How far apart are an object and an image formed by a 75.0cm focal length converging lens if the image is 2.5x larger than the object and is real.
m=vu=2.51u+1v=1fm = \frac{v}{u} = 2.5 \\ \frac{1}{u} + \frac{1}{v} = \frac{1}{f}m=uv=2.5u1+v1=f1
Multiplying by v on both sides and replacing vu\frac{v}{u}uv by m
m+1=vf3.5=v75v=262.25 cmu=vm=262.252.5=105 cmu+v=105+262.25=367.25 cmm + 1 = \frac{v}{f} \\ 3.5 = \frac{v}{75} \\ v = 262.25 \; cm \\ u = \frac{v}{m} = \frac{262.25}{2.5} =105 \; cm \\ u +v = 105 +262.25= 367.25 \; cmm+1=fv3.5=75vv=262.25cmu=mv=2.5262.25=105cmu+v=105+262.25=367.25cm
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