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Kavorka
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If you know the focal length of a lens and use the lens equation 1/do + 1/di = 1/f by assigning some arbitrary do and solve for di, can you then find the magnification just from the focal length from -di/do?
Kavorka said:If you know the focal length of a lens and use the lens equation 1/do + 1/di = 1/f by assigning some arbitrary do and solve for di, can you then find the magnification just from the focal length from -di/do?
sophiecentaur said:I guess that makes assumptions about the accommodation that the user has, to focus close images.
jtbell said:Just to avoid confusion...
After looking at the first post I think it's clear that the OP is referring to linear magnification whereas the rest of us are referring to angular magnification!
jtbell said:Just to avoid confusion...
After looking at the first post I think it's clear that the OP is referring to linear magnification whereas the rest of us are referring to angular magnification!
sophiecentaur said:Now you have introduced something that I don't quite get. Are we just talking trigonometry here?
Andy Resnick said:Much of the OP is unclear.
jtbell said:Linear magnification is the simple ratio of image size to object size, with a +/- sign depending on whether the image is upright or inverted. Via similar triangles on a ray diagram, this equals the ratio of image distance to object distance, with a sign flip if you're using the Gaussian convention for signs. Most textbooks write this as something like m = hi/ho = -di/do.
Angular magnification is the ratio of the angle subtended by the image as seen by the device's user, to the angle subtended by the object when viewed directly. Strictly speaking you have to use trig to get the angles from the object and image sizes and locations, but at the introductory level we always use the small-angle approximation θ ≅ tan θ ≅ sin θ, using whichever function is easier to use from the ray diagram.
I take it to mean, "given f and do, can you use the resulting di to find the (linear) magnification using -di/do?" I would say "yes", but the resulting m will be true only for that value of do, which fits with rest of your statement, about a "range of magnifications".
Kavorka said:<snip>Thus, you're getting magnification directly from focal length through the thin lens equation.
jtbell said:<snip>Angular magnification is the ratio of the angle subtended by the image as seen by the device's user, to the angle subtended by the object when viewed directly. Strictly speaking you have to use trig to get the angles from the object and image sizes and locations, but at the introductory level we always use the small-angle approximation θ ≅ tan θ ≅ sin θ, using whichever function is easier to use from the ray diagram.
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sophiecentaur said:But it strikes me that, when you deliberately place a virtual image at infinity with a simple lens, the linear magnification would be infinite and the angular magnification would be a more realistic figure to quote.
Lens magnification refers to the degree to which a lens can make an object appear larger or smaller. It is typically expressed as a ratio of the size of the object's image to the size of the object itself.
Lens magnification is calculated by dividing the focal length of the lens by the focal length of the image. The focal length is the distance from the lens to the point where the image is focused.
Yes, lens magnification can be determined from the focal length. As mentioned earlier, lens magnification is calculated by dividing the focal length of the lens by the focal length of the image.
No, lens magnification can vary depending on the type of lens and its focal length. Telephoto lenses, for example, have a higher magnification compared to wide-angle lenses.
Lens magnification does not directly affect image quality. However, using a high magnification lens may result in a narrower depth of field and potential distortion, which can impact the overall image quality.