8
\$\begingroup\$

In the context of practical implications for photography, is there a general difference in the qualities of emitted vs. reflected light (other than emitted light being brighter in many situations)?

I am not talking about a specific light source and how its light changes when reflected on a specific object, but the two concepts, for general understanding of light.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ It also matter on which object the light is bouncing off. If its non metallic light will be bounced off in all directions. \$\endgroup\$
    – GoodSp33d
    Feb 11, 2013 at 11:03
  • \$\begingroup\$ @Stroker: Not true. Consider light reflected from a polished plastic surface or a glass window. \$\endgroup\$ Feb 11, 2013 at 13:56
  • 2
    \$\begingroup\$ Pretty much all materials have diffuse and specular components, and the mixture is a function of wavelength. That's why researchers model materials with a complex BRDF distribution function. Check out graphics.tu-bs.de/media/publications/BergerTRriBRDF.pdf and ipgp.fr/~jacquemoud/publications/bousquet2005a.pdf \$\endgroup\$ Feb 11, 2013 at 16:38
  • 3
    \$\begingroup\$ I think we better just leave this here. Better to not play ping pong with another site. There are certainly valuable answers that relate to photography for this question, and those could be of value to our community. \$\endgroup\$
    – jrista
    Feb 11, 2013 at 20:59
  • 1
    \$\begingroup\$ Emitted light will have dark bands on the absorption frequencies of the material. Reflected light does not affect absorption bands, keeping whatever is there from the emitter. \$\endgroup\$ Feb 14, 2013 at 1:57

7 Answers 7

10
\$\begingroup\$

With emitted light you work in the additive colour domain, and reflected light is subtractive colours. See the differences in RGB (screens) and CMYK (printers), e.g. to see yellow you can emit R+G, or subtract B.

Emitting a colour is easier to control than subtracting, because reflected light depends on the product of the (emitting) light source and the material BRDF.

BRDF in General

BRDF applied to leaves, where it is interesting how visible light and NIR light acts differently, which is useful in agricultural engineering. I used this theory myself to successfully make an algorithm that can detect scab on apple leaves.

The strength is controllable in both cases, so you can't see that emitted light is stronger. The perceived strength is a function of the wattage you burn off from the light source, distance and spread, and the integral in your viewer sensitivity spectrum of the light spectrum. This is why LED light seems stronger for the same amount of wattage as a halogen light. The halogen does emit more light, but a lot of it is outside the visible spectrum, and is thus not integrated.

If your reflective surface is very diffuse, you get more spread , and are less likely to see a imprint of the light source. this is the reason diffuse lighting is easier to achieve by emitting light into a dome with a white diffuse coating. Perfect white is hard to achieve so this does have some loss, and the distance form the emitting light to the subject will get longer, too. so to get the same strength as you'd get with emitting light directly, you increase the power. You can also reflect the light on a mirror surface and achieve a longer distance, without polarisation or diffusing it. This is often used to increase the FOV of the light source and the physical dimension does not allow it.

Polarisation you can control in both cases. You can get a polariser sheet to place in front of your light. Or you can carefully choose your reflective surface to give the polarisation you want.

\$\endgroup\$
1
  • \$\begingroup\$ This answer refers more to the differences in area of the sample rather than the nature of the eminent electromagnetic radiation (the light). There are specular emitters and reflectors. There are diffuse emitters and reflectors. Reference to the action of the eminent is also not part of the discussion about the difference but of our manipulation of it into meaningful imaging. Detailed research on the nature of the surface is not discussing the nature of the light itself. \$\endgroup\$
    – Stan
    Jun 11, 2016 at 23:33
5
\$\begingroup\$

No, light is still light, it doesn't get different just because it is reflected.

For some reflections, like in a glass surface or a water surface, the reflected light will be polarised, but that is specific to that kind of reflection, that doesn't happen for all reflected light.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ When light reflects off a material denser (say glass) than the external medium(say air), it undergoes a polarity inversion. \$\endgroup\$
    – GoodSp33d
    Feb 11, 2013 at 11:01
  • \$\begingroup\$ True, light is still light, but I think the notion of polarization is key. You touched on that, but maybe some further clarification and the potential benefits & consequences to photography (i.e. why a polarizing filter works when reflected light becomes polarized?) \$\endgroup\$
    – jrista
    Feb 11, 2013 at 21:00
  • \$\begingroup\$ @2-Stroker: Yes, but if I understand it right, that only matters if the light source was already polarised, and you use a polarising filter on the camera. Otherwise the polarity doesn't make a difference. \$\endgroup\$
    – Guffa
    Feb 11, 2013 at 22:06
2
\$\begingroup\$

One big difference is that emitted light comes from a point source generally -- the filament of a lamp, or an LED, or the sun -- and reflected light, assuming a non-shiny surface, does not have a point source (unless it's a perfect reflection of a point light source, such as a lamp in a mirror). This can produce different types of shadow edges and different glare effects, which very diffuse reflected light would probably not produce. There can even be prismatic effects from a point light source shining through various materials (water, glass). Overall, it's like the difference between shooting in direct sunlight and shooting on an overcast day.

\$\endgroup\$
2
\$\begingroup\$

There are only two different "kinds" of light that we now know about.

The first is divergent light that is relatively random and what our eyes have evolved to use to interpret the world around us. It obeys the "inverse-square law."

The second is monochromatic, collimated light, which is highly parallel and does not comply with the "inverse-square law." Lasers produce this type of "light."

\$\endgroup\$
1
\$\begingroup\$

It may be polarized and it may be a different color as some light may be absorbed during the reflection, but it is otherwise still light. Reflected light is often used in the photographic sense to provide diffusion and indirect lighting (such as a bump flash). The biggest characteristic change in terms of lighting comes from the area the light is emitted from as a point light will produce hard shadows while a diffused or area light will produce soft shadows and ambient light will produce no shadows (or almost no shadows in practice since there is basically always some directional light).

\$\endgroup\$
0
\$\begingroup\$

Reflected light tends to be diffuse, where emitted light can be either diffuse (with appropriate filter), or spot.

\$\endgroup\$
0
\$\begingroup\$

Transmitted light can have colors that can't be reproduced by reflected light.

Practical implications:

Prints have different color gamut than computer displays, so images often do not look the same on print and on display

Digital camera calibration that uses reflective color targets does not cover the whole color range

You are probably more likely to get out of gamut colors (areas of identical color with no detail) with transmitted light like light shining through petals of a flower, traffic lights etc.

\$\endgroup\$
1
  • \$\begingroup\$ Mirrors can reflect every colour that is shown upon it. You have chosen one instance of reflected light. \$\endgroup\$
    – Stan
    Jun 13, 2016 at 4:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.