DoubleLensExPost - = -50 cm 2 The ray coming in parallel to...

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Lens #1 Lens #2 f = -50 cm f = +20 cm 2 1 10 cm O p = 40 cm Where does this double-lens system form an image of an object that is 40 cm to the left of lens #1?
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Lens #1 f = +20 cm C C F F I 1 1 1 O Stage #1: Find the image from lens #1 alone. Object 1 is a “real” object because light diverges onto lens #1. (Real side of Lens #1) (Virtual side of Lens #1)
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Lens #1 f = +20 cm 2 O I = O 1 1 1 Lens #2 f = -50 cm 2 Stage #2 - The image from stage #1 never actually forms because lens #2 intercepts the light first. However, the image that would have formed becomes the object for stage #2. It is a virtual object (light is converging on lens #2. (Virtual side of both lenses) (Real side of both lenses) (Real side of #1 & virtual side of #2)
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Lens #1 f = +20 cm 2 O I = O 1 1 1 Lens #2 f = -50 cm 2 The ray of light incident on the center of Lens #2 passes through undeflected.
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Lens #1 f = +20 cm 2 O I = O 1 1 1 Lens #2 f
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Unformatted text preview: = -50 cm 2 The ray coming in parallel to the axis reflects out along a line which . .. Lens #1 f = +20 cm 2 O I = O 1 1 1 Lens #2 f = -50 cm 2 F for Lens #2 The ray coming in parallel to the axis reflects out along a line which projects back through one of the focal points of Lens #2, which happens to be at the location of the original object in this particular example. Lens #1 f = +20 cm 2 O I = O 1 1 1 Lens #2 f = -50 cm 2 F for Lens #2 I 2 The final image (seen by an observer to the right of this diagram) forms where the two rays intersect. Lens #1 f = +20 cm 2 O I = O 1 1 1 Lens #2 f = -50 cm 2 F for Lens #2 I 2 The diverging lens (#2) causes the rays of light from Lens #1 to diverge (a little) from their paths toward image #1, so that the final image (#2) forms farther to the right....
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DoubleLensExPost - = -50 cm 2 The ray coming in parallel to...

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