An object is placed at 4 cm distance in front of a concave mirror of radius of curvature 24 cm. Find the position of image. Is the image magnified?

Given,

Radius of curvature (R) = 24 cm (negative)

Object distance (u) = 4 cm (negative)

Focal length = $\dfrac{1}{2}$ x Radius of curvature

Substituting the values in the formula above, we get,

$\text{Focal length} = \dfrac{1}{2} \times (- 24) \\[0.5em] \text{Focal length} = - 12 \text{ cm}$

Hence, focal length of the concave mirror = -12 cm.

Mirror formula:

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

Now, substituting the values in the mirror formula , we get,

$- \dfrac{1}{4} + \dfrac{1}{v} = - \dfrac{1}{12} \\[0.5em] \dfrac{1}{v} = - \dfrac{1}{12} + \dfrac{1}{4} \\[0.5em] \dfrac{1}{v} = \dfrac{ - 1 + 3 }{12} \\[0.5em] \dfrac{1}{v} = \dfrac{2}{12} \\[0.5em] \dfrac{1}{v} = \dfrac{1}{6} \\[0.5em] v = 6 \text{ cm} \\[0.5em]$

The image is formed 6 cm behind the mirror.

Computing linear magnification:

$\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] \Rightarrow m = - \dfrac{6}{- 4} \\[0.5em] \Rightarrow m = \dfrac{6}{4} \\[0.5em] \Rightarrow m = 1.5 \\[0.5em]$

As, the length of image is 1.5 times the image of object, hence, the image is magnified.