Step twice as far from a lamp and it does not look half as bright, it looks a quarter as bright. Light from a point source spreads over an ever-larger sphere, so its intensity drops with the square of the distance. This inverse square law is one of the most important and most testable relationships in the topic, and a single equation, $I_1 d_1^2 = I_2 d_2^2$, lets you compare the brightness at any two distances.
A single light source radiates equally in all directions into the space around it.
Before reading: the same amount of light energy spreads over a bigger and bigger sphere as you move away. If you double the distance, by what factor do you think the area of that sphere grows, and so what happens to the brightness? Write your prediction.
The same energy spread over an ever-larger sphere
Light from a point source spreads out over the surface of a growing sphere, and because that surface area grows with the square of the distance, the same light energy is shared ever more thinly, so intensity falls as $1/d^2$.
Intensity $I$ is the power landing on each square metre (W m⁻²). A point source sends its light out equally in all directions, so at distance $d$ the light is spread over a sphere of surface area $4\pi d^2$. Since the total power $P$ is fixed, the intensity is $I = P / (4\pi d^2)$, it is inversely proportional to $d^2$. Double the distance and the sphere's area quadruples, so the intensity drops to a quarter; triple the distance and it drops to a ninth.
$$I = \frac{P}{4\pi d^2} \quad\Rightarrow\quad I \propto \frac{1}{d^2}$$
Light from a point source spreads over a sphere of area $4\pi d^2$, so intensity $I = P/(4\pi d^2)$ is inversely proportional to the square of the distance: $I \propto 1/d^2$. Doubling the distance quarters the intensity.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? If you move from $1\ \text{m}$ to $2\ \text{m}$ from a point light source, the intensity becomes:
One equation to compare brightness at two distances
We just saw that $I \propto 1/d^2$. That raises a question: how do we calculate the intensity at a new distance? This card answers it → with the working form $I_1 d_1^2 = I_2 d_2^2$.
Because the product of intensity and distance-squared is the same everywhere around a point source, you can compare any two points with a single equation: $I_1 d_1^2 = I_2 d_2^2$.
Since $I = P/(4\pi d^2)$ and $P$ is constant, the quantity $I d^2$ is the same at every distance. So for any two points 1 and 2 from the same source:
$$I_1 d_1^2 = I_2 d_2^2$$
Given any three of the four quantities, rearrange to find the fourth. For example, to find the intensity at a new distance, $I_2 = I_1 d_1^2 / d_2^2$.
A lamp gives a light intensity of $80\ \text{W m}^{-2}$ at a distance of $2.0\ \text{m}$. What is the intensity at $5.0\ \text{m}$ from the same lamp?
For a point source, $I_1 d_1^2 = I_2 d_2^2$. Rearrange to find any unknown, e.g. $I_2 = I_1 d_1^2 / d_2^2$. Moving further away always lowers the intensity.
Add the highlighted equation and method to your notes before the check below.
Quick check: Intensity is $36\ \text{W m}^{-2}$ at $1.0\ \text{m}$ from a point source. What is it at $3.0\ \text{m}$?
Pattern A, Plan then calculate
In your book: (1) Design a practical to test how light intensity depends on distance, using a small lamp and a light sensor (light meter or phone app). State the independent and dependent variables, two controlled variables, and what graph you would plot to reveal an inverse square relationship. (2) Then solve: a security light reads $200\ \text{W m}^{-2}$ at $1.5\ \text{m}$, what is the intensity at $6.0\ \text{m}$?
Two truths, one lie: Two truths and a lie about the inverse square law for light. Which statement is the lie?
Dot point: Solve problems involving the inverse square law using I₁d₁² = I₂d₂² to compare the intensity of light at points 1 and 2
Turning a curve into a straight line to confirm the law
We just solved numerical problems. That raises a question: how would a practical actually confirm the inverse square relationship from data? This card answers it → by plotting intensity against $1/d^2$ to get a straight line.
A graph of intensity against distance is a steep curve, but plotting intensity against $1/d^2$ straightens it into a line through the origin, the clean signature that confirms an inverse square law.
If you plot measured intensity $I$ against distance $d$, you get a curve that falls steeply then flattens, hard to interpret. But because $I \propto 1/d^2$, plotting $I$ against $1/d^2$ gives a straight line through the origin. A straight-line fit is strong evidence that the data obey the inverse square law, and the gradient equals $P/(4\pi)$. This linearising trick, replotting against $1/d^2$, is the standard way to test and confirm the relationship experimentally.
A plot of intensity against distance is a curve; plotting intensity against $1/d^2$ gives a straight line through the origin, confirming the inverse square law. This linearising trick is how the relationship is tested experimentally.
Pause, copy the highlighted method into your book before the check below.
Last check: Decide if the statement is true or false.
Plotting light intensity against $1/d^2$ gives a straight line through the origin, which confirms the inverse square law.
Odd one out: Three of these are sensible controlled variables when investigating how light intensity varies with distance. Which one is the odd one out?
Dot point: Conduct a laboratory experiment to analyse the relationship between light intensity and distance from a light source
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 6(5 marks) 1. A point light source produces an intensity of $120\ \text{W m}^{-2}$ at a distance of $3.0\ \text{m}$. Explain why intensity decreases with distance, then calculate the intensity at $9.0\ \text{m}$ using the inverse square law.
1 mark: light spreads over a sphere of area $4\pi d^2$ · 1 mark: same power over larger area ⇒ $I \propto 1/d^2$ · 1 mark: correct use of $I_1 d_1^2 = I_2 d_2^2$ · 1 mark: substitution $I_2 = 120 \times 3.0^2 / 9.0^2$ · 1 mark: $I_2 = 13\ \text{W m}^{-2}$
ApplyBand 5(3 marks) 2. A light sensor reads an intensity of $180\ \text{W m}^{-2}$ at $2.0\ \text{m}$ from a small lamp. Using $I_1 d_1^2 = I_2 d_2^2$, calculate the intensity at $6.0\ \text{m}$, and state the assumption the calculation relies on.
1 mark: correct rearrangement $I_2 = I_1 d_1^2 / d_2^2$ · 1 mark: $I_2 = 20\ \text{W m}^{-2}$ · 1 mark: assumes the lamp behaves as a point source radiating uniformly, with no absorption or reflection
Dot point: Solve problems involving the inverse square law using I₁d₁² = I₂d₂² to compare the intensity of light at points 1 and 2
AnalyseBand 6(4 marks) 3. You are to conduct a laboratory experiment to analyse the relationship between light intensity and distance from a light source. State the independent and dependent variables, name two controlled variables, and explain what graph you would plot and what result would confirm an inverse square relationship.
1 mark: IV distance from source, DV measured intensity · 1 mark: two appropriate controlled variables (lamp power, background light, sensor orientation) · 1 mark: plot intensity against $1/d^2$ · 1 mark: a straight line through the origin confirms the relationship
Dot point: Conduct a laboratory experiment to analyse the relationship between light intensity and distance from a light source
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (5 marks): Light from a point source spreads out equally in all directions over the surface of a sphere of area $4\pi d^2$. As the distance $d$ increases, the same total power is spread over a larger area, so the power per unit area, the intensity, decreases. Because the area grows with $d^2$, the intensity is inversely proportional to the square of the distance, $I \propto 1/d^2$. Using $I_1 d_1^2 = I_2 d_2^2$: $I_2 = I_1 d_1^2 / d_2^2 = 120 \times (3.0)^2 / (9.0)^2 = 120 \times 9 / 81 = 13\ \text{W m}^{-2}$. Tripling the distance (3.0 m → 9.0 m) reduces the intensity to one ninth, as expected.
Q2 (3 marks): Rearranging $I_1 d_1^2 = I_2 d_2^2$ gives $I_2 = I_1 d_1^2 / d_2^2$. Substituting, $I_2 = (180 \times 2.0^2)/6.0^2 = (180 \times 4.0)/36 = 20\ \text{W m}^{-2}$. This is consistent with the inverse square law: the distance has tripled, so the intensity has fallen by a factor of $3^2 = 9$, and $180/9 = 20$. The calculation assumes the lamp behaves as a point source radiating uniformly in all directions, so that the same power is spread over the surface of a sphere whose area grows as $d^2$. It also assumes no light is absorbed, scattered or reflected between the lamp and the sensor, and that the sensor faces the lamp squarely at both positions.
Q3 (4 marks): The independent variable is the distance $d$ from the lamp to the sensor, set to a series of measured values along a metre rule. The dependent variable is the light intensity recorded by the sensor at each distance. Controlled variables include the power of the lamp, which must be kept constant so the emitted power does not change, and the level of background light, which is minimised by darkening the room and is the same for every reading; the orientation of the sensor, kept square-on to the lamp, is a third. Plotting intensity $I$ directly against distance $d$ gives a curve that falls steeply and then flattens, which is difficult to distinguish from other decreasing relationships by eye. Instead I would plot $I$ against $1/d^2$. Because $I \propto 1/d^2$, this linearises the data, and a straight line passing through the origin is the result that confirms an inverse square relationship. A line that is straight but does not pass through the origin would indicate a constant background offset that had not been fully eliminated.
You can now explain why intensity falls as $1/d^2$, solve problems with $I_1 d_1^2 = I_2 d_2^2$, and describe how a $I$-versus-$1/d^2$ graph confirms the law. The same inverse square idea returns for gravitational and electric fields, so you have built a tool you will use again and again.
↩ Back to focus-area hubAt the start you predicted how the sphere's area grows when you double the distance. The area grows by a factor of four ($2^2$), so the same light is spread four times as thinly and the brightness drops to a quarter. That is exactly why tripling the distance crashes the intensity to a ninth, the inverse square law.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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