Free Pearson Edexcel International A Level Physics lesson

Refraction, total internal reflection and polarisation

Use the normal, the medium and the permitted direction of vibration.

What you’ll learn

  • Use Snell’s law with angles measured from the normal.
  • Identify both conditions for total internal reflection.
  • Measure refractive index and explain plane polarisation.

Measure angles from the normal

A wave changes direction at an oblique boundary when different parts of its wavefront enter the new medium at different times and therefore advance at different speeds. The normal is perpendicular to the boundary, and incidence and refraction angles are measured from it. For light the refractive index is n = c/v, so a higher index means lower speed. The frequency remains fixed by the source at a stationary boundary while wavelength changes.

Snell’s law connects the two media as n₁ sin θ₁ = n₂ sin θ₂. Entering a higher-index medium gives a smaller refracted angle and bends the ray toward the normal. At normal incidence the ray does not change direction even though speed and wavelength change. Do not mistake an unchanged direction for unchanged speed, and do not use angles measured from the surface without first converting them to angles from the normal.

For a transparent block with parallel entry and exit faces, the emerging ray can be parallel to the incident ray but shifted sideways. Refraction has still occurred at both faces. In a calculation, label the incident medium and take the inverse sine only after finding a physically meaningful sine value. A requested value greater than one signals that no propagating refracted ray at that angle exists under the assumed model, leading to the critical-angle analysis.

n = c/v; n₁ sin θ₁ = n₂ sin θ₂; source frequency unchanged at a stationary boundary

Symbols and units

  • n₁ and n₂ are dimensionless refractive indices; θ₁ and θ₂ are angles from the normal.
  • c is vacuum light speed and v is medium light speed, both in m s⁻¹.

How the equation explains the idea

  1. Index n = c/v measures the reduction in speed. Snell’s law gives sin θ₂ = (n₁/n₂) sin θ₁.
  2. If n₂ > n₁, the sine and angle decrease, so the ray bends towards the normal. At normal incidence both angles are zero; the speed can still change.
A sine plot separates refraction from angle scale

Conditions. Air-to-glass refraction with nair = 1.00 and nglass = 1.50, angles measured from the normal. sin r is limited to 2/3 so sin i never exceeds 1.

sin r / dimensionless; sin i / dimensionless.

The slope is 1.50 = nglass/nair. For sin r = 0.4, sin i = 0.6; r ≈ 23.6° and i ≈ 36.9°. The angle ratio i/r is not generally 1.50.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

From air into clear plastic

  1. Light enters plastic of index 1.50 from air at 30.0° to the normal.
  2. sin θ₂ = sin 30.0°/1.50 = 1/3.
  3. θ₂ = 19.5°, so the ray bends toward the normal.

The refracted angle is 19.5° from the normal, not from the surface.

Keep this idea. Label both media and the normal before substituting into Snell’s law.

Find the last transmitted ray

When light travels from a higher refractive index to a lower one, the refracted angle exceeds the incidence angle. Increasing incidence eventually makes the transmitted ray graze the boundary at 90° to the normal. The corresponding incidence is the critical angle. Snell’s law then gives sin C = nlow/nhigh. The common expression sin C = 1/n assumes the second medium is approximately air or vacuum.

At incidence greater than C there is no propagating refracted ray in the simple geometrical model, and total internal reflection occurs. At exactly C, the limiting refracted ray lies along the boundary; distinguish this threshold from incidence above it. Incidence from a lower-index medium does not give total internal reflection however large the angle becomes, because the necessary index ordering is absent.

An optical fibre uses a higher-index core surrounded by lower-index cladding so suitable rays remain guided after repeated reflections. Bending the fibre changes the incidence geometry and can allow some rays to escape. Total internal reflection is not a promise of zero loss in a real fibre: absorption, scattering and imperfect geometry can still reduce transmitted power. Medical viewing and communication use this guidance while balancing bend radius, wavelength and coupling into the fibre.

sin C = nlow/nhigh; for higher index to air: sin C = 1/n; TIR requires θincident > C

Symbols and units

  • C is critical incidence angle from the normal; nhigh and nlow are dimensionless indices with nhigh > nlow.

How the equation explains the idea

  1. At the critical angle the transmitted angle is 90°, so Snell’s law gives nhigh sin C = nlow × 1.
  2. Therefore C = arcsin(nlow/nhigh). For glass 1.50 to air 1.00, C ≈ 41.8°. Total internal reflection requires incidence greater than C and travel from higher to lower index.

Worked example

A core and its cladding

  1. A fibre core has index 1.50 and its cladding has index 1.40.
  2. C = sin⁻¹(1.40/1.50) ≈ 69.0°.
  3. A ray reaching the core–cladding interface at 75° to the normal satisfies 75° > C and is totally internally reflected.

Use the two-material index ratio, not 1/1.50, for this cladding.

Keep this idea. Total internal reflection needs the correct index ordering and an incidence angle above the pair’s critical angle.

Measure index and identify a polarisation plane

To measure the refractive index of a solid, trace a transparent block and send a narrow ray through a flat face. Mark the incident and emergent paths with well-separated points, remove the block, join the points and draw a normal at the entry position. Measure several incidence and refraction angles. A graph of sin i against sin r has slope nblock/nair, allowing a best-fit estimate instead of relying on a single angle.

Keep the block aligned with its outline and make rays narrow enough to identify their centres. Small angles produce small separations and large fractional angular uncertainty; grazing rays can become difficult to trace. Repeat sensible intermediate angles, inspect residual scatter and avoid forcing an intercept without justification. A ray box is often suitable; if a laser is used, keep it below eye level, terminate its beam and avoid reflective objects or direct viewing.

Plane-polarised light has its electric-field oscillations restricted to one plane containing propagation. Unpolarised light contains changing transverse directions. One polariser selects a direction; a second analyser rotated through 90° can ideally reduce the transmitted light to a minimum. Longitudinal waves have no choice of sideways oscillation direction and therefore cannot be polarised in this way. Polarising filters help reduce some reflected glare, but the observation establishes transverse behaviour, not a complete description of every property of light.

Graph sin i (vertical) against sin r (horizontal): gradient = nblock/nair; crossed polariser axes give minimum transmission.

Symbols and units

  • i and r are incidence and refraction angles from the normal; their sines and refractive indices are dimensionless.

How the equation explains the idea

  1. For air to a block, nair sin i = nblock sin r. Rearranging gives sin i = (nblock/nair) sin r.
  2. On vertical sin i versus horizontal sin r axes, gradient is nblock/nair. Reversing the axes gives the reciprocal. Polariser transmission is a separate observation of transverse oscillation.

Worked example

Find index from a graph

  1. Two widely separated points on a fitted line are (sin r, sin i) = (0.20, 0.30) and (0.50, 0.75).
  2. Gradient = (0.75 − 0.30)/(0.50 − 0.20) = 1.50.
  3. With air index approximately one, the solid index is 1.50.

Use the fitted line’s gradient and the stated axis order.

Keep this idea. A sine graph gives index, while crossed polarising filters test the allowed transverse directions.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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