A camera is equipped with a lens with a focal length of 39 cm. When an object 2.6 m (260 cm) away is being photographed, what is the magnification?
1/a+1/b=1/f→b=1/(1/f−1/a)=1/(1/0.39−1/2.6)=0.4588 (m)1/a+1/b=1/f\to b=1/(1/f-1/a)=1/(1/0.39-1/2.6)=0.4588\ (m)1/a+1/b=1/f→b=1/(1/f−1/a)=1/(1/0.39−1/2.6)=0.4588 (m)
M=−b/a=−0.4588/2.6=−0.176M=-b/a=-0.4588/2.6=-0.176M=−b/a=−0.4588/2.6=−0.176 or
M=ff−b=0.390.39−2.6=−0.176M=\frac{f}{f-b}=\frac{0.39}{0.39-2.6}=-0.176M=f−bf=0.39−2.60.39=−0.176 . Answer
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