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 noninverted (NI) from O, and (i) on the same side of the mirror as object O or on the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

$M=+0.20\text{cm}$

$f=30\text{cm}$

From mirror properties, the type of mirror can be decide. Then to find focal length, use magnification ratio and formula for focal length. Then, by using focal length the object distance and image distance can be found. Then, from image distance it can be identify which type of image it is.

The formula is as follows:

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

where,

r= radius of curvature,

f= focal length,

p= object distance from mirror,

i= image distance.

Magnification ratio is as follows,

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

The positive value of $m$is given, which means image distance must be negative because object distance is always positive.

As image distance is negative, it means the image is virtual and a virtual image is possible only for convex mirrors.

The focal length is given as$f=30\text{cm}$ .

As the mirror is convex, the focal length should be negative.

So,$f=-30\text{cm}$.

Use the following formula to find the radius of curvature,

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

It is known,

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

role="math" localid="1663141272017" $\begin{array}{rcl}\frac{1}{f}& =& \frac{1}{p}\left(1-\frac{1}{m}\right)\\ p& =& f\left(1-\frac{1}{m}\right)\\ p& =& -30\left(1-\frac{1}{0.2}\right)\\ p& =& 120\text{cm}\end{array}$

So object distance is 120 cm.

Image distance is as follows

$\begin{array}{rcl}i& =& -mp\\ i& =& -0.2\times 120\\ i& =& -24\text{cm}\end{array}$

Magnification ratio is$M$ .

Magnification ratio is +0.20 as given in the table.

Since image distance is negative, image is virtual.

As magnification is positive so image is non- inverted.

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

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