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6.1.3 Inverse square law and light intensity

Light Intensity Falls With Distance

Definition

Inverse square law

The relationship in which light intensity is inversely proportional to the square of the distance from the light source.

  1. Light from a lamp spreads over a larger area as distance increases, so the energy received per unit area decreases.
  2. The inverse square law states that light intensity is inversely proportional to the square of the distance from the source.
  3. The relationship is ** I∝1d2I \propto \dfrac{1}{d^2}I∝d21​ **, where III is light intensity and ddd is distance.
  4. For relative light intensity, use ** I=1d2I = \dfrac{1}{d^2}I=d21​ ** when no constant of proportionality is supplied.
Common Mistake

Distance and light intensity are not directly proportional, so doubling distance does not halve intensity.

Distance Changes Intensity by a Square

  1. If distance is multiplied by a scale factor nnn, light intensity is divided by n2n^2n2.
  2. Doubling distance gives ** 122=14\dfrac{1}{2^2} = \dfrac{1}{4}221​=41​ ** of the original intensity.
  3. Tripling distance gives ** 132=19\dfrac{1}{3^2} = \dfrac{1}{9}321​=91​ ** of the original intensity.
  4. Halving distance gives ** 1(1/2)2=4\dfrac{1}{(1/2)^2} = 4(1/2)21​=4 ** times the original intensity.
Example
  • A lamp is moved from 10 cm10\,\mathrm{cm}10cm to 30 cm30\,\mathrm{cm}30cm, so distance is multiplied by 333.
  • The new relative intensity is 132=19\dfrac{1}{3^2} = \dfrac{1}{9}321​=91​ of the original intensity.

Calculate Relative Light Intensity

  1. Convert all distances to the same unit before squaring them.
  2. At d=0.20 md = 0.20\,\mathrm{m}d=0.20m, ** I=1(0.20)2=25 m−2I = \dfrac{1}{(0.20)^2} = 25\,\mathrm{m^{-2}}I=(0.20)21​=25m−2 ** as a relative value.
  3. At d=0.40 md = 0.40\,\mathrm{m}d=0.40m, ** I=1(0.40)2=6.25 m−2I = \dfrac{1}{(0.40)^2} = 6.25\,\mathrm{m^{-2}}I=(0.40)21​=6.25m−2 **.
  4. The second value is one quarter of the first because the distance doubled.
Example
  • A lamp is 25 cm25\,\mathrm{cm}25cm from a plant, so the relative intensity is I=1252=1625=0.0016 cm−2I = \dfrac{1}{25^2} = \dfrac{1}{625} = 0.0016\,\mathrm{cm^{-2}}I=2521​=6251​=0.0016cm−2.
  • At 50 cm50\,\mathrm{cm}50cm, I=1502=0.0004 cm−2I = \dfrac{1}{50^2} = 0.0004\,\mathrm{cm^{-2}}I=5021​=0.0004cm−2, which is one quarter as large.

Link Intensity to Photosynthesis

  1. When light is the limiting factor, the rate of photosynthesis is directly proportional to light intensity.
  2. Moving a lamp closer increases light intensity, so chlorophyll absorbs more energy and the rate increases.
  3. The rate does not keep increasing indefinitely because carbon dioxide concentration or temperature can become limiting.
  4. A graph of rate against 1/d21/d^21/d2 tests the relationship more directly than a graph of rate against ddd.
Exam technique

Show the substitution, square the distance before taking the reciprocal, and state the final relationship to the rate of photosynthesis.

Check the Calculation

  1. A larger distance must give a smaller relative intensity, so use this pattern to check your answer.
  2. Keep the same distance unit throughout a comparison because changing units changes the numerical relative intensity.
Self review
  • State the inverse square relationship.
  • What happens to intensity when distance is tripled?
  • Calculate relative intensity at d=5 cmd = 5\,\mathrm{cm}d=5cm.
  • Why can photosynthesis stop rising even when light intensity increases?
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Light from a point lamp spreading across detector screens at distances d, 2d, and 3d, with illuminated areas A, 4A, and 9A and intensities I, I/4, and I/9

The diagram shows how light from a point lamp spreads across larger areas as the distance increases. At distances ddd, 2d2d2d, and 3d3d3d, the illuminated areas are AAA, 4A4A4A, and 9A9A9A respectively.

Light from a lamp spreads over a larger area as it travels away from the source. The same amount of energy is therefore distributed over more area, so the light intensity decreases.

For a point source, the area over which the light spreads is proportional to the square of the distance. This produces the inverse square relationship:

I∝1d2 I \propto \frac{1}{d^2} I∝d21​

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What happens to light intensity as distance from a lamp increases?

6.1.3 Inverse square law and light intensity Revision Guide

  1. GCSE
  2. /Biology
  3. /6.1.3 Inverse square law and light intensity

Revision notes for Edexcel GCSE Biology 6.1.3 Inverse square law and light intensity: explanations and worked examples.

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