17 through 29 22 23, 29 More mirrors. Object stands on the central axis of a spherical or plane mirror. For this situation, each problem in Table 34-4 refers to (a) the type of mirror, (b) the focal distance $\mathbf{\text{f}}$, (c) the radius of curvature $\mathbf{\text{r}}$, (d) the object distance $\mathbf{\text{p}}$, (e) the image distance $\mathbf{\text{i}}$, and (f) the lateral magnification m. (All distances are in centimeters.) It also refers to whether (g) the image is real $\left(R\right)$ or virtual$\left(V\right)$, (h) inverted (I) or noninverted $\left(\mathrm{NI}\right)$from$\mathit{O}$, and (i) on the same side of the mirror as the object $\mathit{O}$or the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

$r=40\mathrm{cm}$

Here, the radius of curvature and image distance is given in the problem. It is given that the mirror is convex. Using that the focal distance, object distance, and magnification can be found. Using all these values, it can be decided if the image is virtual or real and the position of the image.

The formula is as follows:

$\begin{array}{l}\mathbf{r}\mathbf{=}\mathbf{2}\mathbf{f}\\ \frac{\mathbf{1}}{\mathbf{1}}\mathbf{=}\frac{\mathbf{i}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{p}}\\ \mathbf{m}\frac{\mathbf{i}}{\mathbf{p}}\end{array}$

Where,

$\mathit{r}$= radius of curvature,

$\mathit{f}$= focal length,

$\mathit{p}$=object distance from the mirror,

$\mathit{i}$= image distance.

It is given in the table $34-4$, that the type of mirror is convex.

Focal length $f$is calculated as,

$r=2\times f$,

As the mirror is convex, the radius of curvature must be negative, so$r=-40\mathrm{cm}$,

$\begin{array}{l}40=2\times f\\ f=-0\mathrm{cm}\end{array}$

As the mirror is convex, the radius of curvature must be negative

$r=-40\mathrm{cm}$

As the mirror is convex, the image distance must be negative.

So, $i=-40\mathrm{cm}$,

The object distance is calculated by,

$\frac{1}{f}=\frac{1}{i}+\frac{1}{p}$

$$\begin{gathered} \frac{\text{1}}{\text{- 20}}\text{=}\frac{\text{1}}{\text{- 4}}+\frac{1}{p} \\ \frac{1}{p}=\frac{\text{1}}{\text{- 20}}\text{+}\frac{\text{1}}{\text{4}}\text{=}\frac{\text{1}}{\text{5}} \\ p=\text{+5.0cm} \end{gathered}$$

As the mirror is convex, the image distance must be negative.

$i=-40\mathrm{cm}$

The magnification ratio is given as,

$$\begin{gathered} M=\frac{-i}{p} \\ M=\frac{\text{4.0}}{\text{5.0}} \\ M=\text{+ 0.80} \end{gathered}$$

As the magnification is positive, so the image is non-inverted.

For spherical mirrors, virtual images form on the opposite side of the object. Since the image is virtual here, so it is formed on the opposite side of the mirror as the object $O$.

Here the basic formulae can be used to find the radius of curvature, image distance, and magnification ratio. Using that it can be decided that the image is virtual or real and on the opposite side of the same side as the object.