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I need help finding the focal length of a single convex lens. The radius of curvature is 200mm. left side is air and the glass has a index of 1.5. I search on google but there was only formulas for think lens, convex and converge lends. I need the formula for a single convex lens. Any idea? I know the ray transfer matrix.

$\matrix{1 & 0 \cr \frac{n-n`/n`R} & \frac{n`/n}}$
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answered here… – Ken Sep 12 '12 at 22:30
Is this homework? This is more appropriate for – Ben Crowell Sep 13 '12 at 0:18
Thanks for the tip. I went and posted on there. – user101699 Sep 13 '12 at 0:28
up vote 1 down vote accepted

You can put in ${1 \choose 0}$ as a ray parallel to and offset from the axis, multiply it by the transfer matrix, and see where the offset goes to zero. This will be the focal length. I'm not sure because your matrix isn't rendering properly, but it looks like it would be $\frac {-n'R}{n-n'}=\frac{-1.5\cdot 200}{1-1.5}=600 \text {mm}$

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ohh this might work thanks for the guidance. – user101699 Sep 13 '12 at 0:27

If we will not make it complicated simply as the lenght of curvature is 200 mm the focal length will be 100 mm as c = 2f where c = curvature and f = focal length

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