I have a solution to this problem that is very similar to yourwork. However, it

Question

I have a solution to this problem that is very similar to yourwork. However, it says to "Consider an object at infinity, imagedat the person's far point[,,,]" So they use the equation 1/p plus 1/q equals 1/f They make p go into infiinity. Then they say that q (Which is25cm like you have it) is the person's far point. How can that be?The object was let to go to infinity but 25 cm is now the farpoint? I can forward you the solution that says this if youwish. I have a solution to this problem that is very similar to yourwork. However, it says to "Consider an object at infinity, imagedat the person's far point[,,,]" So they use the equation 1/p plus 1/q equals 1/f They make p go into infiinity. Then they say that q (Which is25cm like you have it) is the person's far point. How can that be?The object was let to go to infinity but 25 cm is now the farpoint? I can forward you the solution that says this if youwish.

Explanation / Answer

The far point of a person's eye is the distance beyond whichobjects are not clearly seen. Suppose the far point = x which is less than infinity (faraway) The purpose of a correcting lens is to focus objects that arefar away at the person's far point. The lens equation would become 1 / + -1 / x = 1/ f   or f = -x for a lens that would make distance objects appear to be atthe person's far point. Hope that this useful.

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