(a) The energy flux for 1 lm is 3.556 mW at λ = 0.51 µm and 8.989 mW at λ = 0.64 µm. (b) The band flux is 1.534 lm with the linear rule, and 1.479 lm from the exact curve. With the given values these are 3.56 mW, 8.99 mW and 1.53 lm.
Problem statement
Use the spectral response curve of the eye (Fig. 5.1) to find: (a) the energy flux that corresponds to a luminous flux of 1.0 lm at the wavelengths 0.51 µm and 0.64 µm; (b) the luminous flux that corresponds to the wavelength interval from 0.58 to 0.63 µm, if the energy flux of 4.5 mW is spread evenly over all wavelengths of the interval. The function V(λ) is taken as linear in this interval.
We assume the wavelengths are in µm and the energy flux is Φe = 4.5 mW.
We assume A = 1.6 mW per lumen at 0.555 µm, as printed on the figure. This means a lumen at the peak needs 1.6 mW.
We assume V(λ) is read from the printed curve. This means every V value below has an uncertainty of about 0.01.
Symbols: Φ is the luminous flux in lumens, the light as the eye counts it. Φe is the energy flux in milliwatts, the power in the light. V(λ) is the relative response of the eye, 1 at the peak. A is the energy flux that gives 1 lm at the peak.
The big idea in plain words
A lumen counts light the way the eye does. The eye reacts strongly to green and weakly to deep red and blue. So the same power in the light gives many lumens in green and few in red. To get one lumen from a colour the eye sees badly, you must send more power. The curve V(λ) tells you how badly: you need A divided by V.
The solution in each frame
Lab frame
The source and the eye are at rest with respect to each other, so the wavelengths are the ones in the figure. The full solution is in the next section.
Other frames
No second frame is needed. There is no motion, so no Doppler shift changes the colour.
The solution step by step
The definition of V(λ) says that A / V(λ) is the energy flux of light at wavelength λ that gives one lumen. So for a luminous flux Φ at one wavelength:
Part (a). Read V from the figure. At 0.51 µm the curve is at V = 0.45. At 0.64 µm it is at V = 0.18, more exactly 0.178.
Your sliders now give V1 = 0.45 and V2 = 0.178, so Φe = 3.556 mW and 8.989 mW.
Part (b). The energy is spread evenly. A slice dλ carries the energy flux dΦe = Φe dλ / (λb − λa). The luminous flux of the slice is V dΦe / A. Add all slices:
The part in brackets is the mean of V over the band. If V is a straight line, the mean is the average of the two end values:
For your sliders: V(λa) = 0.837, V(λb) = 0.254, so Φ = 1.534 lm. Integrating the traced curve directly gives 1.479 lm, which is 3.74 percent away from the linear rule.
Integrating the traced curve with 400 steps gives 1.479 lm, and with 4000 steps 1.479 lm.
Five methods
1. Force method
This does not suit the problem. No force or motion is involved, only how the eye counts power.
2. Energy method
This is the natural method. The power of the light, Φe, is split into slices, and each slice is weighted by the eye response V and divided by A. The total weighted power is the luminous flux.
3. Graph method
Read V at the wavelengths straight from the figure. For the band, the area under V between λa and λb, divided by the width, is the mean V. The page draws the straight chord that the linear rule uses.
4. Calculus method
Write the sum over slices as an integral, Φ = (Φe/A(λb − λa)) ∫ V dλ. The page also integrates the traced curve by 400 and 4000 steps to check this.
5. Check method
Units: mW divided by mW per lumen gives lumens. Limits: where V = 1 the energy flux for 1 lm is the smallest, A. Where V is small the needed flux is large. Because V is below 1, the band flux is always less than Φe/A, which the page shows as the grey bar.
Fastest: the linear rule. Most reliable: the integral of the traced curve, because it does not assume a straight line.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| λ1, λ2 | Yes | V at the probe, the energy flux for the chosen luminous flux, the markers on the figure, the first graph |
| Luminous flux (a) | Yes | Both energy fluxes of part (a), in proportion. V stays the same. |
| Factor A | Yes | Every energy flux and every luminous flux |
| Band start, band end | Yes | V at the edges, the pink band, the chord, both band fluxes and the middle graph |
| Band energy flux | Yes | The band flux, in proportion |
| Target band flux (given-answer mode) | Sets the answer | Solves Φe = Φ × 2A / (V(λa) + V(λb)) |
| Scan speed | No | Only how fast the green scan moves along the curve |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Factor | A | 1.6 mW/lm |
| Probe wavelengths | λ1, λ2 | 0.51 and 0.64 µm |
| Response at the probes | V1, V2 | 0.45 and 0.178 |
| Luminous flux, part (a) | Φ | 1 lm |
| Band | λa to λb | 0.58 to 0.63 µm |
| Band energy flux | Φe | 4.5 mW |
| Mean V over the band | 0.526 |
Exam traps
- Reading the curve at the wrong wavelength. V is 0.45 at 0.51 µm and 0.18 at 0.64 µm.
- Multiplying by V when you should divide. A small V means more energy is needed for one lumen.
- Leaving out A. The energy flux is A Φ / V, not Φ / V.
- Mixing W and mW. A = 1.6 mW per lumen, so the answers are in mW.
- Using the linear rule over a wide band. It is good only when the curve is nearly straight between the edges.
- Treating V as a percentage. It is a ratio, and 1 at 0.555 µm.