17 through 29 22 23, 29 More mirrors. Object O 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$\mathit{f}$, (c) the radius of curvature$\mathit{r}$, (d) the object distance$\mathit{p}$, (e) the image distance$\mathit{i}$, and (f) the lateral magnification localid="1663002056640" $\mathit{m}$. (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 $\mathbf{\left(}\mathbf{I}\mathbf{\right)}$or noninverted $\mathbf{\left(}\mathbf{NI}\mathbf{\right)}$from O, and (i) on the same side of the mirror as the object O or on the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

The focal length, $f=-30cm$

The image distance,$i=-15cm$

Here, the focal distance and image distance are given in the problem. Using that the radius of curvature and object distance can be found. Then, by using image distance and object distance, the magnification ratio 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 formulae:

The radius of curvature is,

$r=2f$

The spherical mirror formula is,

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

The magnification is,

$m=\frac{-i}{p}$

Where, $r$is the radius of curvature, $f$is the focal length, $p$is the object distance from the mirror, and $i$is the image distance.

As the focal length given in the problem is negative, this means that the mirror is convex.

As the mirror is concave, from the table 34-4, the focal distance is $f=-30cm$.

Use the following formula to find the radius of curvature,

$\begin{array}{rcl}r& =& 2\times f\\ & =& 2\times \left(-30\right)\\ & =& -60cm\end{array}$

Hence, this is the radius of curvature.

Object distance can be calculated by,

$\begin{array}{rcl}\frac{1}{f}& =& \frac{1}{i}+\frac{1}{p}\\ \frac{1}{-30\text{cm}}& =& \frac{1}{-15\text{cm}}+\frac{1}{p}\\ \frac{1}{p}& =& \frac{1}{-30\text{cm}}+\frac{1}{15\text{cm}}\\ \frac{1}{p}& =& \frac{1}{30\text{cm}}\end{array}$

$p=30cm$

Hence, this is the object distance.

It is given in the problem that the image distance is, $i=-15cm$.

The magnification ratio is given as,

$$\begin{gathered} m=\frac{-i}{p} \\ =-\frac{\left(-15\text{cm}\right)}{30\text{cm}} \\ =-\text{0.50} \end{gathered}$$

Hence, the magnification of the mirror is 0.50.

Since the image distance is negative, the image is virtual.

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 $O$.