Answer:
d) 15.4 cm
Explanation:
Radius of curvature = 50 cm
i = image distance
o = object distance
Focal length (f) is half of radius of curvature
So f = 25 cm
Magnification = m = 2.6
[tex]m=-\frac{i}{o}\\\Rightarrow i=-2.6o[/tex]
Mirror equation
[tex]\frac{1}{i}+\frac{1}{o}=\frac{1}{f}\\\Rightarrow \frac{1}{-2.6o}+\frac{1}{o}=\frac{1}{25}\\\Rightarrow \frac{-1+2.6}{2.6o}=\frac{1}{25}\\\Rightarrow \frac{1.6}{2.6o}=\frac{1}{25}\\\Rightarrow 25=\frac{2.6o}{1.6}\\\Rightarrow 25=1.625o\\\Rightarrow o=\frac{25}{1.625}\\\Rightarrow o=15.38\ cm[/tex]
Object distance is 15.4 cm
