Quick optics conceptual question

[Tokspor]

Can we always just assign objects as having positive distance from a mirror or lens? In other words, if an object is placed on the same side of a lens as our eyes or on the opposite side of a mirror as our eyes, then there wouldn't be any optics to consider, correct?

By the way, by "positive distance," I mean when the object is on the same side of a mirror as our eyes or on the opposite side of a lens as our eyes.

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[RogueUnicorn]

you can't place objects on the opposite side of the mirror. there is a sign convention with distances with relation to the object.

 

[wanderer]

Can we always just assign objects as having positive distance from a mirror or lens? In other words, if an object is placed on the same side of a lens as our eyes or on the opposite side of a mirror as our eyes, then there wouldn't be any optics to consider, correct?

By the way, by "positive distance," I mean when the object is on the same side of a mirror as our eyes or on the opposite side of a lens as our eyes.

What if you have a lens-mirror system where the image formed from the first lens is behind the mirror?

 

[RogueUnicorn]

What if you have a lens-mirror system where the image formed from the first lens is behind the mirror?

compound lenses i was deliberately ignoring.. (the way you teach middle schoolers noble gasses don't react with anything) but yes in that case object is at a negative distance (i believe)

 

[thebillsfan]

also, if you have a two lens system, and the image from the first lens is behind (not in between the two lenses) the second lens then that would be negative object distance for the second lens

 

[Tokspor]

also, if you have a two lens system, and the image from the first lens is behind (not in between the two lenses) the second lens then that would be negative object distance for the second lens

So you're saying if we have a situation like this, where the parentheses are the two lenses, do1 is the distance of object 1, and di1 is the distance of the image produced by it:

do1 () () di1

Then di1 = -(distance of object 2), right? So in this case, only one image would be produced by this system?

 

[RogueUnicorn]

yes.