Chapter_23_concepts

Chapter_23_concepts - The connection between ray diagrams...

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The connection between ray diagrams and the lens (mirror) equation for lenses (mirrors) Converging lens To draw a ray diagram for a vertical object whose base is on the optical axis, draw two light rays from the upper tip of the object. Draw one light ray parallel to the optical axis (ray 1 ) from the tip of the object to the lens If the object distance is greater than the focal length of the lens, f (the object is located to the left of the focal point), draw a second light ray (ray 2 ) from the tip of the object to the lens passing through the focal point on the object side of the lens: Since ray 1 is parallel to the optical axis, it will pass through the focal point of the lens on the image side of the lens after emerging from the converging lens: And, since ray 2 passes through the focal point on the object side of the lens, it will emerge from the lens parallel to the optical axis on the image side of the lens 1 2 optical axis optical axis f 2 optical axis f 1 optical axis f f f 1 2 f
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Since both rays 1 and 2 originated from the upper tip of the object, the point at which they intersect after passing through the converging lens will locate the upper tip of the image, as illustrated below: Note that since the base of the object is on the optical axis, the base of the image will also be on the optical axis. Question : Why is this statement true? Furthermore, since the object is vertical, the image will also be vertical. Question : Why is this statement true? Note that the ray diagram shows that for an any object located beyond the focal point of a converging lens, the image formed by the converging lens will be real (light rays actually intersect at the location of the image), inverted , and will be located beyond the focal point on the image side of the lens (q > f). Connecting this ray diagram with the lens equation The lens equation, 1/q + 1/p = 1/f , gives us the relationship between the image distance, q, and the object distance, p, for any lens of focal length f: 1/q = 1/f – 1/p = (p – f)/pf q = [p/(p – f)] f (i) Since p > f (p – f) > 0 and p > (p – f) , for f > 0 p/(p – f) > 1 q > f , in agreement with our ray diagram above (ii) Since p/(p – f) > 0 and f > 0 q > 0 (image is real ) , in agreement with our ray diagram above
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This note was uploaded on 08/25/2009 for the course PHY 213 taught by Professor Cao during the Summer '08 term at Kentucky.

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Chapter_23_concepts - The connection between ray diagrams...

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