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 m. (All distances are in centimeters.) It also refers to whether (g) the image is real (R) or virtual (V), (h) inverted (I) or non-inverted (NI) 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 radius of curvature is,$-40 \text{cm}$ .

The image distance is, $-10 \text{cm}$.

The radius of curvature is given, and from that, find the focal length and decide whether the mirror is convex or concave. Then, using the basic formulas, find the required values.

Formulas are as follows:

$\mathit{r}\mathbf{=}\mathbf{2}\mathit{f}$

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

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

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

Type of the mirror-

The mirror is a convex type because the radius of curvature is negative which means the focal length is negative.

Therefore, the type of mirror is convex.

The focal length-

$r=2\times f$

Substitute the values for the above equation

$\begin{array}{rcl}-40 \text{cm}& =& 2\times f\\ f& =& -20 \text{cm}\end{array}$

Therefore, the focal length of the image has been obtained.

The radius of curvature-

The radius of curvature is $-40\text{cm}$as given in the problem.

Therefore, the radius of curvature is $-40\text{cm}$.

Formula to find object distance is,

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

Substitute all the value in the above equation.

$\begin{array}{rcl}\frac{1}{-20 \text{cm}}& =& \frac{1}{-10 \text{cm}}+\frac{1}{p}\\ p& =& 20 \text{cm}\end{array}$

Therefore, the object distance is,$+20\text{cm}$

The image distance-$i$

The image distance isrole="math" localid="1663077525559" $i=-10\text{cm}$ as given in the problem.

Therefore, the image distance $i$is,-10 cm.

The magnification ratio,

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

Substitute all the value in the above equation.

$\begin{array}{rcl}M& =& \frac{10 \text{cm}}{20 \text{cm}}\\ & =& 0.50\end{array}$

Therefore, the magnification ratio is, 0.50

Determine whether the image is virtual or real-

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

Therefore, the image is virtual.

Whether inverted or non-inverted-

The image is non-inverted because the magnification of the mirror is positive.

Therefore, the image is non--inverted.

Position of the image-

A virtual image is formed on the opposite side of the mirror from the object.

Therefore, the position of the image is on the opposite side.

The basic formulas can be used to find the radius of curvature, image distance, and the magnification ratio and then decide whether the image is virtual or real, on the same side or opposite side.