# 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