Concave mirror practice

Relevant wiki: Mirrors

$\large \displaystyle \text{Given:} \\ \large \displaystyle \text{Focal length } = \frac{a}{2} \text{ cm}\\ \large \displaystyle \text{Image distance } = a \text{ cm}\\ \large \displaystyle \text{Image height } = 40 \text{ cm}.$

It is clear that the image is placed on $\color{#3D99F6}{\text{Center of Curvature}}$ on the principle axis, so the image formed is just right below the object.

$\Large \displaystyle \color{#69047E}{\text{Proof}}$

$\large \displaystyle \text{Mirror formula: }\\ \large \displaystyle \frac{1}{f} = \frac{1}{v} + \frac{1}{u}\\ \large \displaystyle \implies \frac{1}{\frac{a}{2}} = \frac{1}{a} + \frac{1}{u}\\ \large \displaystyle \implies \frac{1}{u} = \frac{2}{a} - \frac{1}{a}\\ \large \displaystyle \implies \frac{1}{u} = \frac{1}{a}\\ \large \displaystyle \therefore u = a.\\ \large \displaystyle \text{The magnification of mirror is given by the formula }\\ \large \displaystyle m = -\frac{v}{u} = \frac{h'}{h}\\ \large \displaystyle \implies - \frac{v}{u} = \frac{h'}{h}\\ \large \displaystyle \implies - \frac{a}{a} = \frac{h'}{40}\\ \large \displaystyle \implies h' = -40 \text{ cm}$ .

Therefore, the image size is same is object size. Here negative sign indicates that the image is inverted.

$\Large \displaystyle \text{Image size } = \color{#D61F06}{40 \text{ cm}}$