20130113

Presentation: two-lens systems

In the previous presentation we considered vision problems caused by defects in the curvature (and focal length) of the eye. By augmenting the eye with a second lens, here either using a contact lens...

...or glasses, we can compensate for these common vision defects.

This will require us to analyze the two lens systems of a contact lens and the eye, or glasses and the eye using a two-step approach. As a result, we will compress four years of post-graduate optometry school into this presentation, and be able to prescribe corrective optics for common vision defects.

Here we have light passing not just through one lens, but through two lenses. You know, double the trouble, twice the fun.

For any two lens system, the main idea is to just take it one lens at a time. The object 1 in front of lens 1 will produce an (intermediate) image 1, and the thin lens equation is applied to relate do1, di1, and f1.

This image 1 will then become the object 2 for lens 2, which will produce the (final) image 2, and the thin lens equation is again applied to relate do2, di2, and f2.

Keep in mind that the intermediate image 1 from the first lens is subsequently "fed" to the second lens as its object 2.

First, applying the two-step model to contacts and eyes.

The contact lens is the first lens, and the eye is the second lens in this two lens system. The (final) image 2 produced by the eye is a real image on the retina, on back of the eye. If the eye suffers from a visual defect, then by itself it will only be able to see an object 2 located at a distance do2. If used correctly, a contact lens will take an object 1 (located at a nominal distance do1), and produce an image 1 that will be located at a distance do2 = di1 that the eye is able to focus on.

So knowing the nominal object distance do1, and the actual distance do2 an uncorrected eye can focus on, the focal length f1 of the contact lens can be solved for using only one thin lens equation.

The focal length of contact lenses are typically not specified, instead they are rated in terms of refractive power P "diopters," which are merely the inverse of the focal length f (measured in meters).

The sign convention for refractive power is the same as for focal lengths: positive values for converging contact lenses, and negative values for diverging contact lenses.

If you have a prescription from your optometrist, or the box that your contact lenses came in, check out the refractive power P ("diopters") listed there.

Second, applying this two-step model to prescribe contact lenses to correct common vision defects.

We only need to measure your actual far point and your actual near point. If the measured far point is less than the nominal value of infinity (i.e., some measurably finite value), and/or the measured near point is greater than the nominal value of 25 cm...

...then you will need corrective optics, which we will solve for using the thin lens equation.

Let's do an example for a myopic Physics 205B student, whose uncorrected far point is 5.0 m. This is the farthest distance do2 = +5.0 m (positive object distances correspond to being in front of the eye) that this student's unaided eye can focus on, instead of the nominal farthest distance of +∞.

What this means is that the contact lens will take an object located (in front of it) at do1 = +∞, and produce a virtual image (in front of it!) at di1 = –5.0 m. This virtual image 1 becomes the object 2 (do2 = +5.0 m) for the eye, which is able to see 5.0 m in front of it.

Solving for the focal length f1 of the contact lens,

(1/do1) + (1/di1) = (1/f1),

(1/(+∞)) + (1/(–5.0 m)) = (1/f1),

f1 = –5.0 m,

as (1/(+∞)) = 0. Remember that contact lenses are prescribed in diopters rather than focal lengths, such that the refractive power is:

P = (1/f1) = (1/(–5.0 m)) = –0.20 m–1 or –0.20 D.

Note that the negative focal length (and refractive power) value means that contact lenses (and glasses) to correct for myopia are diverging lenses, which take distant objects, and make closer, upright virtual images, which is then the object for the eye, which can see it at this closer distance. (Which ray tracing(s) best match this?)

Now let's do an example for a hyperopic Physics 205B student, whose uncorrected near point is 0.60 m. This is the nearest distance do2 = +0.60 m (positive object distances correspond to being in front of the eye) that this student's unaided eye can focus on, instead of the nominal reading distance of 0.25 m.

What this means is that the contact lens will take an object located (in front of it) at do1 = +0.25 m, and produce a virtual image (in front of it!) at di1 = –0.60 m. This virtual image 1 becomes the object 2 (do2 = +0.60 m) for the eye, which is able to see 0.60 m in front of it.

Solving for the focal length f1 of the contact lens,

(1/do1) + (1/di1) = (1/f1),

(1/(+0.25 m)) + (1/(–0.60 m)) = (1/f1),

f1 = +0.43 m,

Remember that contact lenses are prescribed in diopters rather than focal lengths, such that the refractive power is:

P = (1/f1) = (1/(+0.43 m)) = +2.3 m–1 or +2.3 D.

Note that the positive focal length (and refractive power) value means that contact lenses (and glasses) to correct for hyperopia are converging lenses, which take nearby objects, and make farther away, upright virtual images, which is then the object for the eye, which can see it at this farther distance. (Which ray tracing(s) best match this?)

Now what happens if a Physics 205B student is myopic (nearsighted, can see near, but can't see far) due to defects in curvature of the eye, who as a result of aging develops presbyopia, losing the ability to accommodate and see nearby objects? You would need to prescribe separate diverging and converging lenses to correct for myopia and presbyopia...

...or these two lenses could be combined into bifocal glasses, with the compromise that looking down (at reading distances) would have a diverging lens which only corrects for presbyopia, and looking straight (for far distances) would have a converging lens which only corrects for myopia.