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 cm

u = -10 cm

v = ?

According to the mirror formula,

$\dfrac{1}{v}$ + $\dfrac{1}{u}$ = $\dfrac{1}{f}$

Substituting the values we get,

$\dfrac{1}{v} + \dfrac{1}{-10} = \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 10} + \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{3+2}{30} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{5}{30} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{6} \\[1em] \Rightarrow v = 6\text{ cm}$

Therefore, image is formed 6 cm behind the mirror. Image is virtual and erect.