The image distance is positive. The image distance is negative. The image is inverted. The image is upright. The image height is positive. The image height is negative. The magnification of the image is positive. The magnification of the image is negative. It was made by a converging optic. It was made by a diverging optic. The focal length of the optic is positive. The focal length of the optic is negative.
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All those statements are potentially true. None of them is logically necessary. This can be seen from the lens equation $\frac{1}{d_{o}}+ \frac{1}{d_{i}} = \frac{1}{f}$, remembering the convention that the focal length is positive for a converging lens and negative for a diverging lens. All that remains is to draw a few ray diagrams to get a sense for the inversion of the image. (or using the derived result${}^{1}$ $M=-\frac{d_{i}}{d_{o}}$, where M is the magnification of the lens. ${}^{1}$It's easiest to get THIS result from noting that the ray from the top of the object will pass through the center of the lens undeflected, and also gets mapped to the top of the image. So the ray going from the object to the lens creates a similar triangle to the one formed going from the lens to the image. |
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A real image comes from a concave mirror or a converging lens, and for a single optic the image will be inverted. Whether the real image is the same side of the optic as the source depends on whether you have a mirror or a lens. The source must be further away than the focal length of the optic to produce a real image; the image will be reduced if the source is more than two focal lengths away from the optic, and magnified if the source is between one and two focal lengths away. Since this is homework and some of the questions depend on the conventions you have been taught, I will leave the rest to you. |
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