist
- Lens images, magnification and two-element systems (Q2, 6(b)–(d), 7): Topic 13.4 checklist
For a quicker topic-by-topic check, take the Unit 13 diagnostic.