Physics Help Forum Optics / focal length
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 Jul 25th 2012, 08:23 AM #1 Junior Member   Join Date: Jul 2012 Posts: 1 Optics / focal length I am having a very hard time figuring out this problem and was hoping I could get some help. Problem: What was the focal length when the radius of curvature was 0.70m and index of refraction was 1.8? I believe the formula is the Lens markers equation 1/f=(n-1)1/R but I would need to solve for f not 1/f. n=1.8 and R=0.70m when i plug those number in I keep getting the wrong answer and can not figure out why for the life of me. here is what I did. f=(1.8-1)2/0.70------.8/1(2/0.70) = .56/2 = .28 would be the answer but it is not. Can someone please help me find where I am going wrong?
 Jul 25th 2012, 08:56 AM #2 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 the value of f = R/(n-1) f= 0.7/0.8 for radius of curvature, it consists of R1 and R2 which are the radii of curvature closer and more distal to light source respectively. note that signs of R1 and R2 have different meanings. for a thin lense, 1/f =(n-1)(1/R1-1/R2) __________________ Good results were achieved and the new task is to become a good doctor.

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