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## Spherical Lenses

A "spherical lens" is a lens whose surface has the shape of (part of) the surface of a sphere.  On this page, we will determine some values for the focal length of a spherical lens.

 Figure 1 -- Spherical Aberration A spherical lens doesn't actually bring parallel rays to a common focus (see figure 1).  Rather, the "focus" of the lens depends on the distance from the distance from the center of the lens, with light passing through the edges of the lens coming to a focus closer to the lens than light passing through the center.  This produces the phenomenon of "focus shift", which is well known to photographers:  If your lens suffers from severe "focus shift", and you focus the image with the diaphragm wide open, and then stop down to take the picture, the picture won't be in focus.  With the lens opened wide the image is dominated by light passing through the lens far from the center; consequently, it's the edge rays which you've adjusted the lens to bring to a sharp focus.  The center rays, on the other hand, are coming to a focus behind the film plane; when the lens is stopped down, only those (out of focus) center rays are left.

On the remainder of this page, we'll find an exact formula for the focal length of a spherical lens, and we'll then find an approximate formula for the focal length at the center of the lens.  This problem is somewhat messy, and we're going to break the problem down into easier to handle pieces by cutting the lens in half.  We'll find the focal length of each half lens and then put them together to get the focal length of a convex-convex spherical lens.

### Part I:  Plano-Convex Half-Lens

 Figure 2 -- Plano-Convex Half Lens We'll start by looking at a lens which is spherical on one surface and flat on the other, with parallel rays of light entering on the flat side.  Since the light is entering perpendicular to the flat surface, we can ignore the effect of the planar side of the lens, and just analyze the convex surface.

Before we begin, we should point out something that may not be obvious:  This is not an ideal lens, and the focal length when parallel rays enter the flat side may not be identical to the focal length when parallel rays enter the convex side!  The light is following different paths in those two cases -- the paths have not just been reversed -- and there is no a priori guarantee that the results will be simple mirror images of each other.  Obviously the case where the light enters the flat side is simpler to analyze, as we only need to deal with a single refracting surface, and that is where we are starting.

In figure 3 we've shown the lens as a full half-sphere, with the center of the sphere at the origin.  The flat side of the lens cross section lies on the x axis, and the axis of the lens lies on the y axis.  For each ray passing through the lens, we can define the "focal point" of that particular ray as being the point at which the ray crosses the axis of the lens.

 Figure 3 -- Path Followed by a Single Ray In figure 3 we've shown a single ray passing through the lens, with its focal point at the point marked "focus".  We'll now determine how far from the lens the focal point lies.

A real lens has nonzero thickness, unlike an ideal lens, and so the "focal length" of a real lens is a little ambiguous.  In figure 3 we shown two "focal lengths":  f is the distance from the front of the lens to the focus, and F is the distance from the center of the sphere to the focus.

The point at which the ray passes through the surface of the lens is (x,y).  It's at distance R from the origin (since the lens surface is spherical).  The radius line, labeled R, is perpendicular to the surface of the lens at (x,y).  We've also shown the line tangent to the lens surface at (x,y); that line is, of course, perpendicular to the radius line.

We can find y as a function of x.  We want to find f as a function of x.  To do that, we'll first find the tangent of angle ξ, and then use that to find the length marked F - |y| in figure 3; from that we can determine both f and F.

The ray hits the surface of the lens at angle θ0 with the line perpendicular to the surface, which is also the angle between the radius line and the y axis, and is also the angle between the line tangent to the lens and a horizontal line.  We can read off directly:

1) The ray emerges from the lens at angle θ1 with a line perpendicular to the surface of the lens.  In figure 3, we've shown the refractive indices of the lens and the surrounding environment as Nl and Ne.  From Snell's Law, we have

2) We can see from the figure that  ξ = θ1 - θ0.  From that and the formula for the sine of the sum of two angles (which is derived, somewhat haphazardly, here), we have

3) which we can clean up a bit to produce

 4) Note that this is not defined for all possible values of x.  If x>R then the first square root is imaginary.  That's not a problem; x>R implies we're looking at a point outside the lens!

However, if x/R > Ne/Nl, then the second square root is imaginary.  What does that mean? In that case sin(ξ) is also imaginary.  We can't have light emerging from the lens at an imaginary angle, of course.  What's going on is that the ray is hitting the inside of the lens at a steeper than critical angle, and it's being totally reflected inside the lens.  Once we exceed the angle of TIR nothing comes out -- the surface of the lens acts as a perfect mirror.  At the critical angle, sin(θ1)=1, and the ray comes out traveling tangent to the lens surface; beyond the critical angle the ray would actually be bent back inside the lens.  That doesn't happen of course; instead it's just reflected.

We'll refer to the value of x/R at which the ray hits the lens surface at the critical angle as the cutoff value for x.  The effective diameter of the lens is actually the cutoff value; the parts of the lens outside that distance aren't doing anything for us.

We've found sin(ξ).  What we want, however, is the tangent of ξ.  We won't write that formula out in full; we'll just say:

5) which is all we'll actually need later on.

Finally, looking back at figure 3, we can find the value of f from the triangle with (x,y) at its upper vertex and the focus at one of its lower vertices:

6) By dividing through by R we obtain a slightly more useful form, as it doesn't depend on units we use to measure the diameter of the lens:

7) Using a Python program (here) we've found f/R between x=0 and the point at which cutoff occurs.  In figure 4 we show a plot of f/R versus x/R for a range of values of N=Nl/Ne running from 1.1 to 5.  As N approaches 1 the focal length (at the lens center) goes to infinity.  As N approaches infinity, the focal length approaches 0.  At all focal lengths, as we move farther from the lens center the focal length decreases.

 Figure 4: Figure 4 makes it clear that the focal length varies with the distance from the lens center.  It also shows graphically the effect of lens cutoff, which limits the maximum size of lens of this sort which we can actually build.  For glass with a refractive index of roughly 1.5, we see that the focal length at the center of the lens is about twice the radius of curvature, and cutoff is at about 0.67 times the radius of curvature.  The average focal length of the lens will be quite a bit shorter than the focal length at the center, and probably falls between 1 and 1.5 times the radius of curvature.  The lens "speed", as an F number, is the focal length divided by the diameter.  If the average focal length of our "maximum" lens is as small as 1, with cutoff at about 0.67 the lens size will be about F/0.75.

Finally, we'll find the focal length at the center of the lens, by taking the limit of everything as x goes to zero.  From equation (4) we can see that

8) We define

9) Note that, for the situations we're considering, n≥1.
We can write now (8) as

10) As ξ goes to zero, tan(ξ) approaches sin(ξ), and we can finally write

 11) That's  the focal length at the center of the lens, and, as we shall see later, it is also twice the (approximate) focal length of any small convex-convex lens, such as a magnifying glass.

As we've also seen, the cutoff angle is the point where sin(θ1) = 1.  From equation (2) we see that this is:

 12) And finally, we'll find the focal length at the cutoff point.  From (4) and (12) we have:

13) and then from (5) we have:

14) and finally, from (7) and (14), we obtain

15) ### Part II:  Infinitely Thick Convex Half Lens

 Figure 2-1: Convex Half-lens We're now going to reverse the lens, and find the focal length when parallel rays enter the convex surface.  However, we're not going to allow the rays to come out through the flat side -- we're going to assume the lens is so thick that the rays come to a focus inside the glass.  In other words, we're finding the focal length when the light enters an arbitrarily thick convex half-lens.

We'll find this result useful by itself when we eventually look at how eyes work, and we'll also find it interesting to contrast the result we obtain here with the result we obtained in Part I.

In figure 2-2 we've shown the lens as a full half-sphere, with the center of the sphere at the origin.  The axis of the lens lies on the y axis.  For each ray passing through the lens, we can define the "focal point" of that particular ray as being the point at which the ray crosses the axis of the lens.

 Figure 2-2:  Path followed by a single ray In figure 2-2 we've shown a single ray passing through the lens, with its focal point at the point marked "focus".  We'll now determine how far from the surface of the lens the focal point lies.  We'll proceed very much as we did in Part I above.

We will continue to define n as in equation (9) above:

2-1) From figure 2-2 we can see that

2-2) From Snell's Law we have

2-3) From figure 2-2, where ξ is the angle between the ray and the y axis after it enters the lens, we have

2-4) And so we have

2-5) which, using (2-1), we can write as

 2-6) With equation (5) that gives us tan(ξ).  From figure 2-2 we can see that

2-7) or, in terms of x/R,

 2-8) As we did with light from the other direction, we've used a Python program (here) to find the focal length for all distances from the center of the lens for a number of refractive indices.  We once again see that, as we move away from the center of the lens, the focal length decreases. Note that, for all refractive indices, the plot extends all the way to the edge of the lens (or it should; the program seems to have left off the last point on each curve).  There is no cutoff when the light enters the curved side of the lens.  When moving from a less dense medium to a more dense medium there is no "critical angle" -- light can enter the denser medium at any angle.

Finally, we'll find a formula for the focal length near the center of the lens.  First we find the limit of sin(ξ) as we approach the center of the lens by dropping out the terms in x2:

2-9) We know that, as x approaches 0, tan(ξ) approaches sin(ξ).  So, we also have

 2-10) ### The Non-Ideal Nature of a Single Air-Glass Interface

Very near the center, the curvature of the lens can be fully characterized by the second derivative of the  function which describes its surface.  In other words, very near the center, it doesn't matter what the actual figure of the lens is; viewed through a sufficiently small "pupil" all lenses act like spherical lenses.  Consequently, results obtained for the limiting case of a small lens will apply to all physical lenses, not just spherical ones.

We have just seen a clear demonstration that a single air-glass interface can not form an ideal lens.  With an ideal lens, rays traveling parallel to the lens axis are brought to a single focus at a fixed distance from the lens.  But when we consider an air-glass interface which has been formed into a lens, we realize that's not true:  Comparing equations (11) and (2-10) we see that parallel rays passing into the glass are brought to a focus n times farther from the lens than parallel rays passing out of the glass.  (This doesn't violate the "reversibility of light paths" because the paths followed in the cases are very different -- parallel rays passing in are traveling parallel in the air and are converging in the glass.  Parallel rays passing out are traveling parallel in the glass, and converging in the air.  The "reverse" of the first case is a point source located in the glass being focused to parallel rays as it passes into the air, and the "reverse" of the second case would be a point source in the air being focused to parallel rays as it passes into the glass, and in these cases the light would indeed precisely "backtrack" the rays in the corresponding "converging" case.)

Furthermore, as we have shown here, rays passing through an ideal lens are deflected according to the formula

2-11) where x is the distance from the lens center.  In particular, this means rays passing through the center of the lens are not deflected.  That's patently false for a single air-glass surface, so it should not surprise us that such a surface can't form an ideal lens.

As we shall see later, the story is quite different when there are two glass-air surfaces, as in an ordinary camera lens, or a magnifying glass.

### Part III:  Combining Two Half-Lenses

We've found two different values for the focal length of half of a lens.  We'd now like to combine them by gluing two half lenses together.

 Figure 3-1:  Two lenses back to back For us to say anything intelligent about the result of our "glue job", the light passing through the lens must follow paths we've already analyzed.  In particular, for light coming out of the lens (passing out through the "second" surface), the light must either be traveling parallel to the lens axis (figure 2), which is what we analyzed in Part I, or it must be following lines leading directly away from the "inside" focus of the lens (figure 2-1, with arrows reversed), which is the path we analyzed in Part II.  We can exhibit one case in which such a glued up lens can be understood with the results of Part I:  If we place a point source at the focus of a half-lens, light emerging from the lens will be traveling parallel to the lens axis;  if that light then enters a second half-lens, it will converge on the focus of that half-lens.  This is identical to the case in which two ideal lenses are placed next to each other and light originating at the focus of one travels to the focus of the other.  We have shown this arrangement in figure 3-1.

In this case we have an object (point) at distance f in front of the lens, and it's forming an image at the same distance  behind the lens.  To the extent that this combination is acting as an ideal lens, this implies that the focal length of the lens must be f/2 (see our ideal lens images page, case 1).   Note, however, that we have ignored the thickness of the lens body, which is going to throw this result off a bit -- and we've also ignored the fact that a spherical lens hasn't got a particularly well defined focal length to start with.  This result will hold for small, thin spherical lenses, but really thick spherical lenses will be far from ideal.

Applying this to equation (11), and recalling our definition of n from equation (9), we conclude that the focal length of a small, symmetric convex-convex spherical lens must be

 3-1) As we will see later, this matches the focal length of a tiny convex lens, which gives us confidence that the model here is accurate.

### A Small, Symmetric Convex Lens

Of course, when we glue together two large spherical lenses, the result is going to have severe "spherical aberration":  The focal length will be shorter near the edge than it is in the center, with the result that parallel rays falling on the entire lens won't be brought to a precise focus.  However, if we glue together two small lenses, then the result will bring light traveling parallel to the axis to a (nearly) precise focus.  But will such a lens actually behave as an ideal lens does?  Will it focus a distant scene onto a flat focal plane?  To answer this question we need to know more about how the lens will focus light which is not arriving parallel to the axis.

For scenes which are nearly on-axis, the answer is yes, it will behave as an ideal lens.  For scenes which extend a substantial distance off-axis the answer is more complex.  The analysis is somewhat lengthy, however, and I've moved it off to a separate page on tiny convex lenses.

Page created on 01/15/2009.  Last updated 1/18/2009.