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Locate the missing power
Chapters 1–2. Define the measurement and the beam-radius convention.
Calculate the capture
Chapters 3–5. Derive the aperture fraction and compare two beam sizes.
Use a correction carefully
Chapters 6–7. Set a capture target and expose ambiguities.
Practise and continue
Chapters 11–12. Transfer the calculation and follow related guides.
Full chapter index · Practice and worked checks · Photonics subject root
A power meter can report more light after a beam is narrowed even when the beam’s total power stays exactly the same. A finite detector may simply collect a larger fraction of it.
That distinction is easy to miss when the display shows watts. This guide follows a centred Gaussian beam through a circular receiving area, derives the captured fraction and tests the evidence needed to infer total power. It extends the Photonics subject guide with original numerical models rather than a procedure for operating a laser.
Chapter index
Locate the missing power · Chapters 1–2
1. Locate the boundary of the measurement
2. Define the Gaussian radius before using it
Calculate the capture · Chapters 3–5
3. Add power over rings, not over one line
4. Compare two equal-power beams
5. Separate peak irradiance from enclosed power
Use a correction carefully · Chapters 6–7
6. Set a capture target from the missed fraction
7. Identify what one reading cannot tell you
Check the measurement · Chapters 8–10
8. Keep the limits visible
9. Build a useful measurement check
10. Questions that resolve common confusions
Practise and continue · Chapters 11–12
11. Practise beyond the original beam sizes
12. Choose the next optical measurement route
CHAPTER 1 OF 12 · Locate the missing power
1. Locate the boundary of the measurement
Total beam power P is the power crossing the entire transverse plane. Captured power Pa is the part entering a specified receiving area in that plane. A calibrated meter can correctly measure Pa while a reader incorrectly interprets it as P.
For this model, use a circular active detector of radius a, centred on the beam. Assume it absorbs the intercepted light appropriately, responds linearly and has uniform responsivity across that area. Wavelength, electronic gain and background correction remain fixed.
We calculate incident power inside that circle. Any reflection, filter transmission or later propagation loss would need its own treatment. A detector-plane collection fraction does not automatically describe everything transmitted through an aperture and collected farther downstream.
CHAPTER 2 OF 12 · Locate the missing power
2. Define the Gaussian radius before using it
For an ideal circular Gaussian irradiance profile, let r be distance from the beam centre and w be the radius where irradiance has fallen to exp(−2), approximately 0.1353, of its peak. Irradiance means power per unit area, here W/m².
The model is I(r) = [2P / (πw²)] exp(−2r² / w²). At r = 0, the peak is I0 = 2P / (πw²). At r = w, the local irradiance is I0 exp(−2).
The radius w is not a hard beam edge. It is also not the full width at half maximum or the 1/e radius of irradiance. Newport’s Gaussian beam optics note explains the radius convention and enclosed-power concept. Keep the supplied beam-width definition with the number; using a diameter as w produces a different model.
A thin annulus of radius r and thickness dr has area approximately 2πr dr. Its power contribution is I(r) × 2πr dr. Integrating these annuli from the centre to the detector edge gives:
Pa = integral from 0 to a of [2P / (πw²)] exp(−2r² / w²) × 2πr dr.
Set u = 2r² / w². Then du = 4r dr / w², so the integral becomes P times the integral of exp(−u), from 0 to 2a²/w². Therefore Pa = P[1 − exp(−2a² / w²)].
Define f = Pa / P. The capture fraction is f = 1 − exp(−2a² / w²). It lies between zero and one, tends to zero as a tends to zero, and approaches one as a becomes very large. Those checks also confirm that the Gaussian normalisation integrates to P over the full plane.
Let P = 10 mW and a = 1 mm. Compare a broad beam with w = 2 mm against a narrower beam with w = 1 mm. Both are centred and retain the same ideal profile, wavelength and total power.
| Quantity | Broad beam | Narrow beam |
|---|---|---|
| Total power P | 10 mW | 10 mW |
| Beam radius w | 2 mm | 1 mm |
| Detector ratio a/w | 0.5 | 1 |
| Capture fraction f | 0.393469 | 0.864665 |
| Captured power Pa | 3.93469 mW | 8.64665 mW |
| Power outside the detector | 6.06531 mW | 1.35335 mW |
The detector receives about 2.1975 times as much power in the narrow case. No gain in source power is required to explain it. The assumed receiver captures more of the same total.
If uniform responsivity is 0.50 A/W, the corresponding baseline-corrected photocurrents are about 1.96735 mA and 4.32332 mA. These values assume linear operation; they are not specifications for any real detector.
CHAPTER 5 OF 12 · Calculate the capture
5. Separate peak irradiance from enclosed power
Halving w at fixed P multiplies I0 by four because I0 is proportional to 1/w². In the worked case, peak irradiance rises from about 1,592 W/m² to 6,366 W/m².
Captured power rises by about 2.1975, not four. A finite-area integral and a local peak are different quantities. The annular calculation explains why the same beam change can produce those two different ratios.
At a = w, local irradiance at the detector edge is only about 13.53% of the peak, while the circle contains about 86.47% of total power. Neither percentage can replace the other. One describes a point on the profile; the other adds contributions across an area.
CHAPTER 6 OF 12 · Use a correction carefully
6. Set a capture target from the missed fraction
The missed fraction is ε = exp(−2a² / w²). For a desired maximum missed fraction, rearranging gives a/w ≥ sqrt[−ln(ε) / 2]. This is a model-based radius requirement for centred Gaussian interception, not a universal detector-sizing rule.
| Aperture radius relative to w | Captured fraction | Missed fraction |
|---|---|---|
| 0.5w | About 39.35% | About 60.65% |
| 1w | About 86.47% | About 13.53% |
| 1.5174w | About 99.00% | About 1.00% |
| 2w | About 99.966% | About 0.0335% |
The 99% radius is approximately 1.5174w; its diameter is approximately 3.0349w. Stating radius or diameter explicitly prevents a factor-of-two geometry error.
If the total-power estimate is P = Pa/f, uncertainty in f matters as well as uncertainty in Pa. A severe clipping correction amplifies the consequences of an inaccurate beam model.
CHAPTER 7 OF 12 · Use a correction carefully
7. Identify what one reading cannot tell you
Suppose Pa = 4 mW. It could come from a nearly fully captured 4 mW beam, a 10 mW beam captured at 40%, or another power-profile combination. A single current reading does not identify both total power and beam radius.
An independent profile or width measurement can supply missing information under the stated model. Multiple receiver sizes can test the predicted capture curve if the incident beam remains unchanged between measurements. Two settings are not automatically proof of a Gaussian shape.
An off-centre beam requires integration over a displaced circle. Substituting an offset distance for a in the centred formula does not calculate that case. A dimmer reading might reflect displacement, increased width, lower total power, spectral change or detector drift. Choose an observation that distinguishes these explanations.
Real profiles can be elliptical, multimode or irregular. A circular Gaussian formula may then misrepresent the tails even if a central fit looks good. A rectangular active area also requires the appropriate geometry.
Nonuniform detector response changes the measurement from a simple captured-power integral to a spatially weighted one. NIST’s spatial and spectral responsivity measurements explicitly include mapping sensitivity across detector surfaces. Centring alone cannot guarantee uniform sensitivity.
Clipping can change subsequent diffraction and propagation. The formula here calculates the intercepted fraction at the specified plane; it does not assert that a clipped beam keeps its original Gaussian shape later. Saturation, damage, background error and wavelength mismatch also sit outside the ideal collection model.
First state the target: total incident beam power, power through a particular opening, or power on the active detector. These are legitimate but different measurements. Record the receiving-plane location, active-area dimensions, profile convention and alignment evidence.
Next check the electronic chain separately: baseline, linearity and responsivity at the relevant wavelength. Then examine collection. A plateau as the receiving area increases can support adequate capture, provided the beam and detector behaviour remain stable; it does not prove that every possible systematic effect is absent.
For a learning exercise, use synthetic profiles and numerical integration. Compare the analytic centred result with a numerical area sum, then displace the centre or change the profile. The purpose is to see which assumption fails, without treating an optical-power calculation as an equipment-safety instruction.
CHAPTER 10 OF 12 · Check the measurement
10. Questions that resolve common confusions
Does “beam radius” mean all light lies inside it?
No. In this Gaussian convention it marks a specified local irradiance level. The integrated power inside w is about 86.47%, and the mathematical profile has tails beyond it.
Can a correct meter display a misleading source-power trend?
Yes. The instrument can correctly report captured power while collection fraction changes. The misleading step is assigning that trend entirely to total source power without collection evidence.
Can I correct clipping by subtracting a fixed missing power?
Not generally. The missed amount depends on P and on the profile relative to the receiver. Under this model, a known fraction leads to division by f. A fixed correction requires its own justified conditions.
CHAPTER 11 OF 12 · Practise and continue
11. Practise beyond the original beam sizes
Case A: use P = 5 mW, w = 1.5 mm and a = 1.5 mm. Because a/w = 1, capture is again 0.864665, giving Pa ≈ 4.32332 mW. The fraction survives a common scaling of a and w; the absolute power follows P.
Case B: a centred Gaussian beam has independently known w and a detector captures 3 mW at a = 0.5w. Total power is 3 / 0.393469 ≈ 7.6245 mW. Verify by multiplying the inferred total by f. Then explain why the same inference would be unsupported if w were unknown.
Case C: double both w and a while keeping P fixed. Captured fraction and captured power are unchanged, but peak irradiance falls by a factor of four. This tests whether you are tracking a ratio, an integral or a local intensity.
CHAPTER 12 OF 12 · Practise and continue
12. Choose the next optical measurement route
Source, beam and detector
Return to the Photonics root for the complete source-to-detector chain. Photonics Engineering connects optical interfaces and tested system performance.
Distinguish other causes of a changed current
Use spectral mismatch and equal photocurrents when the wavelength mixture changes. Use responsivity drift, dark readings and gain when the detection chain changes. Neither guide supplies evidence that a finite receiver captures the whole beam; the collection question belongs here.
Carry the reasoning into another subject
The Gold Standard of Education and How X Works connect this example to wider learning. Before interpreting a measurement as a total, ask what fraction of the physical system the instrument actually receives.
Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.