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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Solution

Given: $f = 15$
$u = - 10$
From mirror formula, $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{- 10} + \frac{1}{v} = \frac{1}{15}$
$v = 6 c m$
Image is formed 6 cm to the right of the mirror. Image formed is virtual and erect.

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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.

Given: f=15 u=−10From mirror formula, 1u+1v=1f 1−10+1v=115 v=6 cmImage is formed 6 cm to the right of the mirror. Image formed is virtual and erect.

Given: $f = 15$
$u = - 10$
From mirror formula, $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{- 10} + \frac{1}{v} = \frac{1}{15}$
$v = 6 c m$
Image is formed 6 cm to the right of the mirror. Image formed is virtual and erect.