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An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Given: Distance of the object from the mirror, $u=-20\ cm$
The radius of the curvature $R=30\ cm$
Focal length $f=\frac{R}{2}=\frac{30}{2}=15\ cm$
Size of the object, $h=5\ cm$
Let $h'$ be the size of the image.
From mirror formula,
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
Or $\frac{1}{-20}+\frac{1}{v}=\frac{1}{15}$
Or $\frac{1}{v}=\frac{1}{15}+\frac{1}{20}$
Or $\frac{1}{v}=\frac{7}{60}$
Or $v=\frac{60}{7}$
Or $v=8.57\ cm$
The image is formed $8.57\ cm$ behind the mirror. It is virtual and erect.
Magnification, $m=-\frac{v}{u}=\frac{8.57}{20}=0.428$
Also, $m=\frac{h'}{h}$
Or $m=\frac{h'}{5}$
From above, $\frac{h'}{5}=0.428$
$h'=2.14\ cm$
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