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 focal length is,$f=+20 \text{cm}$ .

The object distance is, $p=+30 \text{cm}$.

The focal length and object distance is given which find the radius of curvature and the image distance. Then, using image distance and object distance, find the magnification ratio.

Formulas are as follows:

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

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

Where, $\mathit{m}$ is the magnification, $\mathit{p}$ is the pole, role="math" localid="1663134096837" $\mathit{f}$ is the focal length.

Type of mirror-

The Mirror is a concave type because the focal length is positive.

Therefore, the type of mirror is concave.

Focal length

Focal length is $f=+20\text{cm}$ as given in the problem

Therefore, the focal length is $+20\text{cm}$.

The radius of curvature-

Use the following formula to find the radius of curvature

$r=2\times f$

Substitute all the value in the above equation

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

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

Object distance-

The object distance is $p=+30\text{cm}$as per the given table.

Therefore, the object distance is $+30\text{cm}$.

Image distance -

The image distance relation with focal length and object distance is expressed as,

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

Substitute all the value in the above equation.

$\begin{array}{rcl}\frac{1}{20 \text{cm}}& =& \frac{1}{i}+\frac{1}{30 \text{cm}}\\ i& =& 60 \text{cm}\end{array}$

Therefore, the image distance is$+60 \text{cm}$

Magnification ratio-

The magnification ratio $M$is,

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

Substitute all the values in the above equation.

$\begin{array}{rcl}M& =& -\frac{60 \text{cm}}{30 \text{cm}}\\ M& =& -2.0\end{array}$

Therefore, the magnification ratio is,$-2.0$

Determine whether the image is virtual or real

Since the image distance is positive, the image is real.

Therefore, the image is real.

Whether inverted or non-inverted-

As magnification is negative, so the image is inverted.

Therefore, the image is inverted.

Position of the image-

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

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

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