1/F = 1/Do + 1/Di – Thin Lens Equation Calculator

HelloIm really confused on the signs for f, do and di in this equation..Please tell me if Im correct.. but here is how i presently understand ifFor definition sake, the “front” of a mirror is where you would normally stand to look in a mirror, and the “front” of a lens (like a camera) is where the objects are that you want to take a picture of.f is (+) for a converging lens/mirror, (-) for diverging. always.do is (+) when its in the “front”di is (+) when it is a real imageis this right?

You are watching: 1/f = 1/do + 1/di

that”s correct. i think it helps to use p and q instead of di, do since i tend to think distance in instead of image and out instead fo object, but maybe that”s just me. I”ve attached a diagram showing where the di and do is positive and negative for lenses and for mirrors since it changes when you”re looking at one or the other.
focal point is no problem, as you said.becareful with do and di.now, if i use light rays coming from left to right, and the mirror/lens is on the right,for mirror:do is positive when object is on the left, in front of mirror, i.e. real object.do is negative when the object is on the right, behind the mirror, i.e. virtual object.how can you create virtual object like this? easy. put a real object in fornt of a convex lens and adjust it so that the image falls behind the mirror. this image becomes virtual object of the mirror.di is positive when image is on the left, in front of mirror, i.e. real image.di is negative when image is on the right, behind mirror, i.e. virtual image.in practice, the easiest way to differentiate them is you can always see virtual image directly by your eyes (that”s why you see yourself in the mirror).for lens:do is + when object is on the left, front of lens, i.e. real object.do is – when object is on the right, behind lens, i.e. virtual object.how can you create virtual object? the same as above.di is + when image is on the right, behind lens, i.e. real image.di is – when image is on the left, front of lens, i.e. virtual image.to be even surer, draw the triangles and image formation diagram, and you can see that these convention is extremely consistent no matter how you vary the configuration. try to derive the mirror formula first, very very easy.if get stuck i may be able to provide drawing … if i have time.
Jul 31, 2005#4

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Brad Barker
4290

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it”s been my experience that the object distance is always positive! for mirrors and lenses combined, there are four possibilities:1) to the left of the mirror (in the “real image” zone)2) to the right of the mirror (in the “virtual image” zone*)3) to the left of the lens4) to the right of the lens* option 2 makes absolutely no sense! you can”t have an object as a virtual image! that”s the friggin” mirror world! so for mirrors, we have object distance can only take on positive values.now, for lenses: if the object is to the left of the lens, it”s positive. but if the object is to the right of the lens, we can rotate our diagram so that it”s back to being on the left side without loss of realism! (a real image formed by a lens would be on the opposite side of the object, etc.)so for both mirrors and lenses, the object distance is always positive.my mnemonic for this, when using “p” and “q” is “p is always positive.”

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