Surface power formula

Surface power formula

PAGE References to Optical Formulas Tutorial:  (first reference is to edition 1 / second reference is to edition 2).

In the Focal Length module we defined the power of a lens in terms of how much the lens changes the direction of the incoming light rays. But the lens was made with two surfaces, at least one of which was curved, and we need to define the power of the surfaces as well as the power of the lens. Since the lens was made of some material and for our purposes we consider it to be in air, the formula that defined the power of the lens was relatively uncomplicated: just punch in 1, divide, and the power to get the focal length; or punch in 1, divide, and the focal length to get the power.

With the surfaces, however, we need to take into consideration the material from which the lens was made and the amount of curvature of the surface.

Does it seem reasonable to you that the less curvature a surface has the less it will bend the ray and therefore the less power it will have? Rephrasing that statement, the more curvature a surface has, the more it will change the direction of a light ray and therefore the more power it will have?

If a ray of light is traveling parallel to the axis of the lens when it gets to the lens it will change direction according to Snell's law. The amount that it will change direction will depend on how far the ray is away from the axis and how much the surface is curved. Looking at the two surfaces shown here, the surface on top is not as curved as the surface on the bottom, so the surface on top does not change the direction of the ray's travel as much as the surface on the bottom.

We classify "how curved" a surface is by its radius of curvature. Our surfaces (at least for now) are going to be classified as spherical surfaces. That means that the surface is a portion of a sphere. Consider a globe, or a basket ball, both of which are round. There is a point in the center of each that is the exact same distance from every point on the surface. This is the center of curvature, and the distance from this point to each point on the surface is the radius of curvature of the ball or the sphere.

A golf ball has a much shorter radius than a baseball, which has a shorter radius than a basketball, which has a shorter radius than a beach ball, which has a shorter radius than the earth!  The surface of the golf ball has much more curvature than the surface of the beach ball. So, the shorter the radius, the more curved the surface; the longer the radius, the less curved the surface. The earth has a very long radius, and you have to be somewhere like the flat plains of the midwest or the middle of a calm ocean to see the fact that the surface curves; and even then, if you are looking at a 2 inch section of surface you cannot see the curvature. A two inch section of a beach ball will have a noticeable curvature, but not nearly as much curvature as a two inch section of a golf ball. (I don't play golf -- is a golf ball over two inches in diameter?)

When we did the Snell's law exercises we found out that a ray will be bent more if it enters a high index material than it was when it entered a low index material. The more you slow the ray down, the more the ray is bent. So we need to be concerned with the index of the material that the surface is made of as well as the curvature of the surface. If we consider only lenses that are in air (no scuba masks here, and we are not going to immerse our lens in the fishbowl for this exercise!) we can describe the relationship between the power of the surface, the index of the material the surface is made from, and the curvature of the surface this way:

        n - 1
D = --------
        r
where:
   D = dioptric power of the surface,
   n = the index of the material that the surface is made from,
   r = the radius of curvature of the surface, in meters
   and where the surface is in air.

 If the lens is in water or imbedded in some other material, then we use the first formula on page (35 / 48) in the textbook. We will not use that form of the formula in this course. If you ever decide to take the ABO Masters exam, or if you continue on reading the textbook after the class is over, you will want to come back to this one again.

[If you are using the first edition of the textbook you probably noticed that I skipped two pages in the book, and that I did not do very many examples of this formula. We are going to go back to the skipped pages]

Lets do a few exercises.

  1. We have a surface made of CR39, and it has a radius of curvature of 5 cm -- that is about 2 inches, and is from a relatively small ball. What is the surface power of this curve?

    2.  
  2. Lets use the same curvature, 5 cm, but make this small ball out of a high index glass -- say, 1.80 index. Now what is the surface power?

    3.  
  3. Go back to the CR39, now make a beach ball from it, with a radius of, oh, half a meter. What is the power of this surface?

    4.  
  4. You guessed it, now we will do the beach ball of the high index glass; radius is 0.5 m, index of refraction is 1.80.
OK, that is the setup. We are going to end out with a table that looks like this:
 
  radius 5 cm = 0.05 m radius 0.5 m
index 1.498    
index 1.80    
  1. We have a surface made of CR39, and it has a radius of curvature of 5 cm. What is the surface power of this curve?
  2. Lets use the same curvature, 5 cm, but make this small ball out of a high index glass -- say, 1.80 index. Now what is the surface power?
    3.  n - 1
    D = --------
            r
    where:
               D =
         n =
         r =   in meters
    1.    D = ?
      2.    n = 1.80
         r = 5 cm = 0.05 m

             1.80 - 1
      D = --------------
              0.05

             0.80
      D = --------------
              0.05

        D = 16.00 D

  3. Go back to the CR39, now make a beach ball from it, with a radius of, oh, half a meter. What is the power of this surface?
    4.  n - 1
    D = --------
            r
    where:
  4. Finally, we will do the beach ball of the high index glass; radius is 0.5 m, index of refraction is 1.80.
    5.  n - 1
    D = --------
            r
    where:
               D =
  radius 5 cm = 0.05 m radius 0.5 m
index 1.498 9.96 D 0.996 D
index 1.80 16.00 D 1.60 D

Is the relationship here clear? The higher the index, without changing the curve, the more power the surface has. The longer the radius the less curvature, so, without changing the index, the longer the radius the less the power.
 

The higher the index, the more the ray slows, the higher the surface power.
The lower the index, the less the ray slows, the lower the surface power.
The shorter the radius, the steeper the curvature, the higher the surface power.
The longer the radius, the flatter the curvature, the lower the surface power.

OK, you do some. DO THEM FIRST. Check my answers second.

  1. A surface is made from crown glass, and has a radius of curvature of 85 mm. If the surface is in air, what is its surface power?
    2.  
  2. A surface with radius of curvature of 15 cm is made from polycarbonate. If the surface is in air, what is its surface power?
    3.  
  3. What is the surface power in air of a curve with radius 1 m, if it is made from a material with index of refraction of 2.0?
Do them first. Take your time. Then click here for my answers.
 
 

OK.  Read page (22 / 48-49)) in the Optical Formulas book, do the exercise, and check those answers in the back of the book.

 Click here for the Nominal Power module if you have not already done it.

Click here for the Lensmaker's module if you HAVE done Nominal Power.


 
 
 

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