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Lecture16

# Lecture16 - Lecture 16 More on Lenses Last time we began a...

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Last time we began a discussion of using refraction to form an image: As long as t is small, we showed that: 1 s + 1 σ = ν -1 ( 29 1 Ρ 1 - 1 2 Lecture 16: More on Lenses t R 1 R 2 s Object Image

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Focal length As with a spherical mirror, we can define the focal length of the lens as the point at which the image of an object infinitely far away would appear: With this definition, the lens-maker’s equation looks just like the mirror equation: 1 + 1 φ = ν -1 ( 29 1 Ρ 1 - 1 2 1 ( 29 1 1 - 1 2 1 s + 1 σ = 1 Note: a lens has two focal points (one on each side of the lens), but both have the same focal length
Converging and Diverging Lenses We can separate lenses into two classes based on what happens to parallel rays of light that pass through them: 1. Converging lens (thickest at the middle): 1. Diverging lens (thickest at the edges):

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Sign Conventions As usual for geometric optics, we need to understand the signs of all the quantities Quantity + when… - when… s Object is real (in front of lens) Object is virtual (behind lens) Image is real (behind lens) Image is virtual (in front of lens) Image is upright Image is inverted R 1 and R 2 Center of curvature is behind lens Center of curvature is in front of lens f Lens is converging Lens is diverging y
Magnification Conveniently, the magnification of an image by a lens is given by the same expression as that for a mirror: So when m is positive, the image is upright and on the same side of the lens as the object When m is negative, image is inverted and on opposite side of lens m = ψ = - σ

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Example An antelope is 20m away from a converging lens with f = 30cm.
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Lecture16 - Lecture 16 More on Lenses Last time we began a...

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