An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
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Focal length of convex mirror, f = +15 cm Object distance, u = - 10 cm According to the mirror formula,$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{15}-\frac{1}{(-10)}$
$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{15}+\frac{1}{10}$
$\Rightarrow \ \frac{1}{\text{v}}=\frac{2+3}{30}$
$\Rightarrow \ \frac{1}{\text{v}}=\frac{5}{30}$
$\Rightarrow \ \text{v}=6 \ \text{cm}$
Magnification $=\frac{-\text{v}}{\text{u}}=\frac{-6}{-10}=0.6$ The image is located at a distance 6 cm from the mirror on the other side of the mirror. The positive and value less than 1 of magnification indicates that the image formed is virtual and erect and diminished.
$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{15}-\frac{1}{(-10)}$
$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{15}+\frac{1}{10}$
$\Rightarrow \ \frac{1}{\text{v}}=\frac{2+3}{30}$
$\Rightarrow \ \frac{1}{\text{v}}=\frac{5}{30}$
$\Rightarrow \ \text{v}=6 \ \text{cm}$
Magnification $=\frac{-\text{v}}{\text{u}}=\frac{-6}{-10}=0.6$ The image is located at a distance 6 cm from the mirror on the other side of the mirror. The positive and value less than 1 of magnification indicates that the image formed is virtual and erect and diminished.
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