When does a critical angle exist?

Only for light going from a denser to a less dense medium; the reverse always has a refracted ray.

What is total internal reflection used for?

Optical fibres, prism reflectors, endoscopes and diamond sparkle, with less loss than silvered mirrors.

S4 Physics refraction paper 3: multi-layer media and dispersion

2026-11-10

Tutor Prep Guide · S4 Physics Refraction and Total Internal Reflection

Ten items apply Snell law to direct comparison of media, reverse critical angle, fibre timing and dispersion.

Direct answers　Items 01 to 03 cover large angles and layered media, 04 to 06 reverse critical angle and depth, 07 to 09 fibre timing, frequency and dispersion, and 10 the 90 degree ray myth.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter	Chapter: S4 Physics Refraction and Total Internal Reflection
Level	S4

02　Questions (10 items)

FoundationLight enters glass (n = 1.5) from air at 60 degrees. Find the angle of refraction.
FoundationA medium refracts a 60 degree ray to 35 degrees. Find its refractive index.
IntermediateLight goes from water (n = 1.33) into glass (n = 1.5) at 30 degrees. Find the refraction angle.
IntermediateA medium has a critical angle of 30 degrees. Find its refractive index and the speed of light in it.
IntermediateLight travels at 1.5 x 10^8 m/s in a medium. Find the critical angle.
FoundationA pool 1.2 m deep looks 0.9 m deep from directly above. Find n for the water.
ChallengeA 1 km fibre has core n = 1.5. How long does light take along the fibre?
ChallengeLight of 600 nm in vacuum and 5.0 x 10^14 Hz enters a medium of n = 1.5. Find the frequency, speed and wavelength.
ChallengeRed (n = 1.51) and violet (n = 1.53) light both hit glass at 50 degrees. How far apart are the refracted angles?
Error redoA student says that under total internal reflection the refracted ray leaves along the surface at 90 degrees. What is correct?

03　Answers and one-line notes

#	Answer	One-line note
1	35.3 degrees	sin r = sin60 / 1.5 = 0.577, so r = 35.3 degrees and grows more slowly than incidence.
2	about 1.51	n = sin60 / sin35 = 0.866 / 0.574 = 1.51, with both angles measured from the normal.
3	26.3 degrees	1.33 x sin30 = 1.5 x sin r gives sin r = 0.443 and r = 26.3 degrees; the two media can be compared directly.
4	n = 2.0; speed 1.5 x 10^8 m/s	n = 1 / sin30 = 2.0 and v = 3.0 x 10^8 / 2.0 = 1.5 x 10^8 m/s; the smaller the critical angle the larger n.
5	30 degrees	n = 3.0 / 1.5 = 2.0, then sin C = 1 / 2.0 = 0.5, so C = 30 degrees.
6	about 1.33	n = real / apparent = 1.2 / 0.9 = 1.33, valid for viewing straight down.
7	5.0 microseconds	v = 2.0 x 10^8 m/s, so t = 1000 / 2.0e8 = 5.0 x 10^-6 s, that is 5.0 microseconds.
8	5.0 x 10^14 Hz; 2.0 x 10^8 m/s; 400 nm	Frequency is unchanged, v = 3.0 x 10^8 / 1.5 = 2.0 x 10^8 m/s, and lambda = v / f = 400 nm, matching lambda0 / n.
9	about 0.4 degrees (red 30.5, violet 30.0)	sin r = sin50 / n gives 30.49 degrees for red and 30.05 for violet, a gap of 0.4 degrees that separates colours in a prism.
10	There is no refracted ray; all the incident energy is reflected back	The 90 degree ray is only the limit exactly at the critical angle; beyond it the refracted ray vanishes and all energy is reflected.

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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