Em = √[ (A Φ / (2π r² V)) × √(μ0/ε0) ] and Hm = Em / √(μ0/ε0). For your sliders Em = 1.164 V/m and Hm = 3.09 mA/m. With the given values and V read from the curve (0.71) these are 1.16 V/m and 3.09 mA/m.
Problem statement
A point isotropic source emits a luminous flux Φ = 10 lm of wavelength λ = 0.59 µm. Find the peak values of the electric and magnetic field strengths of the light at a distance r = 1.0 m from the source. Use the curve of Fig. 5.1.
We assume the source is a point that radiates equally in all directions. This means the power spreads evenly over a sphere of area 4π r².
We assume the light is a plane wave in vacuum at that distance. This means I = (1/2) √(ε0/μ0) Em², and Hm = √(ε0/μ0) Em.
We assume A = 1.6 mW per lumen at 0.555 µm, as in Fig. 5.1. The figure prints "w/lm" but the unit is mW per lumen. This means 1 lm at the peak is 1.6 mW.
Symbols: Φ is the luminous flux in lumens. Φe is the energy flux in watts. V is the relative response of the eye. I is the intensity in W/m². Em and Hm are the peak electric and magnetic field strengths. Z0 = √(μ0/ε0) = 376.73 Ω.
The big idea in plain words
The lumens tell you how bright the lamp looks to the eye. To know the fields you need the real power in the light. The eye sees 0.59 µm light well but not perfectly, so the power is a bit more than A times Φ. That power spreads over a bigger and bigger sphere. The power per area gives the strength of the electric and magnetic fields in the wave.
The solution in each frame
Lab frame
The source and the observer are at rest. The full solution is in the next section.
Other frames
No second frame is needed. Nothing moves, so there is no Doppler shift.
The solution step by step
Step 1: the energy flux. The definition of V says A / V is the energy flux that gives one lumen. So
Step 2: the intensity. The power spreads over a sphere of area 4π r².
Step 3: the electric field. In a light wave the intensity is the average of E × H, which gives I = (1/2) √(ε0/μ0) Em². Solving for Em:
Written in one line, Em² = (A Φ / (2π r² V)) √(μ0/ε0).
Step 4: the magnetic field. In vacuum E and H in a light wave are tied by Z0:
For your sliders: V = 0.708, Φe = 22.6 mW, I = 1.798 mW/m², Em = 1.164 V/m and Hm = 3.09 mA/m.
A separate check: the intensity I = 1.7984 mW/m², and the speed of light times the average energy density of the wave gives 1.7984 mW/m².
Five methods
1. Force method
This does not suit the problem. No force on a charge is asked, only the field strengths.
2. Energy method
This is the main method. Power is conserved as it spreads, so the intensity falls as 1/r². The energy in the wave is shared equally by the electric and the magnetic field, which gives I = (1/2) √(ε0/μ0) Em².
3. Graph method
Read V from the figure, then use the graphs: Em falls as 1/r with distance, and rises as 1/√V when the colour moves away from the peak.
4. Calculus method
The Poynting vector S = E × H is the power per area. Averaged over a cycle it is (1/2) Em Hm, and with Hm = Em / Z0 it gives (1/2) Em² / Z0.
5. Check method
Units: √(W/m² × Ω) is √(V²/m²) = V/m. Hm in A/m follows from V/m divided by Ω. Limits: doubling r halves Em and Hm. Doubling Φ multiplies Em by √2. The electric and magnetic energy densities are equal, as the bar graph shows.
Fastest: the one-line formula Em² = (A Φ / (2π r² V)) Z0. Most reliable: the four steps, because each one has a unit check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Wavelength λ | Yes | V from the curve, then Φe, I, Em, Hm, the marker and the second graph |
| Luminous flux Φ | Yes | Φe, I, Em, Hm (as √Φ for the fields) |
| Distance r | Yes | I (1/r²), Em and Hm (1/r), the probe on the first graph |
| Factor A | Yes | Φe, I, Em and Hm |
| V by hand | Yes (when not 0) | Replaces V from the curve. |
| Target peak E (given-answer mode) | Sets the answer | Solves the luminous flux Φ |
| Scan speed | No | Only how fast the green scan moves along r |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Wavelength | λ | 0.59 µm |
| Response used | V | 0.708 (curve: 0.708) |
| Luminous flux | Φ | 10 lm |
| Distance | r | 1 m |
| Factor | A | 1.6 mW/lm |
| Wave impedance of vacuum | Z0 | 376.73 Ω |
Exam traps
- Using Φ in watts directly. Lumens must be converted with A and V first.
- Dividing by 4π r² twice, or using r instead of r². The area is 4π r².
- Forgetting the factor 2 in I = (1/2) √(ε0/μ0) Em². The peak value is larger than the average value.
- Giving H in A/m when the field is in mA/m. Check the prefix.
- Using V = 1 for any colour. V is 1 only at 0.555 µm.
- Mixing the roles of E and H. Em is in V/m and Hm is in A/m, and Hm is much smaller in numbers.