Edexcel A Level (IAL) Physics-2.12 Equation for the Intensity of Radiation- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -2.12 Equation for the Intensity of Radiation- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -2.12 Equation for the Intensity of Radiation- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
Intensity of Radiation \( I = \dfrac{P}{A} \)
The intensity of radiation describes how much power is delivered per unit area. It tells us how “concentrated” the energy transfer is and is a key idea for waves, sound, light, and electromagnetic radiation.
Definition of Intensity
Intensity is defined as the power per unit area received by a surface perpendicular to the direction of energy transfer.
$ I = \frac{P}{A} $
- \( I \) = intensity (W m⁻²)
- \( P \) = power (W)
- \( A \) = area over which power is distributed (m²)
Meaning of the Equation
- Greater power → greater intensity.
- Larger area → lower intensity (energy is spread out).
- Intensity tells how strong a wave is at a particular location.
- Used for light, sound, microwaves, radio waves, lasers, etc.
Intensity Depends on Distance (for point sources)
If radiation spreads out uniformly from a point source, it spreads over the surface area of a sphere:
\( A = 4\pi r^2 \)
So intensity decreases with distance as:
$ I = \frac{P}{4\pi r^2} $
- Double the distance → intensity becomes one-quarter.
- This is called the inverse square law.
Notes on Using the Equation
- Ensure area is perpendicular to direction of wave travel.
- Convert power to watts and area to square metres.
- Useful in calculating brightness, sound levels, and energy transport.
- Applies to both mechanical and electromagnetic waves.
Example (Easy)
A lamp emits \( 60\,\mathrm{W} \) of light uniformly. What is the intensity at a distance where the light spreads over an area of \( 3.0\,\mathrm{m^2} \)?
▶️ Answer / Explanation
$ I = \frac{P}{A} = \frac{60}{3.0} = 20\,\mathrm{W\,m^{-2}} $
Example (Medium)
A radio transmitter outputs \( 200\,\mathrm{W} \). Find the intensity \( 30\,\mathrm{m} \) away assuming spherical spreading.
▶️ Answer / Explanation
Step 1: Find area
$ A = 4\pi r^2 = 4\pi (30)^2 = 3600\pi $
Step 2: Use intensity formula
$ I = \frac{200}{3600\pi} \approx 0.0177\,\mathrm{W\,m^{-2}} $
Example (Hard)
A laser beam delivers \( 5.0\,\mathrm{mW} \) of power over a circular spot of radius \( 2.0\,\mathrm{mm} \). Find the intensity.
▶️ Answer / Explanation
Convert units:
\( P = 5.0\,\mathrm{mW} = 5.0\times10^{-3}\,\mathrm{W} \)
\( r = 2.0\,\mathrm{mm} = 2.0\times10^{-3}\,\mathrm{m} \)
Find area:
$ A = \pi r^2 = \pi (2.0\times10^{-3})^2 = 1.26\times10^{-5}\,\mathrm{m^2} $
Find intensity:
$ I = \frac{5.0\times10^{-3}}{1.26\times10^{-5}} \approx 397\,\mathrm{W\,m^{-2}} $
Intensity ≈ \( 4.0\times10^2\,\mathrm{W\,m^{-2}} \)