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An object is placed at \(10 \ \text{cm}\)  in front of a concave mirror of radius of curvature \(15 \ \text{cm}\) . Find the position, nature, and magnification of the image.

Answer

Given data,
Radius of curvature \(R=-15\text{ cm}\)
Distance of object \(u=-10\text{ cm}\)
Focal length \(f=\frac R 2\)
\(=-\frac{15} 2\)
\(=-7{\cdot}5\text{ cm}\)
By mirror formulae,
\(\frac 1 v+\frac 1 u=\frac 1 f\) , by putting values
\(\Rightarrow \frac 1 v+\left(\frac{-1}{10}\right)=\left(\frac{-1}{7{\cdot}5}\right)\)
\(\Rightarrow v=-30\text{ cm}\)
\(-\mathit{ve}\)  sign indicates that image is real and inverted and it’s position is \(30\text{ cm}\)  from the mirror.
Magnification,
\(m=\frac{-v} u\)
\(=\frac{-30}{-10}\)
\(=3\)
Thus the image is \(3\) times magnified.
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