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You have two lenses at your disposal, one with a focal length f1 = +31.9 cm, the other with a focal length f2 = -31.9 cm.

If you want to produce an image of the bulb that is enlarged by a factor of 3.00, how far from the wall should the lens be placed?
.............m

2007-11-12 09:47:48 · 1 answers · asked by ? 1 in Science & Mathematics Physics

1 answers

I guess you want to see an enlarged real image on the wall. This means an inverted image thus negative magnification (-3). First, you have to use the converging (f = +31.9) lens. From the thin-lens formula (ref.), with R1=object distance and R2=image distance,
1/F = 1/R1+1/R2; M = -R2/R1 ==> R1 = -R2/M
Substituting, 1/F = 1/R2-1/(R2/M) = (1-M)/R2 ==> R2 = F*(1-M) = 1.276 m
If you really meant positive magnification, it's a virtual image and a wall seems irrelevant. In that case, using M=3 and the same equations, you will see the image at R2 = -0.638 m.

2007-11-13 03:08:42 · answer #1 · answered by kirchwey 7 · 0 1

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