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physicsphreak2
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Homework Statement
Suppose we have a lens of known focal length, f=100 cm.
(a) If light enters the lens from air, but after the lens enters a medium with index 1.33, how far from the lens will focus? (Assume the incident light is collimated.)
(b) What if there is instead air after the lens, except for a 5 cm patch of medium with index 1.5?
Homework Equations
1/f=(n-1)(1/R1-1/R2)
The Attempt at a Solution
I don't have the index of the lens, so maybe I don't need to use the lens-maker's equation?
So instead, I proceed like this: Let's say the light travels distance L in air, and x in the medium. Clearly L+x=f because you travel through the two mediums until you got to the focus. If L+x=f, but some of x is replaced by something of effective length x/n, then you have to travel L=f-(x/n) in the air still. So the total travel distance is L+x=(f-(x/n))+x= f+x(1-1/n).
For (a), then, we get that the light focuses to a point at 100+100(1-1/1.33)=124.8 cm
For (b), then, we get that the light focuses to a point at 100+5-(5/1.5)=101.67 cm.
This makes sense, because I would expect a lens to be less powerful in a higher index medium... but is this right?Also, I'm confused about whether I would DIVIDE by the index (since really we care about diffraction effects, and those are proportional to wavelength, which gets smaller in a medium) or MULTIPLY by the index (because that's what happens to the optical path length, although I don't see why we care about the optical path length for focusing, isn't that more for interference effects?)
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