Why is the angle of incidence measured from the normal?

The normal is perpendicular to the surface, so any incoming direction is described consistently.

When can a critical angle occur?

Only when light goes from a denser to a less dense medium at a large enough angle.

How do real and virtual images differ?

A real image is where rays actually meet and can be caught on a screen; a virtual one is only where extensions meet.

S4 Physics: Optics Practice Paper 3

2026-11-01

Tutor Prep Guide · Secondary Physics

Paper three opens with reflection, refractive index and mirror distances, moves to mirror angles, Snell law, critical angle and lens images, then reverses for index and magnification, ending with the unchanged-angle slip.

Quick answer　40 degrees, 1.5 and 2 x 10^8 m/s, 5 m, 65 degrees, 35.2 degrees, 45.6 degrees, 30 cm and 0.5, 1.31 and 2.30 x 10^8 m/s, 30 cm and 1.5, 22.1 degrees.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10 questions
Suggested time	25 minutes
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Physics: Optics – Tutor Lesson Plan
Level	S4

02　Questions (10 items)

FoundationA ray hits a plane mirror at 40 degrees to the normal. Find the angle of reflection.
FoundationLight travels at 1/1.5 of its vacuum speed in a medium. Find the refractive index and the speed.
FoundationA person stands 2.5 m in front of a plane mirror. Find the distance to the image.
FoundationA ray meets a plane mirror at 25 degrees to the normal. Find the angle between the reflected ray and the mirror.
IntermediateLight goes from air into water (n = 1.33) at 50 degrees. Find the angle of refraction.
IntermediateA medium has refractive index 1.4. Find the critical angle into air.
IntermediateA convex lens of focal length 20 cm has an object 60 cm away. Find the image distance and magnification.
ChallengeA medium has a critical angle of 50 degrees. Find its refractive index and the speed of light inside.
ChallengeA convex lens of focal length 12 cm has an object 20 cm away. Find the image distance and magnification.
Error redoLight enters water (n = 1.33) from air at 30 degrees. Find the refraction angle. A common mistake answers 30; why is that wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	40 degrees (reflection equals incidence)	Both angles are measured from the normal.
2	n = 1.5, speed 2 x 10^8 m/s	n is the ratio of vacuum speed to the speed in the medium.
3	5 m (image distance equals object distance)	The image sits as far behind the mirror.
4	65 degrees (90 – 25)	The incidence angle is from the normal, so subtract from 90 for the mirror.
5	35.2 degrees (sin r = sin 50 / 1.33)	Snell law: n1 sin theta1 = n2 sin theta2.
6	45.6 degrees (sin c = 1 / 1.4)	At the critical angle the refracted ray runs along the surface.
7	Image at 30 cm, magnification 0.5 (inverted, diminished, real)	Use 1/f = 1/u + 1/v, then v/u for magnification.
8	n about 1.31, speed about 2.30 x 10^8 m/s	Invert sin c = 1/n, then divide the vacuum speed.
9	Image at 30 cm, magnification 1.5 (inverted, enlarged, real)	With the object between f and 2f the image is beyond 2f and enlarged.
10	22.1 degrees (sin r = sin 30 / 1.33)	Entering a denser medium bends towards the normal, so the angle must shrink.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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