This higher tier covers Higher The Inverse Square Law and Light Intensity within Photosynthesis for GCSE Biology. Topic 1: Photosynthesis It is section 11 of 14 in this topic. This section is most useful once the core foundation idea is secure, because it adds the detail that pushes answers higher.
Higher The Inverse Square Law and Light Intensity
Light intensity decreases with distance according to the inverse square law:
Intensity ∝ 1 / distance²
This means if you double the distance from the light source, the intensity falls to one quarter (not one half). If you treble the distance, intensity falls to one ninth.
Example calculation: A lamp produces an intensity of 400 arbitrary units at 10 cm. What is the intensity at 20 cm?
- Ratio of distances: 20/10 = 2
- Ratio of intensities: 1/2² = 1/4
- Intensity at 20 cm = 400 × 1/4 = 100 units
This is important for interpreting Elodea bubble-counting experiments. A graph of rate of photosynthesis against 1/d² (not against d itself) should give a straight line through the origin when light is the limiting factor, confirming the inverse square relationship.
Graph analysis at higher tier: On a graph of photosynthesis rate vs light intensity, the rate initially increases linearly. The gradient then decreases and the curve levels off — this plateau indicates that another factor (CO2 or temperature) has become limiting. The plateau itself shifts up when CO2 is increased or temperature is raised (up to the enzyme optimum), confirming which factor was limiting.
Worked method: calculating light intensity with 1/d²
Turn the inverse-square law into marks: real numbers, ratios and graph reading.
- Why you plot against 1/d², not distanceRate of photosynthesis is proportional to light intensity, and light intensity is proportional to 1/d². Plotting rate against 1/d² (not raw distance) gives a straight line through the origin while light is the limiting factor — that straight line is the evidence examiners want in your conclusion.
- Calculating 1/d² for your tableFor each distance in cm, square it, then find the reciprocal. Example: distance = 10 cm, so 1/d² = 1/10² = 1/100 = 0.01. Keep every distance in the same unit and give answers to at least 2 significant figures.
- Comparing two distances with the ratio methodTo find how many times brighter the light gets, square the ratio of the old distance to the new distance: I2 = I1 × (d1/d2)². Halving the distance (ratio = 2) means the intensity increases by 2² = ×4, not ×2.
- Linking your graph to limiting factorsOn a rate vs 1/d² graph, the straight portion near the origin shows light is limiting — more light gives proportionally more oxygen produced. Where the line flattens off, light has stopped being limiting; CO2 concentration or temperature has taken over instead.
- The exam trap examiners flag mostSquaring the distance itself instead of the distance ratio, or plotting rate against distance instead of 1/d², both lose the mark for showing a valid straight-line relationship. Always show the squaring step explicitly in your working.
A lamp is placed 40 cm from pondweed and the light intensity is 800 arbitrary units. The lamp is moved to 20 cm. Calculate the new light intensity.
Distance halves (40 cm → 20 cm), so the ratio d1/d2 = 40/20 = 2. Since intensity ∝ 1/d², I2 = I1 × (d1/d2)² = 800 × 2² = 800 × 4 = 3200 units.
The most commonly lost mark: multiplying by 2 instead of 2² = 4 — forgetting to square the distance ratio.
Practice questions for Photosynthesis
Where does photosynthesis take place in plant cells?
Write the balanced symbol equation for photosynthesis.