Optics question

jlmac2001

Compute the focal length in air of a thin biconvex lens (n=1.5) having radii of 20 and 40 cm. Locate and describe the image of an object 40cm from the lens.

1/f = (n-1)(1/R1-1/R2)
1/f = (1.5-1)(1/20-1/40)
1/f = .5(.025)
f = 1/.0125 = 80cm (convering lens because f is positive)

Mt=-f/x0=-80/40=-2cm (Inverted image because Mt is negative)

Did I use the correct equations?

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arcnets

Originally posted by jlmac2001
f = 1/.0125 = 80cm
Looks O.K. to me.
Mt=-f/x0=-80/40=-2cm
I should expect, if the object is only 40cm from the lens while the lense's focal length is 80cm, then the image should be upright and virtual, and magnified by a factor of 2.

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