This is a very basic question on optics. How are we able to view an object kept in a room with a bulb?

From what I understand, light rays from bulb will hit the object and some colour will be absorbed and rest will be thrown off. but how does the thrown off light reach eyes?

I thought a light ray will reflect with same angle back (as the angle between incident ray and normal) then how is it guaranteed that a reflected ray will reach eyes?

enter image description here


I think your illustration answers your own question fairly well. It's not guaranteed at all that a reflected ray will reach the eye however I'm sure that you can cast a ray between the light source and the eye that bounces off the object. Note that the object you have drawn has perfectly specular reflections, whereas most objects around you will have a degree of surface texture that will result in surface normals that vary wildly over small distances, resulting in a degree of lambertian reflection. In this regime, there are many different paths a photon can take between the light source and the eye. Throw in refraction, subsurface scattering and the rest and you will get a better picture of everyday environments, where light disperses and bounces around everywhere.

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    $\begingroup$ +1 Right. If the object was a good mirror, and the reflection was not to the eye, the object would be invisible. Stealth technology works that way. Don't let the radar get reflected back to the antenna. $\endgroup$ – Mike Dunlavey Oct 8 '11 at 17:03
  • $\begingroup$ It's not obvious to me why in all cases, we should be able to cast a ray between the light source and the eye that bounces off the object, if I have to stick to the same reflection angle(as incidence angle). But, yeah it may be true considering that there will be surface aberrations resulting in many normals etc etc..any more detailed pointer on this? any link etc? Thanks! $\endgroup$ – xyz Oct 8 '11 at 17:07
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    $\begingroup$ @Mike Dunlavey: That's only true if you're against a black background (which is presumably the case in the radar scenario); otherwise reflecting away means you're a black spot on the background. $\endgroup$ – Kevin Reid Oct 9 '11 at 1:58
  • $\begingroup$ @p2pnode: Basically, unless a surface is a good mirror, it will reflect light in many/all directions. A substance like chalk is more like a Lambertian Reflector. Most surfaces are to some degree more like a mirror, to some degree more like chalk, and to some degree more like soot (not reflective). $\endgroup$ – Mike Dunlavey Oct 9 '11 at 3:03

The other answers have covered that most objects scatter light (diffuse reflection) and therefore are visible similarly from any angle. But also consider a perfectly shiny object, with only specular reflection, as you assumed in your original question. It would not be invisible! Consider a small sphere, such as a ball-bearing ball. It looks “shiny”. But if you examine it carefully, you will see that its appearance is a distorted image of the room around it, whose walls are probably diffuse-reflective.

A useful tool for thinking about this is to recognize that it's possible to follow the path of a hypothetical light-ray backwards from your eye through many reflections. If the ray ends at a light source, then you must be seeing light in that direction; if the ray ends at something absorbing then you aren't. In computer graphics, this is known as ray tracing.

Let's eliminate all diffuse reflection. Suppose you (and an ideal lamp) were in a room with perfectly mirrored (i.e. specular) walls and a mirrored sphere. Furthermore, suppose that you absorb all light — you are perfectly black. What will you see?

The space around you will seem to be an infinite number of images of everything in the room (the "house of mirrors" effect). What you see will will be either perfectly white (for backwards rays which reach the lamp) or black (for those which reach your body).

However, you will still be able to recognize the mirrored sphere, because of its characteristic distortion of its reflected image! The edges will be especially visible, because the angle of reflection varies especially rapidly there.

On the other hand, if the walls were a single color and perfectly uniform brightness, you would only see a single spot on the sphere, the reflection of the lamp, and all else would be the same black as the walls. And if it was a box instead of a sphere, you would only see the reflection if you held it at the right angle — the box would be invisible by way of being the same color as the walls it is reflecting.

Here's a ray-traced simulation of the first scene I described. The light is a sphere and "your body" is a box. All reflections in this image have the angle of incidence equal to the angle of reflection.


For the sake of editability, here is the POV-Ray scene file which produced the above image:

#include "colors.inc"
global_settings { max_trace_level 30 }
#declare Specular = finish {
  ambient 0
  reflection 1.0
  diffuse 0
  brilliance 0
camera {
  location  <-1, 1, -7.0>
  look_at   <0, 0, 0>
box {
  <-1,-4,-1>, <1,1,1>
  finish { reflection 0 diffuse 0 brilliance 0 }
  translate <-1, 1, 7.0>
sphere {
  <0,4,0>, 2
  pigment { color White }
  finish { ambient White }
plane { +y, -10 texture {Specular} }
plane { -y, -10 texture {Specular} }
plane { +x, -10 texture {Specular} }
plane { -x, -10 texture {Specular} }
plane { +z, -10 texture {Specular} }
plane { -z, -10 texture {Specular} }
sphere { <2,0,0>,1 texture {Specular} }
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  • $\begingroup$ ""But also consider a perfectly shiny object, with only specular reflection, as you assumed in your original question. It would not be invisible!"" Such objects are in fact invisible! You dont see a mirror, You see the reflected image. $\endgroup$ – Georg Oct 9 '11 at 11:07
  • $\begingroup$ @Georg: I disagree, but we are disagreeing about defining “invisible”, not about the physics of the matter. $\endgroup$ – Kevin Reid Oct 9 '11 at 11:19
  • $\begingroup$ I agree to this reason to disagree :=) $\endgroup$ – Georg Oct 9 '11 at 11:21
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    $\begingroup$ Cool simulation. Well done. $\endgroup$ – Marty Green Oct 9 '11 at 11:22

Most surfaces scatter light, for two reasons. First, they're not flat, so at each point that a ray hits, there will be some degree of divergence. Second, electromagnetic scattering is strong in most materials. So rather than be absorbed, pass through, or reflect, the light will often be scattered in a new direction.

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Usually objects reflect the light in all directions, that's why you see them from any angle. This is called a diffuse scattering.

OTOH some surfaces tend to scatter most of the light in specific directions only. For instance metallic surfaces reflect the light as you drew in the picture. This is called specular scattering, and such surfaces are called glossy. This type of scattering is observed due to a crystalline structure of the matter.

So that most surfaces scatter the light in all directions, whereas the intensity of the scattered light depends on the angle, and it's usually the highest for the angle that you drew. The extremal cases are:

  • Matt surfaces, gas/mist and etc. Those have no explicit specular scattering.
  • Ideal flat metallic surfaces (mirror). Those have no diffuse scattering.

BTW, the bulb will not reveal a mirror.

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  • $\begingroup$ ""and such surfaces are called glossy. This type of scattering is observed due to a crystalline structure of the matter."" Nonsense $\endgroup$ – Georg Oct 9 '11 at 11:05
  • $\begingroup$ @Georg: You're right, crystalline structure is not directly related to the phenomenon. I confused with the X-Ray scattering (Bragg's planes). At the wavelength of the visible light the things are simpler: it's just due to the conducting electrons $\endgroup$ – valdo Oct 9 '11 at 14:00

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