A diverging lens formed an image that is 20.0 cm in front of it. If the focal length
of the lens is 25 cm, where is the object placed?
1a−1b=−1f→1a−10.2=−10.25\frac{1}{a}-\frac{1}{b}=-\frac{1}{f}\to\frac{1}{a}-\frac{1}{0.2}=-\frac{1}{0.25}a1−b1=−f1→a1−0.21=−0.251
1a=1→a=1(m)\frac{1}{a}=1\to a=1(m)a1=1→a=1(m) . Answer
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