Point Sources and the Inverse-Square Law
For an isotropic point source of radiated power P with negligible absorption, intensity at distance r equals P divided by the area 4 pi r squared. Points at the same radius have the same intensity.
Why this shows up in the exam
Estimating bulb intensity at a distance · Comparing detector readings on spherical surfaces · Finding field amplitudes from source power and efficiency
Learn the idea
An isotropic point source spreads fixed power over spherical area, so intensity falls as one over distance squared. The same emitted power must cover larger spheres as it travels outward. Doubling radius makes the sphere four times larger, leaving one-fourth the intensity at each point.
🧠 Memory hook: Twice as far means one-fourth the intensity.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- I = P/(4 pi r²) — intensity of an isotropic point source
- I₂/I₁ = (r₁/r₂)² — inverse-square comparison
- P_radiated = eta P_input — radiated power when source efficiency is eta
How to approach it
- 1Check whether the source is isotropic or otherwise specified
- 2Find the relevant radius and radiated power
- 3Use inverse-square ratios before doing full arithmetic
Common slip-ups that cost marks
- •Dividing by 4 pi r instead of 4 pi r squared
- •Changing intensity when only position changes on the same sphere
- •Using electrical input power without applying stated efficiency
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
An electromagnetic wave in vacuum has wavelength 1 m. Take c = 3 x 10^8 m/s. Find its frequency in units of 10^8 Hz.
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