Explain normal-incidence light transmission
Keep direction unchanged while recognising changes of speed and wavelength across a transparent boundary.
Exam scope and assessment notes
Core and Extended · Physics 0972, 2026–2028. Follow the selected Light section; ray diagrams use an ideal thin-lens or plane-boundary model. Written work is self-checked. Human teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
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.
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.
Pause and explain
A normally transmitted ray enters a slower region. What can change?
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
- Both incidence and refraction angles are 0°, so there is no bending.
- Initially λ = v/f = 6.0 × 10⁻⁷ m.
- After entry λ = 4.0 × 10⁻⁷ m at unchanged frequency.
Answer Straight transmission; wavelength 600 nm → 400 nm, with unchanged frequency.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →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.
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
- The approaching ray lies along the normal.
- i = r = 0°.
- 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.
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
- The frequency is unchanged.
- Wavelength is proportional to speed for that fixed frequency.
- 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.
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
- Normal transmission establishes direction.
- Some reflection can coexist at the boundary.
- 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.
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.
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Make it stick
Explain straight transmission with changed speed and wavelength but unchanged stationary-boundary frequency.
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