If you look at the lenses that we drew in the beginning of the module you will see why the rays that emerge from this lens diverge from each other, instead of converging. We defined divergence as negative vergence. The lenses that we looked at earlier in this lesson added plus vergence to the rays entering it. This lens will add negative vergence to the rays entering it.

In this case the rays emerge from the lens diverging as if they had come from the focal point, where in the other lens they actually crossed at the focal point. In the lens that added plus vergence the focal point was real because the rays actually went through that point. In this new case the focal point is virtual because, although the rays act as if they came from the focal point, most of them never actually pass through the focal point.

You can start a fire by placing a piece of paper at the real focal point of the lens that adds plus vergence. This lens is a plus lens. You cannot start a fire with a virtual focal point, such as we have with the minus lens, which adds minus vergence.

You have probably used the word virtual before. "Virtual reality" or "virtual games" are all phrases that are used to describe something that seems very real, even though it is not. Well, a virtual image from a lens that adds negative vergence still appears to be very real; it just isn't!

When you look at a some object, an image is formed on the light sensitive film in your eye, and your brain interprets the result as the object. The rays diverge from the object and are absorbed by the special light-sensitive nerve cells in your eye. The processing that occurs in the nerve cells of the eye and brain interpret the object to be at the point that the rays diverged from.

When a plus lens changes the vergence of the rays that come from a distance, the result is that the eye interprets the position of the object (in this case the sun) to be at the point that the rays are diverging from. (In these drawings the light is a long way away, and the lens is several feet away from the eye.  And I am using the sun for illustration;  you and I both know that you know better than to look directly at the sun, through any lens or any other way.)

When a minus lens changes the vergence of the rays that come from the distance, the result is that the eye interprets the position of the object to be at the point where the rays appear to be diverging from, even though the rays never actually went through that point.

Well, what about the focal length?

OK, with the plus lens, which added plus power, we had a real focal point, and the power of the lens was 1 divided by the focal length in meters. We will now say, by definition, that the focal length of a plus lens is positive, and the power in diopters is positive.

Then the lens that adds minus vergence is a minus lens, the focal length is negative, and the power is 1 divided by the negative focal length in meters, so the power of this lens is negative.

From this moment on, whenever you specify a power for a lens, it must have a + or a - sign. If it does not have a sign it will be wrong, whether it is right or not!

Suppose that I have a lens with a virtual focal length of 1 meter. What is the lens power?

We are going to use f to represent focal length and D to represent diopters. A virtual focal length is negative, so the focal length is -1 m. D = 1f = 1(-1), so D = -1 diopters.

If the virtual focal length is 0.5 m, then the power is D = 1f = 1(-0.5), so D = -2 diopters.

A minus lens has a focal length of 40 cm. What is its power?
       A minus lens has a negative focal length, so f = -40 cm.
       The focal length must be in meters, so f = -40 cm = -0.40 m.
       D = 1f = 1(-0.4)m
       D = -2.50D

Right?

What is the focal length of a -6.25D lens in mm?
       D = -6.25
       f = 1D = 1(-6.25)
       f = -0.16 m = -160 mm.

Make sure that you keep track of the minus sign and the decimal point!

We talked about 40 inches being almost equal to one meter. So, if you are given the focal length in inches, or if you are given the power and asked for the focal length in inches, you can use the equation D = 40f or f = 40D.

In this case, the negative lens still has a negative focal length, ant the positive lens still has a positive focal length.

Please read pages (29-31 / 43-46) in the Optical Formulas Tutorial, do the exercises, and check your answers. Then read pages 385-top of page 390 in Systems for Ophthalmic Dispensing, and do exercises 1-4 on page 398.

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