# Calculators

## Optical Lens Calculator

In science fiction movies, we often hear the phrase "visual range". Visual range is limited by optical characteristics, including diffraction. Light diffracts when it passes through an aperture (such as a telescope lens), and this puts limits on the clarity with which pictures or videos can be taken.

So how far is "visual range"? We can determine this by using something called Rayleigh's Criterion. The first minimum for the diffraction pattern of light with wavelength λ passing through a circular aperture of diameter d (assuming Fraunhofer conditions) is given by the simple equation:

 sin θ = 1.22λ d

This means that two distant dots will be marginally resolvable from each other if they are separated by the angle θ, which is known as Rayleigh's Criterion. At a lesser angle, they will appear to merge into a single dot, while at a larger angle, they will appear as two distinct dots. The appearance of two dots at the Rayleigh's Criterion separation angle is that of two dots half-overlapping onto each other. You can use this as a lower limit for pixel separation in a computer image; if the pixels are any closer together, they overlap and the added resolution becomes useless.

The wavelength of visible light is from roughly 400 nm to 700 nm, so 550 nm is a useful average approximation. Therefore, using this criterion, we can compute the maximum resolution of useful image that one could acquire of an object of known width, using a lens of any given diameter, at any given distance, assuming the apparatus itself is perfect and all other optical conditions are also perfect (obviously an idealized assumption). In order to further simplify the equation, we note that the angle is very small. At small angles, sin θ is approximately equal to θ (if we use radians), so the equation becomes even simpler:

 θ = 1.22 λ d

For example, given a lens of 10mm diameter and a visible light wavelength of 550 nm, Rayleigh's criterion would tell us that θ = 1.22(550E-9)/(10E-3) = 6.71E-5 rad. At a distance of 10 metres, this means you would be able to resolve two dots which are 0.671mm apart, or you could take a 1000x1000 pixel digital picture of an object which is 671mm wide (remember, this is for an imaginary perfect camera in perfect optical conditions; obviously, a realistic camera under realistic conditions would be far less effective). As an aside, this is why astronomers like to use large telescopes. If we increased the lens diameter in the previous example by ten times, we would find that the Rayleigh Criterion angle becomes one tenth as large, thus allowing us to take pictures of objects ten times farther away with equal resolution.

With the above in mind, this calculator can be used to determine the maximum visual range where you can take a picture of a given resolution (in pixels) of an object of fixed width (in metres). You can also adjust the lens diameter and the light wavelength. The default values are set to get a 4000 pixel wide picture of an object the size of the Moon:

## Input Values

 Lens diameter (metres) Wavelength (nm) Width of target (metres) Desired resolution (pixel width)

## Calculated Values

 Minimum resolvable angle (rad) Maximum range (km)

Notes:

• It may be instructive to see what your maximum visual range would be to get similarly sharp pictures (or even pictures one tenth as sharp) of a much smaller object such as a spaceship. You may find that a startlingly large lens is required in order to achieve the kind of range you want.
• One may argue that it is possible to detect objects at much greater range than this calculator would suggest. This is true, but only because in order to detect an object, we need only identify a single light source. That is much easier than acquiring a picture with visual details.
• It cannot be stressed enough that this calculator assumes perfect performance, so realistic ranges would be smaller.
• Optical interferometers can be used to create a "virtual telescope" of sorts, by combining multiple light streams from multiple telescopes and then using computers to reconstruct an image from the resulting interference patterns (look up "thin slit diffraction" in your physics text). In that case, you would use the radius of the telescope array instead of the radius of the individual telescopes. However, this method presents many new difficulties: the sensitivity is orders of magnitude lower when compared to a conventional telescope (no small problem), and individual array elements must be aligned so precisely that the incoming light will be in phase; since light wavelengths are measured in nanometres, this requires nanometre-scale positioning accuracy.