Two surfaces of a thin biconvex lens have radii of curvature R1=.07m and R2=.05m

Question

Two surfaces of a thin biconvex lens have radii of curvature R1=.07m and R2=.05m. The image of an object placed at a distance p=.12m appears upright and has lateral magnification of M=2.3. At what distance should you keep the lens from the object to laterally magnify it 1.5 times?

The two answers are p=.0708m or p=.3538m.

I already found the index of refraction to be n=1.1374 for the glass. Any thoughts on how do do this?

Explanation / Answer

Given u=12cm, m=2.3=v/u => v=2.3u and image is virtual (upright) Using 1/f = 1/u - 1/v= 1/12 - 1/(2.3*12) we get f = 21.23 cm Now using m=v/u = 1.5 => v=1.5u we have CASE 1 : REAL IMAGE 1/f = 1/u + 1/v = 1/u + 1/1.5u= 5/3u Since f=21.23 we get u = 5f/3 = 35.384cm = 0.03539 cm CASE 2: VIRTUAL IMAGE Simialr to above, using 1/f= 1/u - 1/v and using v=1.5u We get u = 7.08cm = 0.0708m Hope you liked my detailed explanation :)

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