S4 Physics — Refraction and Total Internal Reflection (tutor guide)

S4 Physics — Refraction and Total Internal Reflection (tutor guide) | Snell law, critical angle, optical fibre

2026-11-01

Tutor lesson-prep guides · Secondary Physics

Students can quote n = sin i / sin r but freeze when asked when total internal reflection happens. This lesson derives the critical angle from the limit case, so they calculate 41.8° themselves.

Objectives — Students can: (1) apply Snell law to find a refracted angle and state which way the ray bends; (2) derive the critical angle from sin c = 1/n and decide whether total internal reflection occurs; (3) use v = c/n and the wavelength change to explain optical fibres and objects seen in water.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Sine and inverse sine	What is sin 30°?	0.5
Frequency is fixed	Does frequency change when light enters water?	No; speed and wavelength change
Dense and less dense	Bending towards the normal happens going which way?	Less dense into denser

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Hook	Show a straw that looks bent in water; ask for a ray diagram	Draw ray and normal
8-22 min	Snell law	Write n1 sin i = n2 sin r and substitute 45°	Obtain r = 28.1°
22-32 min	Critical angle	Push r to 90° and invert to sin c = 1/n	Obtain c = 41.8°
32-42 min	TIR conditions	State both conditions; test 45° inside glass	Decide if TIR occurs
42-54 min	Applications	Fibres, endoscope, v = c/n = 2.0 x 10^8 m/s	Explain shorter wavelength
54-60 min	Exit check	Collect and mark three questions	Work independently

03 Teaching script (read it out)

Scene 1: Which way does the ray bend

Light goes from air (n = 1.00) into glass (n = 1.50) at 45°. sin r = sin 45° / 1.50 = 0.7071 / 1.50 = 0.4714, so r = 28.1°.
One sentence only: less dense into denser bends towards the normal, denser into less dense bends away. Frequency never changes; speed and wavelength do.

Task: change the angle to 30° and say whether r grows or shrinks.

Scene 2: The critical angle is just the limit of r

Increase the angle of incidence and the refracted ray eventually runs along the surface. Set r = 90°, so sin r = 1, and Snell law becomes n1 sin c = n2.
Glass to air: sin c = 1 / 1.50 = 0.6667, c = 41.8°. Water to air: sin c = 1 / 1.33, c = 48.8°. The smaller n, the larger the critical angle.

Ask: why does 45° inside glass reflect totally here, and where is the boundary?

Scene 3: Speed, wavelength and the fibre

Speed in the medium v = c / n = 3.00 x 10^8 / 1.50 = 2.0 x 10^8 m/s. Frequency is unchanged, so wavelength = 600 nm / 1.50 = 400 nm.
An optical fibre works by total internal reflection: light hits the core boundary above the critical angle and stays inside, so it can turn corners without leaking. Endoscopes use the same idea.

Reminder: TIR needs both conditions, denser to less dense and angle above critical. Otherwise it is plain refraction.

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Refracted angle	sin r = n1 sin i / n2	28.1°
Calculation	Critical angle	sin c = 1/n	41.8°
Decision	Does TIR occur	denser to less dense and i > c	Yes at 45° in glass
Explanation	Speed and wavelength	v = c/n, f fixed	2.0 x 10^8 m/s, 400 nm

Sanity check (teach this): The refracted angle is always smaller than the angle of incidence going into a denser medium. If r comes out larger than i, the refractive indices were swapped. A critical angle must lie between 30° and 45° for n = 1.50.

05 Independent practice (3 questions)

Question	Answer	What to watch
Light enters glass of n = 1.60 at 30°, find r	r = 18.2°	Divide sin i by n first
Critical angle of n = 1.60 glass against air	c = 38.7°	sin c = 1/n
Speed and critical angle in a medium of n = 2.00	1.5 x 10^8 m/s and 30°	v = c/n then sin c = 1/n

06 Common mistakes and what to say

Mistake	Student writes	Say this
Swapping n1 and n2	Writing 1.50 sin 45° = 1 x sin r	Put each n with the sine of its own medium.
Assuming the critical angle is 45°	Answering that anything above 45° reflects	It is inverse sine of 1/n, which is 41.8° here.
Thinking frequency changes	Saying frequency drops inside water	Frequency is set by the source; speed and wavelength change.
Confusing TIR with mirror reflection	Drawing TIR like a mirror	In TIR no light leaves the medium at all.

07 Homework (paper, section and question numbers)

Secondary Physics chapter index — All chapters
DSE physics practice paper — Optics section

08 Exit check (last 5 minutes)

Question	Answer	Decision
45° into n = 1.50, find r	28.1°	Three correct means move on to lenses
Critical angle of n = 1.50 against air	41.8°	One wrong means re-derive sin c next lesson
Speed of light in n = 2.00	1.5 x 10^8 m/s	Two wrong means repeat act three

09 Tutor FAQ

Must students memorise critical angles?

No. Remember sin c = 1/n and use the calculator; know that larger n gives a smaller c.

Why does a fibre not leak light?

The angle at the core boundary stays above the critical angle, so total internal reflection traps the light.

Can the refracted angle exceed 90°?

No. A sine above 1 means total internal reflection and no refracted ray exists.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.