An object is kept at a distance of 5cm in front of a convex mirror of focal length 10cm. Calculate the position and magnification of the image and state its nature.
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u = -5cm, f = 10cm We know that$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}+\frac{1}{-5}=\frac{1}{10}$
$\frac{1}{\text{v}}=\frac{1}{10}+\frac{1}{5}=\frac{3}{10}$
$\therefore \text{v}=\frac{10}{3}\text{cm}=3.33\text{cm}$
The position of the image is 3.33cm behind the convex mirror. Magnification, $\text{m}=-\frac{\text{v}}{\text{u}}\frac{-3.33}{5}=0.66$ The image is virtual and erect.
$\frac{1}{\text{v}}+\frac{1}{-5}=\frac{1}{10}$
$\frac{1}{\text{v}}=\frac{1}{10}+\frac{1}{5}=\frac{3}{10}$
$\therefore \text{v}=\frac{10}{3}\text{cm}=3.33\text{cm}$
The position of the image is 3.33cm behind the convex mirror. Magnification, $\text{m}=-\frac{\text{v}}{\text{u}}\frac{-3.33}{5}=0.66$ The image is virtual and erect.
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