Why can eagles see more clearly farther away than we can?
This article will explain how the eye works and why birds of prey such as eagles can see distant objects more clearly further away than we can.
You see the world around you because light scatters off of the objects in your surroundings and enters your eyes. The light from whatever you are looking at directly is focused onto a region of the retina called the fovea (see Figure 1). More specifically onto the center of the fovea called the foveola. Refraction at the air/cornea boundary is responsible for most of the focusing because that is where the light entering your eye experiences the greatest refractive index change from $n_\text{air} = 1.00$ to $n_\text{cornea} = 1.38$. The aqueous humour ($n =1.33$), vitreous humour ($n=1.33$) and lens ($n = 1.38-1.41$ depending on part of lens[note]Cameron, Skofronick, Grant, Physics of the Body, p. 265, Medical Physics Publishing(1992)[/note]) all have similar indices of refraction to the cornea and do not refract the light as strongly. The shape of the lens is changed for final adjustment in focusing.
The fovea is composed of tiny cells called rods and cones that send a signal to your brain through the optic nerve when light hits them. The cones allow us to see colour. There are three different types of cones that are sensitive to green, yellow or violet light[note] Wikipedia. Cone Cell (online). ]http://en.wikipedia.org/wiki/Cone_cell [12 March 2010].[/note]. The foveola contains only cones which are very small in diameter and densely packed. Light from objects that you are not directly looking at enters your eye but is focused on the retina outside the foveal region. Outside the fovea there are less cones and more rods. Rods are cells that are very sensitive to light intensity but not to the colour (wavelength) of light and are slower in response[note] Montag, Ethan. Centre for Imaging Science. Rods and Cones (online). http://www.cis.rit.edu/people/faculty/montag/vandplite/pages/chap_9/ch9p1.html [12 March 2010].[/note]. Rods allow you to see in the dark but you can not see colour very well in the dark.
Figure 1. The human eye. Light entering the eye from two point sources shown is focused onto the retina. The focusing occurs mostly due to refraction at the cornea. Point sources that you are directly looking at get focused onto the fovea. Point sources off to the side are focused outside of the foveal region.
The iris closes and opens depending on the intensity of light in the surroundings. The pupil is the hole formed by the iris through which light passes to go to the retina (Figure 2).
Figure 2. The pupil is the opening through which light goes to your retina. It is formed by the iris which contracts or dilates depending on the intensity of light in your surroundings.
In bright sunlight your pupil is around 1 mm in diameter, whereas it can go up to 8 mm in a dark room[note] Wandell, Brian. Stanford Vision and Imagine Science and Technology. A Brief Organized List (online). http://white.stanford.edu/~brian/numbers/node1.html [12 March 2010].[/note]. Because the pupil acts like a circular aperture light will diffract when passing through the pupil. According to Huygen’s principle each point at the pupil will act as a point source for a secondary wave and the waves from all these points will interfere and form a diffraction pattern at the back of the eye consisting of a central maximum surrounded by fainter rings. Note that except in Figure 3 only the central maximum will be shown for clarity.
Figure 3. Light from a point source is diffracted by the pupil. As a result the light being focused onto the retina is not a perfect point but consists of a central maxima and rings. Note the picture is not to scale.
The angle the central maximum makes from center to edge with respect to your pupil is[note]Knight, R.D , Physics for scientists and engineers: a strategic approach (2nd ed.), p.685, Pearson Addison-Wesley, 2008.[/note]
$$\Theta _1 = \dfrac{1.22 \lambda}{D}$$
where $λ$ is the wavelength of light from the point source (we are assuming the light is monochromatic) and $D$ is the diameter of your pupil. $θ_1$ is in radians, $λ$ and $D$ are in metres. This formula requires an upper level optics course to derive[note] Wikipedia. Airy Disk (online). http://en.wikipedia.org/wiki/Airy_disk [12 March 2010].[/note],[note]Saleh, B.E.A, Teich, M.C., Fundamentals of Photonics, p.138, John Wiley & Sons, 1991.[/note]. Note here that $λ$, the wavelength of light, changes once it enters the eye. The wavelength of light inside the eye is $λ_\text{eye}$ and we will take the index of refraction of the eye to roughly be that of the vitreous humour ($n = 1.33$)[note]Robin Wood Ent. Refraction Index of Various Substances for 3D modelers (online). http://www.robinwood.com/Catalog/Technical/Gen3DTuts/Gen3DPages/RefractionIndexList.html [12 March 2010].[/note] so that
$$\lambda _\text{eye} = \dfrac{\lambda _\text{air}}{1.33}$$
Imagine two green points on a piece of paper. The points act like point sources of green wavelength light. If the paper is placed far enough from your eyes you lose the ability to resolve the points, meaning you can not distinguish if there are two points or just one. The limit where two objects can no longer be resolved is when the center of the central maximum from one of the sources is right on top of the edge of the central maximum from the other source (Figure 4). This statement is called the Rayleigh criterion.
Figure 4. The distance, d, between the centers of the the central maximum determines resolvability. Here r is the radius of the central maxima.
The angle $θ$ that the two sources make with respect to your pupil is equal to the angle that the two resulting central maxima make with respect to your pupil. Rayleigh’s criterion as stated above is equivalent to saying if the angle $θ$ is greater than or equal to $θ_1$ you will be able to resolve the two points (Figure 5).
Figure 5. Light from two point sources enters your eye making an angle $θ$ with respect to your pupil. Two central maxima, due to the diffraction of light entering the pupil, fall on the retina at the back of the eye also subtending an angle $θ$ at your pupil. The angle subtended by a central maximum from center to edge is $θ_1$. According to Rayleigh’s criterion, the point sources are resolvable if $θ ≥ θ_1$.
Birds of prey
Figure 6. Birds of prey such as owls, eagles and hawks have large pupil size compared to humans[note]http://en.wikipedia.org/wiki/File:Eagle_Owl_IMG_9203.JPG [/note].
Birds of prey such as the eagle, owl and hawk are known for their remarkable ability to see clearly at large distances. Their eyes operate along similar principles as our eyes, however these birds have a larger pupil size compared to humans for the same light conditions[note]The Journal of Experimental Biology 211, 2752-2758 [/note].
Having a larger pupil size means that the minimum angle of resolution $θ_1$ is smaller (Figure 7). This means that for a fixed separation between the point sources the points sources can be further away and still resolvable for a larger pupil size.
Figure 7. The angle two point sources make with your eye, $θ$, is determined by the separation between the point sources and the distance to your eye. The minimum angle of resolution, $θ_1$, is inversely proportional to pupil size. For a pupil size (case A) where the point sources are barely resolvable, the point source would still be resolvable for a larger pupil size (case B).
Cone Density
Even if two point sources were resolvable according to Rayleigh’s criteria there is another important issue to consider in order for us to actually be able to resolve the point sources. If the two central maxima from the two point sources fall on the same cone as shown in Figure 8, then we would not be able to distinguish the two point sources as different. This is because a cone only senses that light is hitting it. It cannot distinguish where that light came from. Assuming dense packing so that the cones are touching we need the angle subtended by a cone to be less than the minimum angle of resolution given by Rayleigh’s criterion. In other words we need the diameter of the cone to be less than the diameter of the central maximum. If there is to be no ‘waste’ of cones by having way more than is needed than the diameter of the cone will be in the same range as the diameter of the central maximum and not order of magnitudes smaller.
Digital cameras
Digital cameras work along much the same principle as our eyes. They have an aperture and the light passing through the aperture is focused on an array of pixels which create electrical signals (like our cones). The lens does not change shape like in our eyes but can move back and forth. Cameras have a typical pixel size of 10 μm across[note]Cambridge In Colour. Tutorials: Diffraction and Photography (online). ]http://www.cambridgeincolour.com/tutorials/diffraction-photography.htm[12 March 2010][/note].
Figure 9. The design of a digital camera is similar to the eye[note]Picture of CCD chip from http://en.wikipedia.org/wiki/Charge-coupled_device [/note],[note]Picture of aperture from http://en.wikipedia.org/wiki/Aperture[/note]. The pixels in the CCD chip are analagous to cones in the eye[note]Picture of cones from http://en.wikipedia.org/wiki/Eye credited to Nation Kingdom. Note that this is not actually a picture of cones, but was similar enough and not copyrighted.[/note].
Imagine trying to draw a horse but you only have four squares that you can shade dark or bright. That would not give a very discernible horse but if you had millions of tiny squares then you could make a recognizable horse. More pixels means better resolution. Is more pixels always better? We mentioned how that there is a fundamental limit imposed by physics on being able to resolve two point sources, a camera can not get around that. However if the pixels are so large that the central maximum from the point sources falls on the same pixel then we would benefit from having smaller pixels.
Considerations for choosing how many pixels are needed are that as we discussed previously for a 2.5 mm pupil size our minimum resolution angle for $\lambda_\text{air} = 550 \text{ nm}$ is $θ_1 = 2.0\times 10^{-4} \text{ rad}$. This means that on a photograph if the image from one pixel makes an angle less than this with your eye we would not be able to discern any pixelation. For a larger photograph the image from each pixel makes a larger angle with your eye than for a smaller photograph for the same distance from you to the photograph. The number of pixels you need to produce good quality prints will depend on what size photographs are intended and how far away you plan on the photograph being viewed. To produce a picture as large as your field of view (like your eye sees) you would need about 600 megapixels[note]Clark Vision. Notes on the Resolution and Other Details of the Human Eye (online). http://www.clarkvision.com/articles/human-eye/index.html [12 March 2010][/note].
Interpretation:
Light diffracts as it enters the pupil resulting in a minimum angle of resolution possible. Having a larger pupil size partially accounts for eagles being able to resolve objects at large distances.
Picture of eagle taken from[note]http://en.wikipedia.org/wiki/Eagle[/note].
