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 $\mathit{m}$. (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 $\left(I\right)$or noninverted $\left(NI\right)$from $\mathit{O}$, and (i) on the same side of the mirror as the object $\mathit{O}$or on the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

$m=+1.0$.

$p=+10cm$.

Here, the object distance and lateral magnification are given in the problem. Using that the focal distance, the radius of curvature, and image distance can be found and decide if the image is virtual or real and also find the position of the image

The formula is as follows:

$\begin{array}{rcl}r& =& 2f\\ \frac{1}{f}& =& \frac{1}{i}+\frac{1}{p}\\ m& =& \frac{-i}{p}\end{array}$

It is given that the lateral magnification is $m=1.0$, thus the mirror is flat.

For flat mirrors, the focal length is $\infty$,

$f=\infty$.

Use the following formula to find the radius of curvature,

$\begin{array}{rcl}r& =& 2\times f\\ r& =& 2\times \infty \\ r& =& \infty \end{array}$

It is given in the tablethat the object distance is,

$p=+10cm$.

Image distance can be calculated by,

$$\begin{gathered} \frac{1}{f}=\frac{1}{i}+\frac{1}{p} \\ \frac{{\displaystyle 1}}{{\displaystyle \infty }}=\frac{{\displaystyle 1}}{{\displaystyle i}}+\frac{{\displaystyle 1}}{{\displaystyle 10}} \\ \frac{{\displaystyle 1}}{{\displaystyle i}}=-\frac{1}{10} \\ i=-10cm \end{gathered}$$

The image distance is $i=-10cm$.

The magnification ratio is given as,

$M=\frac{-i}{p}$,

It is given in the tablethat the magnification ratio is,

$M=+1.0$.

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.

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 and on the opposite side as the object.