Intensity (I)

Astrophysics & Cosmology - OCR A-Level Physics

Key Definition

Intensity (I)
The power received per unit area at a given distance from a source. Units: $W m^{-2}$.

I = \frac{L}{4\pi d^2}

Light from a star spreads out uniformly over a sphere of area $4\pi$$d^{2}$ as it travels distance d.
IntensityThe powerThe rate of energy transfer. Measured in watts (W). transmitted per unit area perpendicular to the wave direction. Measured in W m⁻². Proportional to amplitude squared. follows an inverse square lawA relationship in which a physical quantity is inversely proportional to the square of the distance from its source.: doubling the distance quarters the intensityThe powerThe rate of energy transfer. Measured in watts (W). transmitted per unit area perpendicular to the wave direction. Measured in W m⁻². Proportional to amplitude squared..
If we can measure I (from a detector on Earth) and know L (from a standard candleAn astronomical object whose absolute luminosityThe total powerThe rate of energy transfer. Measured in watts (W). radiated by a star across all wavelengths. Measured in watts (W). is known or can be determined independently, allowing its distance to be calculated from its observed intensityThe power transmitted per unit area perpendicular to the wave direction. Measured in W m⁻². Proportional to amplitude squared..), we can calculate d.
This is one of the primary methods for measuring astronomical distances.

Worked Example

The Sun has luminosity $3.85 \times 10^{26}$ W and is $1.50 \times 10^{11}$ m from Earth. Calculate the solar intensity at Earth.

Show Solution

$I = L / (4\pi d^2).$

$I = 3.85 \times 10^{26} / (4\pi \times (1.50 \times 10^{11})^2).$

$I = 3.85 \times 10^{26} / (4\pi \times 2.25 \times 10^{22}).$

$I = 3.85 \times 10^{26} / (2.827 \times 10^{23}).$

$I = 1360 W m^{-2} \approx 1.36 kW m^{-2}.$

Answer

$$I \approx 1360$ $W m^{-2}$ (the solar constant)$