By computing the amount of refraction occurring at each surface of a lens, we can determine the precise direction of any given ray when it emerges from the lens. By extending this direction back into the lens until it meets the incident ray's original path, we can construct a plane. This plane is the PRINCIPAL PLANE of the lens. We may trace the paths of all entering rays as if all refraction occurred at this plane.
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Diagram 2.1: Construction of the principal plane.In thin lens ray tracing, we make two assumptions: 1. the thickness of the lens does not affect the power of the lens; and 2. the lens has the same medium on both sides. We will discuss what happens when these two assumptions are not true in the next lesson.
We will draw converging lenses with base in prisms at top and bottom of the principal plane, and diverging lenses with base out prism at top and bottom of the principal plane.
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Diagram 2.2: Refraction through a converging and a diverging lens.
The AXIS is the ray which travels perpendicular to the principal plane, and is therefore not deviated. The OPTICAL CENTER of a lens is the point where the axis crosses the principal plane. We may now state the rules for thin lens ray tracing.
CONVERGING LENSES:
1. The light ray traveling parallel to the axis will be refracted at the principal plane, and emerge to travel through the secondary focal point. (f ').2. The light ray traveling through the primary focal point (f') will be refracted at the principal plane, and emerge parallel to the axis of the lens.
3. The light ray that travels through the optical center of the lens will not be refracted.
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Diagram 2.3a: Refraction through a converging lens.DIVERGING LENSES:
1. The light ray traveling parallel to the axis will be refracted at the principal plane, and emerge as if it had come from the secondary focal point (f').2. The light ray traveling toward the primary focal point (f) will be refracted at the principal plane, and emerge parallel to the axis of the lens.
3. The light ray that travels through the optical center of the lens will not be refracted.
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Diagram 2.3b: Refraction through a diverging lens.There are two items that must be kept in mind. First, the object has rays reflecting from all points on it and going in all directions. We are drawing only the principal ones. Second, these rules are valid only for rays that are traveling "close" to the axis with respect to the focal length of the lens. When the rays come in obliquely and not close to the axis, aberrations are generated which are discussed in the lesson on aberrations. The rules would be true for other rays only if the lens surfaces were parabolic instead of spherical.
We will use the convention that object distance is positive if the object is on the incident side of the lens. Image distance is positive if the image is on the refracted side of the lens. Image size and object size are positive if the image or object is above the axis or erect. If the image or object is below the axis or inverted then the image or object size is negative. Object size and distance will always be positive for single lens systems There are examples of virtual objects in a later lesson, when we consider multiple lens systems.
Using the triangles formed by the axis, the object, the image, and ray 3 in diagram 2.3a, it can be seen that I/O = q/p, as in the mirror discussion. In diagram 2.4, we use ray 2 to demonstrate that the formula 1/f = 1/p + 1/q also holds true.
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<a = <b
tan <a = O/(p-f) = tan <b = I/f
O/(p-f) = I/f
O/I = (p-f)/f
but since O/I = p/q
(p-f)/f = p/q
p/f - 1 = p/q
1/f - 1/p = 1/q
1/f = 1/q + 1/p
Diagram 2.4: Derivation of the distance formula
EXERCISES1. In diagrams 2.5-2., trace at least two of the three principal rays, and show the image formed by each concave or convex lens. Perform the mathematical calculations necessary to verify that the images in the drawings are correct.
scale: 1 square = 1 cm.
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Diagram 2.5scale: 1 square = 1 cm.
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Diagram 2.6scale: 1 square = 1 cm.
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Diagram 2.7scale: 1 square = 1 cm.
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Diagram 2.8scale: 1 square = 1 cm.
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Diagram 2.9scale: 1 square = 1 cm.
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Diagram 2.102. Describe the changes in the size and placement of the image in a convex lens as the object approaches the lens from a distance.
scale: 1 square = 1 cm.
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Diagram 2.11scale: 1 square = 1 cm.
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Diagram 2.12scale: 1 square = 1 cm.
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Diagram 2.13
3. Describe the changes in the size and placement of the image in a concave lens as the object approaches the lens from a distance.
Reading assignment: Stoner & Perkins, Optical Formulas Tutorial 2ed, page 192-194. After reading this section and doing the exercises above, note the 'rules' in the box on page 191.
The answers are here. Do them yourself first. If you just look at my answers and say "Yep, I would have gotten that" then you will have learned nothing. Does not matter to me. I already know how to do them. May very well matter to you.
Copyright 2001, Ellen Stoner, MALS, ABOM