How should Snell law be remembered?

n1 sin i = n2 sin r with each n bound to its own medium, and both angles from the normal.

Can a refractive index be below 1?

No; entering any medium from air gives n above 1, so a value below 1 means the formula was inverted.

S4 Physics refraction paper 1: refractive index and Snell law

2026-11-10

Tutor Prep Guide · S4 Physics Refraction and Total Internal Reflection

Ten items start from n = c / v and n1 sin i = n2 sin r, with angles measured from the normal.

Direct answers　Items 01 to 03 cover basic refraction and speed, 04 to 06 reverse work, wavelength and 45 degrees, 07 to 09 relative index and denser to less dense, and 10 the inverted sine.

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 30 degrees. Find the angle of refraction.
FoundationLight travels at 2.0 x 10^8 m/s in a medium. Find its refractive index.
FoundationLight in water (n = 1.33) meets the surface at 40 degrees. Find the angle in air.
IntermediateA ray in glass (n = 1.5) leaves at 15 degrees from the normal. Find the angle of incidence.
IntermediateRed light of 600 nm enters glass of n = 1.5. What is its wavelength, and does the frequency change?
IntermediateLight hits glass (n = 1.5) at 45 degrees from air. Find the refraction angle.
ChallengeA ray goes from medium A to B with 50 degrees incidence and 30 degrees refraction. Find the relative refractive index.
ChallengeLight goes from glass (n = 1.5) into water (n = 1.33) at 35 degrees. Find the refraction angle.
ChallengeLight travels at 2.25 x 10^8 m/s in water and 3.0 x 10^8 m/s in vacuum. Find n for water and say whether the frequency changes.
Error redoA student computes sin r / sin i and gets 0.667 for 60 degrees in and 35.26 degrees refracted. What is the correct value?

03　Answers and one-line notes

#	Answer	One-line note
1	19.5 degrees	1.00 x sin30 = 1.5 x sin r, so sin r = 0.5 / 1.5 = 0.333 and r = 19.5 degrees, bent towards the normal.
2	1.5	n = c / v = 3.0 x 10^8 / 2.0 x 10^8 = 1.5, and refractive index has no unit.
3	28.9 degrees	sin r = 1.33 x sin40 / 1.00 = 0.855, so r = 28.9 degrees, bent away from the normal.
4	22.8 degrees	sin i = 1.5 x sin15 = 0.388, so i = 22.8 degrees; the factor multiplies the less dense side.
5	400 nm; the frequency stays the same	lambda = 600 / 1.5 = 400 nm; frequency is set by the source, so the speed drops and the wavelength shortens.
6	28.1 degrees	sin r = sin45 / 1.5 = 0.471, giving r = 28.1 degrees; larger incidence deviates more.
7	about 1.53	n = sin i / sin r = 0.766 / 0.5 = 1.53, with the incidence side on top.
8	40.3 degrees	1.5 x sin35 = 1.33 x sin r gives sin r = 0.647 and r = 40.3 degrees, larger because it leaves the denser medium.
9	n = 1.33; the frequency is unchanged, while speed and wavelength fall	3.0 / 2.25 = 1.33; frequency is fixed by the source, so speed and wavelength drop together.
10	1.50	Refractive index must exceed 1, so incidence goes on top: sin60 / sin35.26 = 1.50. A value below 1 means it was inverted.

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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