A concave mirror produces an image of a 20 cm high object on a...

Given that the image is formed on a screen, it is a real and inverted image.

Height of the object, $h_o = 20 \text{ cm} = 0.2 \text{ m}$

Height of the image, $h_i = -50 \text{ cm} = -0.5 \text{ m}$

Image distance, $v = -5.0 \text{ m}$

Magnification, $m = \dfrac{h_i}{h_o} = \dfrac{-0.5}{0.2} = -2.5$

Using the relation $m = -\dfrac{v}{u}$:

$-2.5 = -\dfrac{-5.0}{u} \Rightarrow u = -2.0 \text{ m}$

The distance between the mirror and the object is $2.0 \text{ m}$.

Using the mirror formula $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$:

$\dfrac{1}{f} = \dfrac{1}{-5.0} + \dfrac{1}{-2.0} = -0.2 - 0.5 = -0.7 \text{ m}^{-1}$

$f = -\dfrac{1}{0.7} \approx -1.43 \text{ m}$

The focal length of the mirror is $1.43 \text{ m}$.