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IGCSE · Cambridge (CIE) · 0654

Use the sine ratio for light refraction

Calculate index or a transmitted angle from normal-referenced incidence and refraction angles.

Refraction of light · Cambridge IGCSE 0654 · P3.2.2

Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.

Before you begin

  • Measure both angles from the local normal.
  • Use sine and inverse sine in degree mode.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which operation belongs in the refraction index equation?

What you will learn

  • Recall and use n = sin i / sin r.
  • Use degree-mode trigonometry and the correct region order.
Original air-to-material ray geometry uses supplied relative index 1.50, incidence 30° and refraction about 19.47°. The speed falls from a supplied 3.0 × 10⁸ to 2.0 × 10⁸ m/s. The transmitted ray bends toward the local normal, with stationary-boundary frequency unchanged. The drawing gives no energy fraction.Air → slower materialSupplied relative index: 1.50i = 30°r ≈ 19.5°Bends toward the normalFrequency stays unchanged
Original air-to-material ray geometry uses supplied relative index 1.50, incidence 30° and refraction about 19.47°. The speed falls from a supplied 3.0 × 10⁸ to 2.0 × 10⁸ m/s. The transmitted ray bends toward the local normal, with stationary-boundary frequency unchanged. The drawing gives no energy fraction.

Use the matching speed and angle ratio

For index of the transmitted region relative to the incident region, n = sin i / sin r. The angles are measured from the same local normal on their respective sides. For air into glass, the usual air-as-reference approximation gives the glass index. A larger than one ratio corresponds to the slower transmitted region and an oblique ray closer to the normal.

For i = 30° and r = 20°, n = sin 30°/sin 20° = 0.5000/0.3420, approximately 1.46. Do not divide the angles 30/20: the relation uses their sines. State the calculator mode before obtaining numerical sine values and keep enough digits until the final answer.

Rearrange for an angle and check the branch

For a supplied relative index n = 1.50 and i = 30°, sin r = sin i/n = 1/3, so r ≈ 19.47°. The acute physical transmitted angle agrees with bending toward the normal. For a reversed region change the corresponding relative index is reciprocal; do not retain an air-to-glass ratio while relabelling the incident side.

Normal incidence gives sin i = sin r = 0, so the direct measured-angle ratio is 0/0 and cannot determine index from that one path. Speed or a suitable oblique observation is needed. If rearrangement for high-to-low transmission requests sin r greater than 1, there is no propagating refracted ray in this school model: test the critical-angle condition instead.

Sources for this lesson

Pause and explain

For n = 1.50 and i = 30°, what is sin r?

Put the idea to work

Worked example

Use n = 1.50 for air-to-glass light at i = 30° to find r.

Show the worked solution
  1. sin r = sin i/n.
  2. sin r = sin 30°/1.50 = 1/3.
  3. In degree mode r = sin⁻¹(1/3) ≈ 19.47°.

Answer r ≈ 19.5° to the normal.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Find n from i = 30° and r = 20°.

Give me a hint

Use sine values rather than the bare degree numbers.

Compare my reasoning
  1. sin 30° = 0.5000 and sin 20° ≈ 0.3420.
  2. n = 0.5000/0.3420.
  3. n ≈ 1.46, dimensionless.

n ≈ 1.46.

Look for these in your work

  • Use degree sines.
  • Retain the reference order.
2 · Independent

Find r for a relative index 1.50 and incidence 45°.

Give me a hint

Divide sin 45° by the index, then use inverse sine.

Compare my reasoning
  1. sin r = sin 45°/1.50 ≈ 0.471405.
  2. r = sin⁻¹(0.471405).
  3. r ≈ 28.13°.

r ≈ 28.1°.

Look for these in your work

  • Rearrange before calculation.
  • Check bending toward the normal.
3 · Transfer

A normally entering ray has i = r = 0°. A student declares n = 1 from those angles. Explain the problem.

Give me a hint

Check the denominator before interpreting a ratio.

Compare my reasoning
  1. Both sines are zero.
  2. The expression 0/0 is not a defined measured ratio.
  3. Use speeds or an oblique path to determine index.

The normal path gives an indeterminate ratio, not n = 1.

Look for these in your work

  • Identify the undefined ratio.
  • Suggest adequate evidence.

Common mistakes

  • Use sines, not i/r.
  • Use degree mode for supplied degree angles.
  • Normal incidence cannot determine n from a 0/0 measured-angle ratio.
Recall without your notes

What can you explain now?

Can i = r = 0° alone determine n by dividing their sines?

Compare with the explanation

No.

The ratio is indeterminate; it is not evidence for n = 1.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Calculate with the sine ratio, explain its reference and detect a normal-incidence limitation.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Sciences — Co-ordinated (Double)

Course scope and assessment

This 2028–2029 pathway maps all 152 detailed subject-content sections in the published 0654 syllabus to original teaching and staged practice. It adapts teaching from the related single-science pathways and excludes clear 0610-only topics such as classification systems, human excretion, tropic responses, food supply, population dynamics, detailed pollution and biotechnology. Some explanations may include wider single-science context. A 0654-specific objective-by-objective review, Core/Supplement audit, human subject-teacher review, supervised laboratory experience and cumulative assessment remain in progress. Written practice is self-checked and does not predict an official grade. The 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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