An object 5 cm high is placed at a distance 60 cm in front of a concave mirror of focal length 10 cm. Find (i) the position and (ii) size, of the image.

Given,

Object height (O) = 5 cm

focal length (f) = 10 cm (negative)

Object distance (u) = 60 cm (negative)

Mirror formula:

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

Substituting the values in the formula above, we get,

$- \dfrac{1}{60} + \dfrac{1}{v} = - \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} = - \dfrac{1}{10} + \dfrac{1}{60} \\[0.5em] \dfrac{1}{v} = \dfrac{ - 6 + 1}{60} \\[0.5em] \dfrac{1}{v} = - \dfrac{ 5}{60} \\[0.5em] \dfrac{1}{v} = - \dfrac{1}{12} \\[0.5em] v = - 12 \text{ cm} \\[0.5em]$

The image distance (v) = 12 cm infront of the mirror.

$\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] \dfrac{I}{5} = - \dfrac{12}{60} \\[0.5em] I = -\dfrac{12 \times 5}{60} \\[0.5em] \Rightarrow I = - 1 \text { cm}\\[0.5em]$

Hence, length of image = 1 cm

Negative sign shows that the image will be inverted.