I can not find a good explanation on how the subjective speckle pattern is formed (in terms of interference).

Objective speckle. The image below explains well how the speckle is formed: the light from each point on the illuminated surface adds constructively/destructively to form a single speckle. The phases in yellow help a lot in understanding how this happens.

enter image description here

Subjective speckle. It is clear that the speckle pattern is imaged on the detector using a lens. Each point in the image can be considered to be illuminated by a finite area in the object. The size of this area is determined by the diffraction-limited resolution of the lens which is given by the Airy disk (taken from here). Subjective patterns are caused by the interference of waves from the various scattering regions of a resolution element of the lens. In this region, the response functions of the randomly de-phased waves are added, resulting in the formation of speckle patterns (taken from here).


What exactly happens in that "finite" area? Is it the same as a resolution element? Which rays interfere? How is this even possible if all the rays travel the same distance from the object point to the image point? Aren't they all in phase? What causes randomly de-phased waves?


Check out speckle patterns here.

  1. Different paths that light takes will cause different phases at a given point A at a screen. (That's geometry.)
  2. If there's no imaging system - the light hits an object, it gets scattered/reflected off the object, and meets at the screen. The waves interfere and form a speckle pattern depending on the roughness of the object. That's an objective speckle.
  3. If you take a lens and image a point A at the object. A is not a point it's an area, defined by the Airy disc. The the disc's diameter is proportional to the distance u between the object and the lens.
  4. The Airy disc that can be resolved in the image is proportional to the distance v between the lens and the image.
  5. The wavefronts captured within A do not necessarily interfere within the area A'.

Say the v is smaller than u, i.e. the Airy disc A is smaller than A'. The wavefronts that interfere in A' must, therefore, origin from the surrounding area of A. Different wavefronts with different phases result in a different intensity. Because the resulting pattern is depending on the imaging system, it is called subjective.

Moreover, the Airy disc depends also on the aperature (or the lens diameter) the subjective pattern also changes when changing the aperature.


Objective and subjective speckle patterns have a lot in common. In both cases, a plane wave undergoes a random phase change at plane A, and is then imaged at another plane B. Planes A and B are far enough apart so that Fourier optics applies.

The intensity pattern at plane B is a speckle pattern. If A is the reflective surface of an object, the speckle pattern is known as objective. If A is the surface of a lens (or another intermediate surface of the imaging system), the speckle pattern is known as subjective.

The mathematical description is the same in both cases. The transfer function at plane A is a random phase change, i.e. $A(\vec{x}) = e^{i \phi(\vec{x})}$, where $\phi(\vec{x})$ varies randomly with position $\vec{x}$ in the plane. The intensity at plane B is a Fourier transform of $A(\vec{x})$.

It's possible to have a more complicated setup - plane wave reflects off a rough surface A, then passes through a rough lens B, and then is imaged at plane C. In this case we would have both objective and subjective speckles.

  • $\begingroup$ Sergei, thanks for the explanation, but I am still not able to capture the concept of which rays from the object (rough surface) contribute to which points on the image (speckle). For the objective speckle it is clearly demonstrated on the picture above - the waves from all points on the object leave in phase, but arrive out of phase. Now, in case of subjective speckle, if we placed a lens between the object and the sensor above, the waves are leaving and arriving to the same point (conjugate) in phase, so how the interference happens then? $\endgroup$ – Nazar Sep 15 '16 at 18:15

I've got a different take on this. I'm very near sighted, and I notice that the speckle pattern around the light dot from a cheap 650 nm laser doesn't really change in sharpness even if I'm looking at it from many feet away without glasses where I cannot actually see such detail.

If the pattern is caused by interference, then the interference must be happening in my eye. But I don't think that's what's happening. There are three types of cone cells (the color sensitive photoreceptors) in our eyes. Only one type is sensitive to 650 nm light. This is what the distribution of those red-sensitive cells looks like:


Since the 650 nm light only triggers the red receptors, I think what I'm really seeing is the distribution of those receptors. To help verify this, I checked if my camera captures the speckling. It does not, even though it can capture other patterns that aren't from laser light and are just as fine, because it uses a different method of distinguishing colors.

  • $\begingroup$ Here you were talking about the subjective speckle, so you should not expect to see the the same thing and the camera sees. I think a lot depends on the camera aperture and the relative size of the sensor pixels. With your theory, taking a narrow band non-laser light should be also perceived as a speckle pattern by your only red cones. $\endgroup$ – Nazar Aug 4 '17 at 13:00
  • $\begingroup$ What could be used as a narrow band red light source in the 650 nm and longer wavelength range? Below 650 nm, the green sensitive cone cells start picking it up: upload.wikimedia.org/wikipedia/commons/thumb/1/1e/… $\endgroup$ – John Aug 4 '17 at 19:57

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