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 non-inverted localid="1663128936002" $\left(\mathit{N}\mathit{I}\right)$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 object distance is, $+40\text{cm}$

The magnification,$m=-0.7$

The object distance and the magnification ratio are given which find the image distance and the object distance. Then, using the image and object distance find the focal length. From the focal length, decide the type of mirror, and from the image distance, decide whether the image is real or virtual.

Formulae:

The radius of curvature is,

$r=2f$

The spherical mirror expression is,

role="math" localid="1663129482481" $\frac{1}{i}+\frac{1}{p}=\frac{1}{f}$

The magnification is,

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

Where,$m$ is the magnification, $p$ is the pole, $f$is the focal length.

Write the formula for the magnification below.

$\begin{array}{rcl}M& =& \frac{-i}{p}\\ i& =& -Mp\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}i& =& -\left(-0.7\right)\times 40\text{cm}\\ & =& 28\text{cm}\end{array}$

Now the focal length is defined as follows.

role="math" localid="1663129856144" $\begin{array}{rcl}\frac{1}{f}& =& \frac{1}{i}+\frac{1}{p}\\ & =& \frac{1}{28\text{cm}}+\frac{1}{40\text{cm}}\\ & =& \left(0.0357+0.025\right){\text{cm}}^{-1}\\ & =& 0.0607{\text{cm}}^{-1}\end{array}$

$\begin{array}{rcl}f& =& \frac{1}{0.0607{\text{cm}}^{-1}}\\ & =& 16.5\text{cm}\\ & \approx & +16\text{cm}\end{array}$

Hence, the mirror is concave because the focal length is positive.

The focal length is$+16\text{cm}$

Use the following formula to find the radius of curvature,

$\begin{array}{rcl}r& =& 2\times f\\ & =& 2\times 16.5\text{cm}\\ & =& 33\text{cm}\end{array}$

Hence, the radius of curvature is$33\text{cm}$

The object distance is $p=+40\text{cm}$as given in the problem.

From the part (a), you can say that the image distance is,

$i=28\text{cm}$

As given in the problem, the magnification ratio is$M=-0.70$

Since image distance is positive, the image is real.

As the magnification is negative and image is real, so the image is inverted.

A real image is formed on the same side of the mirror as the object.