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Specifically, does the focal length change? How can this be rationalized?

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Think of the question in reverse. Suppose you start with two lenses (say with one flat side each) and stick them together. Do you know how the focal length of the resulting object is related to the two initial focal lengths? (Of course, if you don't know the answer to this, then this won't help at all. I'm guessing that this is a homework question, though, and I wouldn't be surprised if this information is in your textbook.) –  Ted Bunn Feb 14 '11 at 20:30
It's not a homework question, I'm just a chemistry student looking at diagrams of spectrometers :) –  Radu Feb 14 '11 at 20:34
It's been a long time since I've done optics but 1/f' = 1/f1 + 1/f2? So then the focal length of a plano convex lens is twice that of the biconvex lens. If you place a light source at 2f from a biconvex lens and then proceed to cut it, you will get collimated light. Is this more of less right? –  Radu Feb 14 '11 at 20:42
My mistake, then. Apologies. The formula you've got is right: when you place two lenses right next to each other, the "powers" (reciprocal focal lengths) add. –  Ted Bunn Feb 14 '11 at 20:49
To address the rationalization part of the my question, is it sound to state that in general, more curvature -> 'better' (closer) focusing. –  Radu Feb 14 '11 at 21:08

2 Answers 2

up vote 5 down vote accepted

It might help to think about the symmetry of a sliced biconvex lens.

enter image description here

If the biconvex lens can focus light from point F1 to F2, both distance $f$ from the lens, then when you cut the lens in half, each half will have a focal length equal to $f$.

The focal length of the biconvex lens is $f/2$.

The general lensmaker's equation, from Wikipedia, is

enter image description here

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This entry about the Lens' maker equation in Wikipedia may help you:


Can calculate the ratio of focal lenghts before and after cutting by doing $R_2\rightarrow\infty$ and $d\rightarrow d/2$.

The focal power of the second surface that now is plane will be smaller (actually 0) and, unless this is compensated by a huge decrease in thickness, the lens will have longer focal distance. That's what intuitively seems to happens with the dimensions used in real-life lenses.

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Makes sense, I ditched the d term entirely (thin lens), and I get the same result as with the formula I found above. Thanks! Now that I think about it, it does kind of make sense that having more curvature will result in the light focusing closer. –  Radu Feb 14 '11 at 21:02

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