Total Internal Reflection, Optical Fibres and Dispersion · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Trapped light and spread colours

Science · Waves, Light & Sound · ages 19-20
Name ______________________   Date ____________

  1. Light incident from air of index 1.00 into water of index 1.33 can undergo total internal reflection at that boundary.

    Circle one:   True   False

  2. At an ideal transparent boundary, which conditions give total internal reflection? Measure incidence from the normal.

    • Lower to higher index, above the critical angle of the reverse direction
    • Higher to lower index, above the critical incidence angle
    • Higher to lower index, below the critical incidence angle

  3. Light reaches a boundary from index n1 toward a lower index n2. Which equation gives its critical angle θc?

    • sin θc = n1 ÷ n2
    • sin θc = n2 ÷ n1
    • sin θc = n1 - n2

  4. Why can a transparent glass prism separate white light into colours?

    • The glass absorbs each colour and emits it at a new angle
    • The glass changes each colour into a new source frequency
    • Index varies with wavelength, so colours refract differently

  5. Light reaches an ideal boundary from a material of index 2.00 toward air of index 1.00. Calculate the critical angle in degrees.

    Answer: ______________

  6. In the ray model of a step-index fibre, what does lower-index cladding allow at the core boundary?

    • Total internal reflection for suitable incident rays
    • Total internal reflection for every possible launch direction
    • Complete removal of absorption from the core material

  7. At an ideal higher-to-lower index boundary, the critical angle is 40 degrees. A ray arrives at 50 degrees from the surface. What is its limiting refracted angle from the normal, in degrees?

    Answer: ______________

  8. Parts of each pulse follow guided paths taking 10 ns and 14 ns in a multimode fibre. Any resulting limit on data rate must be caused by leakage, rather than by pulse broadening that makes neighbouring signals overlap.

    Circle one:   True   False

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Answer key

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Trapped light and spread colours W1-mt_ZNSZMLkOD_-s1

  1. False · The incident index is lower. This direction does not satisfy the required higher-to-lower index condition.
  2. Higher to lower index, above the critical incidence angle · Both conditions are needed: the incident material has higher index, and incidence exceeds its critical angle.
  3. sin θc = n2 ÷ n1 · At critical incidence, θ2 = 90 degrees. Snell's law becomes n1 sin θc = n2, giving sin θc = n2 ÷ n1.
  4. Index varies with wavelength, so colours refract differently · Different wavelengths have different refractive indices in the glass. Snell's law therefore gives different refracted directions.
  5. 30 · sin θc = 1.00 ÷ 2.00 = 0.5. In degree mode, arcsin(0.5) = 30 degrees.
  6. Total internal reflection for suitable incident rays · The cladding provides the required lower index. A ray must also meet the core boundary above the critical angle.
  7. 90 · The incidence angle from the normal is 90 - 50 = 40 degrees, exactly θc. Its limiting refracted angle is therefore 90 degrees.
  8. False · Different guided travel times can broaden pulses until neighbouring signals overlap. This makes the signals harder to separate and can limit data rate without requiring leakage.
Worksheet · LightMySky