An object of 1.2 cm height is placed 30 cm before a concave mirror of focal length of 20 cm to get a real image at a distance of 60 cm from the mirror. What is the height of the image formed?

Given, height of the object = 1.2 cm
Focal length f = -20 cm (focal length is negative for a concave mirror)
Distance of the image v = -60 cm
We know that
$$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$$
Here, u is object distance, v is image distance, f is focal length
$$\dfrac{-1}{60} + \dfrac{1}{u} = \dfrac{-1}{20}$$

$$\dfrac{1}{u} = \dfrac{1}{60} - \dfrac{1}{20}$$

$$\dfrac{1}{u} = \dfrac{-1}{30}$$

=> u = -30 cm
We know that,
$$\dfrac{\text{Height of the image}}{\text{Height of the object}} = \dfrac{-v}{u}$$

$$\dfrac{\text{Height of the image}}{1.2} = \dfrac{-(-60)}{-30}$$

Height of the image = -2.4 cm