Geometric Optics: Unit Diagnostic (Physics 2 Unit 13)

What this diagnostic is for

Use this check to find which Unit 13 (Geometric Optics) topics to revisit. It has eight multiple-choice and three short-answer questions. These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, and your total does not predict an exam score. What matters is which questions you miss.

How to take it:

Work without notes, in about 30 minutes. Write a reason for each answer; sketch ray diagrams.
Scientific calculator in degree mode. c = 3.00 × 10⁸ m/s; n for air = 1.00. Angles are measured from the normal.
Signs: f > 0 for a concave mirror or converging lens, f < 0 for convex or diverging; sᵢ > 0 real, sᵢ < 0 virtual. |M| = |hᵢ/hₒ| = |sᵢ/sₒ|.
Open each answer only after committing to your own.

Question 1 (multiple choice · 13.1)

A ray reflects from a plane mirror. The angle between the incident ray and the reflected ray is 64°. What angle does the incident ray make with the surface of the mirror?

(A) 32°
(B) 58°
(C) 64°
(D) 26°

Answer and explanation

Answer: (B). The rays make equal angles on opposite sides of the normal, so θᵢ = θᵣ = 64° ÷ 2 = 32° from the normal. The angle with the surface is 90° − 32° = 58°.

(A) is the angle of incidence, measured from the normal, not the surface.
(C) is the angle between the two rays.
(D) is 90° − 64°, mixing two different angles.

If you missed this: read “Reflection and the law of reflection” in the Topic 13.1 study guide.

Question 2 (multiple choice · 13.1)

Under a lamp, a matte (slightly rough) photo looks clear from every seat, but a glossy photo shows a glare spot from certain seats only. Which is the best explanation?

(A) The glossy surface reflects light at a larger angle than the angle of incidence, concentrating it.
(B) The matte surface does not obey the law of reflection.
(C) The matte surface absorbs all the light; the glossy one reflects all of it.
(D) The normal varies across the matte surface, so light reflects in many directions; on the glossy surface it is nearly constant, so light reflects one way.

Answer and explanation

Answer: (D). Diffuse (matte) against specular (glossy) reflection. The law of reflection holds at every tiny patch of both; what differs is how the normal varies across the lit area.

(A) No surface breaks θᵢ = θᵣ. The glare is bright because the reflected rays all leave one way.
(B) is the common mistake: rough surfaces still obey θᵢ = θᵣ patch by patch.
(C) If the matte photo absorbed all the light, you could not see it at all.

If you missed this: read “Specular and diffuse reflection” in the Topic 13.1 study guide.

Question 3 (multiple choice · 13.2)

A beam of rays parallel to the principal axis strikes a convex spherical mirror whose radius of curvature is 30 cm. Which describes the reflected rays?

(A) They spread out as if they came from a point on the axis 30 cm behind the mirror.
(B) They meet at a point on the axis 15 cm in front of the mirror.
(C) They spread out as if they came from a point on the axis 15 cm behind the mirror.
(D) They stay parallel to the axis after reflection.

Answer and explanation

Answer: (C). A convex mirror diverges light. Parallel rays appear to come from the focal point behind it, and f ≈ R/2 = 15 cm.

(A) uses the radius of curvature instead of half of it.
(B) describes a concave mirror.
(D) describes a plane mirror, whose focal point is infinitely far away.

If you missed this: read “Curved mirrors and the focal point” in the Topic 13.2 study guide.

Question 4 (multiple choice · 13.2)

An object 3.0 cm tall stands on the axis 25 cm in front of a concave mirror of focal length 10 cm. Which describes the image?

(A) 2.0 cm tall, real and inverted
(B) 2.0 cm tall, virtual and upright
(C) 4.5 cm tall, real and inverted
(D) 1.2 cm tall, real and inverted

Answer and explanation

Answer: (A). 1/sᵢ = 1/10 − 1/25 = 3/50, so sᵢ = +16.7 cm: real, so inverted. |M| = 16.7/25 = 0.667, so hᵢ = 2.0 cm.

(B) has the wrong type: positive sᵢ means real.
(C) uses sₒ/sᵢ, the ratio upside down.
(D) uses f/sₒ as the magnification.

If you missed this: read “The mirror equation and sign conventions” and “Magnification” in the Topic 13.2 study guide.

Question 5 (multiple choice · 13.3)

A ray of light passes from medium A into medium B and bends away from the normal. Which statement is correct?

(A) n_B < n_A, and light travels slower in B than in A.
(B) n_B > n_A, and light travels slower in B than in A.
(C) n_B < n_A, and light travels faster in B than in A.
(D) n_B > n_A, and light travels faster in B than in A.

Answer and explanation

Answer: (C). Bending away from the normal means entering a lower n. Since n = c/v, lower n means faster light.

(A) and (D) break n = c/v: larger n, smaller speed.
(B) describes bending toward the normal.

If you missed this: read “Why light bends” and “The index of refraction” in the Topic 13.3 study guide.

Question 6 (multiple choice · 13.3)

A ray in air strikes the flat surface of a clear plastic (n = 1.48) with an angle of incidence of 52°. What is the angle of refraction?

(A) 35.1°
(B) 32.2°
(C) 24.6°
(D) 52°

Answer and explanation

Answer: (B). Snell’s law: (1.00) sin 52° = 1.48 sin θ₂, so sin θ₂ = 0.788 ÷ 1.48 = 0.532 and θ₂ = 32.2°, bent toward the normal as expected.

(A) divides the angle by n. Snell’s law works with the sines of the angles.
(C) uses 38°, the angle with the surface, as the angle of incidence.
(D) assumes no bending, true only along the normal.

If you missed this: read “Snell’s law” in the Topic 13.3 study guide.

Question 7 (multiple choice · 13.4)

In a ray diagram, a ray from the top of an object passes through the focal point on the object’s side of a converging lens and then reaches the lens. Where does it go after the lens?

(A) It bends so that it crosses the axis at twice the focal length on the far side.
(B) It passes through the focal point on the far side of the lens.
(C) It continues in a straight line without bending.
(D) It leaves parallel to the principal axis.

Answer and explanation

Answer: (D). This is the parallel ray reversed. A parallel ray refracts through the far focal point, and light paths are reversible, so a ray through the near focal point leaves parallel.

(A) is not a principal-ray rule.
(B) is what happens to a ray that arrives parallel to the axis.
(C) is the rule for the ray through the centre of the lens.

If you missed this: read “The three principal rays” in the Topic 13.4 study guide.

Question 8 (multiple choice · 13.4)

A thin lens forms an upright image three times the size of an object placed 4.0 cm from the lens. What is the focal length of the lens?

(A) +6.0 cm (converging)
(B) +3.0 cm (converging)
(C) −6.0 cm (diverging)
(D) +12 cm (converging)

Answer and explanation

Answer: (A). Upright means virtual, so sᵢ = −3 × 4.0 cm = −12 cm. Then 1/f = 1/sₒ + 1/sᵢ = 1/4.0 − 1/12 = 2/12, so f = +6.0 cm.

(B) uses sᵢ = +12 cm, a real (inverted) image.
(C) A diverging lens only gives reduced images.
(D) takes the image distance as the focal length.

If you missed this: read “Worked example 2: a magnifying glass” and the sign-convention table in the Topic 13.4 study guide.

Question 9 (short answer · 13.2)

An object 2.0 cm tall stands on the axis 24 cm in front of a concave mirror of focal length 16 cm.

(a) Describe the paths of two principal rays from the top of the object after they reflect. (b) Calculate the image distance and image height, and describe the image fully. (c) The object is moved to 16 cm from the mirror. What happens to the image? Explain.

Worked answer

(a) Any two of: the parallel ray reflects through F; the ray through F reflects parallel; the ray to the vertex reflects at the same angle to the axis on the other side.
(b) 1/sᵢ = 1/16 − 1/24 = 1/48, so sᵢ = +48 cm. |M| = 48/24 = 2.0, so hᵢ = 4.0 cm. The image is 48 cm in front of the mirror: real, inverted, enlarged.
(c) The object is at F: 1/sᵢ = 0. The reflected rays leave parallel and never meet, so no image forms.

If you missed this: read “The three principal rays” and the image table under “Magnification” in the Topic 13.2 study guide.

Question 10 (short answer · 13.3)

A laser in a tank of clear liquid points up at the flat surface. When the beam meets the surface at 43.0°, the light that leaves skims exactly along the surface.

(a) Calculate the index of refraction of the liquid. (b) Calculate the speed of light in the liquid. (c) Calculate the angle in the air when the beam meets the surface at 30.0°. (d) Describe what happens when the beam meets the surface at 55.0°.

Worked answer

(a) θ₂ = 90°, so 43.0° is the critical angle. sin θc = n_air/n, so n = 1.00 ÷ sin 43.0° = 1.47.
(b) v = c/n = (3.00 × 10⁸ m/s) ÷ 1.466 = 2.05 × 10⁸ m/s.
(c) 1.466 × sin 30.0° = 1.00 × sin θ₂, so θ₂ = 47.2°, bent away from the normal.
(d) 55.0° is beyond the critical angle: total internal reflection. All the light reflects back into the liquid at 55.0°.

If you missed this: read “Total internal reflection” in the Topic 13.3 study guide.

Question 11 (short answer · 13.4)

A diverging lens has a focal length of magnitude F, so f = −F. A real object is a distance sₒ from it.

(a) Starting from the thin-lens equation, show that the image distance is sᵢ = −sₒF/(sₒ + F). (b) Use your result to explain why the image is always virtual and always reduced. (c) Evaluate sᵢ and the image height for a 1.8 cm object with sₒ = 24 cm and F = 12 cm.

Worked answer

(a) 1/sᵢ = 1/f − 1/sₒ = −1/F − 1/sₒ = −(sₒ + F)/(sₒF). Inverting, sᵢ = −sₒF/(sₒ + F).
(b) sₒ and F are positive, so sᵢ < 0 always: virtual. |M| = |sᵢ|/sₒ = F/(sₒ + F) < 1 always: reduced.
(c) sᵢ = −(24)(12)/(36) = −8.0 cm (8.0 cm from the lens, on the object’s side). |M| = 12/36 = 1/3, so hᵢ = 0.60 cm, upright.

If you missed this: read “The thin-lens equation and sign conventions” in the Topic 13.4 study guide.

Your next step

Count a short answer as wrong if any part is missing, and any guess as wrong.

Topic	Question(s)	If you missed it, read
13.1 Reflection	1, 2	Topic 13.1 study guide
13.2 Images formed by mirrors	3, 4, 9	Topic 13.2 study guide
13.3 Refraction	5, 6, 10	Topic 13.3 study guide
13.4 Images formed by lenses	7, 8, 11	Topic 13.4 study guide

How to use your result

Start with the topic where you missed most. Read its study guide, then do its practice questions.
Look at why you missed it. An angle from the surface, or a dropped minus sign on f or sᵢ, needs a habit, not a re-read. Mirrors and lenses share one equation, so fixing a sign slip fixes both.
Missed nothing? Go to the Unit 13 mixed review, which combines topics.
Retake this check after a week of revision.