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This is probably a naive question :) I was looking at a far off street light (also applies to a star) and thinking about the spherical EM wave it was emitting. From afar, the wave-front will be pretty much planar, right? Shouldn't that mean that the EM field should fill up my eyes? In other words, why do we see the object with a specific angular size while the wave we perceive has the same intensity over the area of our eyes?

I hope my question makes sense :)

Best M.

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We'll assume the star/lamp is far enough away that the light rays coming from it are effectively parallel. start by replacing your eye with a sheet of photographic film, or a CCD or something similar.

Film

As you say, the light intensity will be constant over the whole area of the film.

In your eye the equivalent of the sheet of film is the retina, but your eye contains a lens that focusses the light. So the light rays now behave like this:

Eye

A lens bends all parallel rays so they converge in spot at the focal length $f$, and this is where your retina is. So the distant lamp/star does produce a uniform illumination over the whole of the lens, but the lens brings all this light to a single spot on the retina. That's why you see the star/lamp as a bright spot not a uniform wash of light.

John Rennie
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  • The cornea does most of the focussing. The lens in @JohnRennie 's answer is the effective lens of the optical system (cornea and lens & various fluids) of the eye. – Farcher Sep 07 '16 at 09:30
  • @Farcher: yes, agreed. I've presented a description that is simplified for clarity. – John Rennie Sep 07 '16 at 09:31
  • If the eye cannot focus effectively then the incoming light is spread out over more of the retina and in an extreme case may well illuminate most of the retina. This is then no different than looking at the light scattered from a sheet of white paper the paper being the receiver of the light equivalent to the retina with no focussing mechanism in front of it. – Farcher Sep 07 '16 at 09:35
  • Thanks @JohnRennie, your explanation made my day! You see, I was trying to draw parallels with massive antenna arrays and wanted to understand the differences/similarities. From your explanation, I gather that the angular difference between the incoming, planar, wavefronts emanating from various objects/reflectors will allow the eye to resolve the objects by projecting them in different spots. – malekcellier Sep 08 '16 at 07:38
  • @user1713952: What arrays record is the Fourier transform of the image. If the image really was a point the array would just record a uniform intensity. However when objects have a non-zero angular size, no matter how small, the array records a non-uniform distribution of light and when you Fourier transform this you get the image. This approach is called Fourier optics. – John Rennie Sep 08 '16 at 07:44
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If our eyes were just big photosensitive plates, then yeah. You'd be facing directly towards the light, the whole wavefront hits/excites the whole photosensitive plate at the same time, and the whole thing is excited. This would make everything look like a blur, because all light from all directions would be present everywhere on your eye.

Instead, our eye uses lenses. Ideally, you want a lens to behave as follows: If you have an infinitely far away object, the lens focuses its parallel rays (or, alternatively, flat wavefront) into a point behind the lens - the focus. In this way the lens turns a direction into a point. Parallel rays coming from a single direction all get focused onto that point.

This sounds magical, that everything in a flat wavefront coming from a given direction can be mapped onto a point, using as dumb an object as a piece of glass, but that's how it works! Indeed it's a bit too magical. In physics it comes about from the "paraxial ray approximation" - that the light rays are close to the axis of your eye or microscope or whatever, and that they move for the most part straight down the axis. You get lots of complicated effects when you try to account for the fact that light is a wave and that things aren't always paraxial.