70 - Radiation Intensity

70 - Radiation Intensity

Relate radiation power to illuminated area and test how intensity changes with distance from a small source.

Intensity Is Power Spread Out

Imagine a small lamp sending radiation onto a detector. The lamp does not just have a "brightness"; it transfers energy each second. That rate of energy transfer is power, PP, measured in watts.

The detector cares about how much of that power arrives on each square metre.

Intensity

The power per unit area incident on (or passing through) a surface at right angles to the radiation; unit W m⁻².

Intensity

I=PAI=P4πr2 I = \frac{P}{A} \qquad I = \frac{P}{4\pi r^{2}}

II: intensity (W m⁻²); PP: power (W); AA: area the radiation passes through (m²). The second form is for a small source radiating uniformly in all directions with no absorption, at distance rr.

A useful physical test is this: if the same power is spread over a larger area, each square metre gets less. If the same power is concentrated onto a smaller area, each square metre gets more.

Worked example: a ray box transfers 0.48 W0.48\,\mathrm{W} of optical power through a rectangular aperture of width 4.0 cm4.0\,\mathrm{cm} and height 3.0 cm3.0\,\mathrm{cm}. Calculate the intensity at the aperture.

Decision point 1: the area is not 4.0×3.0=124.0 \times 3.0 = 12 in exam units. Convert lengths to metres first:

4.0 cm=0.040 m3.0 cm=0.030 mA=0.040×0.030=1.20×10−3 m2\begin{aligned} 4.0\,\mathrm{cm}=0.040\,\mathrm{m} \\ 3.0\,\mathrm{cm}=0.030\,\mathrm{m} \\ A=0.040\times0.030=1.20\times10^{-3}\,\mathrm{m^2} \end{aligned}

Decision point 2: use the power through that area, not the total electrical power supplied to the ray box unless the question says all of it becomes radiation through the aperture.

I=0.481.20×10−3=4.0×102 W m−2\begin{aligned} I=\frac{0.48}{1.20\times10^{-3}}=4.0\times10^2\,\mathrm{W\,m^{-2}} \end{aligned}

For Edexcel calculate, show the substitution and keep the unit with the final value. A number without W m−2\mathrm{W\,m^{-2}} has lost the physical meaning.

Misconception contrast: intensity is not "total energy" and it is not "power somewhere nearby". It is power per unit area at a particular place.

When a flat detector is tilted, use its projected area perpendicular to the rays. If its normal makes angle θ\theta to the rays, P=IAcos⁡θP=IA\cos\theta. For example, tilting a detector through 60° halves the intercepted power because its projected area halves, even if the beam intensity is unchanged.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account