Skip to main content
ScotlandPhysicsSyllabus dot point

How does light refract and totally internally reflect, and what produces line spectra?

Refraction and the refractive index, critical angle and total internal reflection, and the emission and absorption line spectra produced by electron energy-level transitions.

An SQA Higher Physics answer on refraction and spectra, covering refraction and the refractive index, the critical angle and total internal reflection, and the emission and absorption line spectra produced by electron energy-level transitions.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this key area is asking
  2. Refraction and refractive index
  3. Critical angle and total internal reflection
  4. Emission and absorption line spectra
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to use the refractive index to relate angles of incidence and refraction, find the critical angle and explain total internal reflection, and explain how emission and absorption line spectra arise from electron transitions between energy levels.

Refraction and refractive index

Going into a denser medium (larger nn), light slows down and bends towards the normal, so the refraction angle is smaller than the incidence angle. The wavelength shortens in step with the speed, but the frequency, and hence the colour, is unchanged.

Critical angle and total internal reflection

When light travels from a denser to a less dense medium, the refracted ray bends away from the normal. At a particular angle of incidence, the critical angle θc\theta_c, the refracted ray travels along the boundary. Beyond it, all the light is reflected back inside.

Emission and absorption line spectra

Electrons in an atom can only occupy certain discrete energy levels. When an electron moves between levels, the energy difference is carried by a photon.

An emission spectrum is bright lines on a dark background (from a hot, low-pressure gas); an absorption spectrum is dark lines on a continuous background (white light passed through a cooler gas, which removes the same wavelengths the gas would emit). These line patterns are used to identify elements in stars and gas clouds.

Examples in context

Optical fibres carry internet and telephone data as pulses of light that bounce along the core by total internal reflection, losing almost no signal even over long distances. Endoscopes use fibre bundles to see inside the body for the same reason. Diamonds sparkle because their very high refractive index gives a small critical angle, so light entering is totally internally reflected many times before it escapes. Astronomers read the absorption lines in starlight to identify the elements in a star and, through the redshift of those lines, its motion. Sodium street lamps glow orange because of the characteristic emission lines of sodium.

Try this

Q1. A medium has a refractive index of 1.331.33. Calculate its critical angle with air. [2 marks]

  • Cue. sin⁡θc=1n=11.33=0.752\sin\theta_c = \frac{1}{n} = \frac{1}{1.33} = 0.752, so θc=48.8∘\theta_c = 48.8^{\circ}.

Q2. State what happens to the speed and wavelength of light as it passes from air into glass. [1 mark]

  • Cue. Both the speed and the wavelength decrease (the frequency stays the same).

Q3. Explain why each element produces a unique line spectrum. [2 marks]

  • Cue. Each element has its own unique set of electron energy levels, giving a unique set of transition energies and so a unique set of spectral line frequencies.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20184 marksA ray of light passes from air into a glass block with a refractive index of 1.50. The angle of incidence in air is 40 degrees. Calculate the angle of refraction in the glass.
Show worked answer →

Use the refractive index relationship between the two media.

Relationship: n=sin⁡θ1sin⁡θ2n = \frac{\sin\theta_1}{\sin\theta_2}, where medium 1 is air (n=1n = 1) and medium 2 is glass.

Substitution: 1.50=sin⁡40∘sin⁡θ21.50 = \frac{\sin 40^{\circ}}{\sin\theta_2}, so sin⁡θ2=sin⁡40∘1.50=0.6431.50=0.428\sin\theta_2 = \frac{\sin 40^{\circ}}{1.50} = \frac{0.643}{1.50} = 0.428.

Answer: θ2=sin⁡−1(0.428)=25.4\theta_2 = \sin^{-1}(0.428) = 25.4 degrees.

Markers reward the correct form of the relationship, the rearrangement, and the smaller refraction angle in the denser medium.

SQA Higher 20223 marksExplain how an emission line spectrum is produced when atoms are excited, and explain why each element produces its own characteristic set of lines.
Show worked answer →

An emission line spectrum is produced when electrons in excited atoms drop from a higher energy level to a lower one. The energy lost in each transition is emitted as a photon, with photon energy equal to the difference between the levels, E2−E1=hfE_2 - E_1 = hf. Each photon frequency (and so colour) appears as a bright line.

Each element has a unique set of energy levels, so it produces a unique set of transition energies and therefore a unique pattern of spectral lines, which acts like a fingerprint to identify the element.

Markers reward electrons dropping between levels emitting photons, the relationship E2−E1=hfE_2 - E_1 = hf, and the unique-energy-levels reason for characteristic lines.

Related dot points

Sources & how we know this