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Old Jul 25th 2012, 09:23 AM   #1
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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?
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Old Jul 25th 2012, 09:56 AM   #2
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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)
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