discrepancy between kaplan and EK approach to optics?

[bema20]

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sooo here is one of the kaplan questions from the sectionals:


A thin converging lens of focal length 100 mm is used as a magnifying glass. If the object viewed is 80 mm from the lens, how far from the lens will the image be?

So I know that i have to do 1/f = 1/i + 1/o and the tricky part is figuring out which is negative and which is positive. According to the approach that EK teaches in their book, you're supposed to assume that the positive side is where your eye is and that the object is on the other side (and thus negative). So I assumed from this that f = 100 mm (positive), and o = -80 mm.

Alas! Kaplan's explanation states: "since we are talking about a single lens system, the object distance is positive" and "the fact that the lens is a converging lens implies that focal length is positive" and the answer is 400 mm, which is the answer you get when you use all positive numbers.

For lenses, do you assume that the object is always placed on the other side of the eye (or what would be the point of a lens?) and thus negative distance? Why doesn't the EK method work??

 

[Prateek]

Hmm the way I would do this question is:

f = 100mm
o = 80mm
i = ?

1/100 = 1/80 + 1/i

i = -400mm Meaning the object is 400mm BEHIND the lens.

The Kaplan method is similar to the TPR method and I think you should just stick to this since it's much simpler:

o = ALWAYS positive
f = "Positive if converging" and "Negative if diverging"

 

[lorenzomicron]

in physics, we take the distance behind the lens to be negative, and the distance in front of the lens to be positive.