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IGCSE · Cambridge (CIE) · 0653

Explain normal-incidence light transmission

Keep direction unchanged while recognising changes of speed and wavelength across a transparent boundary.

Section P3.2.2 · Refraction of light

Refraction of light · Cambridge IGCSE Combined Science 0653 · P3.2.2

Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.

Before you begin

  • Identify a ray along the normal.
  • Use v = fλ for a regular wave.

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

Quick readiness check

What is the incidence angle for a ray along a boundary normal?

What you will learn

  • Describe transmission at normal incidence.
  • Distinguish speed, frequency and wavelength changes.
Original normal-incidence comparison keeps direction and frequency unchanged while supplied light speed falls from 3.0 × 10⁸ to 2.0 × 10⁸ m/s. At 5.0 × 10¹⁴ Hz the wavelengths are 600 and 400 nm. Straight transmission does not establish a 100% transmitted-energy fraction.Normal transmission · no bendingBefore: 600 nm · 3.0 × 10⁸ m/sAfter: 400 nm · 2.0 × 10⁸ m/sSame frequency: 5.0 × 10¹⁴ Hzi = r = 0°
Original normal-incidence comparison keeps direction and frequency unchanged while supplied light speed falls from 3.0 × 10⁸ to 2.0 × 10⁸ m/s. At 5.0 × 10¹⁴ Hz the wavelengths are 600 and 400 nm. Straight transmission does not establish a 100% transmitted-energy fraction.

No bending does not mean no speed change

A ray incident along the normal on a transparent boundary can pass straight through without changing direction. Its incidence and refraction angles are both 0°. Entering a slower region still changes its propagation speed. A direction-only sketch therefore cannot establish equal speeds in the two regions.

At a stationary boundary the frequency of transmitted light remains the same as the source frequency. With v = fλ, a lower speed at unchanged frequency gives a shorter wavelength. On returning normally to the original region, the ray remains straight and its wavelength returns to the original value under the stated conditions.

Retain the boundary conditions

For a supplied light wave with frequency 5.0 × 10¹⁴ Hz, speed 3.0 × 10⁸ m/s gives wavelength 6.0 × 10⁻⁷ m. A supplied transmitted speed 2.0 × 10⁸ m/s gives 4.0 × 10⁻⁷ m at that same frequency. These are original supplied values, not a claim of measured dispersion for a particular glass.

Normal transmission may coexist with some reflection. The reflected direction is back along the incident line while the transmitted direction continues forward. Neither 'straight through' nor unchanged frequency means all incident energy must pass; intensity would need separate information.

Sources for this lesson

Pause and explain

A normally transmitted ray enters a slower region. What can change?

Put the idea to work

Worked example

At normal entry a supplied wave speed falls from 3.0 × 10⁸ to 2.0 × 10⁸ m/s with f = 5.0 × 10¹⁴ Hz. Describe direction and wavelengths.

Show the worked solution
  1. Both incidence and refraction angles are 0°, so there is no bending.
  2. Initially λ = v/f = 6.0 × 10⁻⁷ m.
  3. After entry λ = 4.0 × 10⁻⁷ m at unchanged frequency.

Answer Straight transmission; wavelength 600 nm → 400 nm, with unchanged frequency.

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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

Describe i, r and bending for normal transmission from air to glass.

Give me a hint

Measure the ray against the normal, not the surface.

Compare my reasoning
  1. The approaching ray lies along the normal.
  2. i = r = 0°.
  3. The transmitted direction stays straight despite the changed speed.

i = r = 0°; no bending, with a possible speed change.

Look for these in your work

  • Use the correct angles.
  • Separate speed from direction.
2 · Independent

At fixed frequency, transmitted speed is three quarters of incident speed. Compare wavelengths.

Give me a hint

Use λ = v/f with the same f.

Compare my reasoning
  1. The frequency is unchanged.
  2. Wavelength is proportional to speed for that fixed frequency.
  3. The transmitted wavelength is three quarters of the incident value.

Transmitted wavelength = 0.75 × incident wavelength.

Look for these in your work

  • Hold frequency fixed.
  • Compare consistent quantities.
3 · Transfer

A straight transmitted ray is used to claim 100% of the incident light energy passes. Is the claim established?

Give me a hint

The direction sketch omits energy measurements.

Compare my reasoning
  1. Normal transmission establishes direction.
  2. Some reflection can coexist at the boundary.
  3. No supplied intensity ratio proves complete transmission.

No; direction alone does not establish the transmitted energy fraction.

Look for these in your work

  • Identify what is observed.
  • State what extra evidence is missing.

Common mistakes

  • No bending does not establish equal speeds.
  • Frequency stays fixed at the stationary boundary.
  • Straight transmission is not an energy-fraction measurement.
Recall without your notes

What can you explain now?

Does shorter transmitted wavelength require a different source frequency here?

Compare with the explanation

No.

The frequency is unchanged while speed falls.

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

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Make it stick

Explain straight transmission with changed speed and wavelength but unchanged stationary-boundary frequency.

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 Science — Combined

Course scope and assessment

This pathway maps all 121 detailed subject-content sections in the published 0653 syllabus for exams in 2028 and 2029. It adapts original lessons from the related single-science pathways and keeps Combined Science's narrower scope: Biology B1–B16, Chemistry C1–C12 and Physics P1–P5. A 0653-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 separate 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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