17 through 29 22 23, 29 more mirrors. Object $\mathit{O}$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, and (i) on the same side of the mirror as the objector the opposite side. Fill in the missing information, where only a sign is missing, answer with the sign.

1. The lateral magnification of the image,$M=+0.10\text{cm}$
2. The focal length of the mirror,$f=20\text{cm}$.

From the given magnification value, the type of mirror can be determined. For concave and convex mirrors both, the lateral magnification can be positive. But if the value of the magnification is less than one, this implies that the image is smaller and at a closer distance to the mirror. This is only possible if the given is convex. The convex mirror is a curved mirror with the reflective surface bulging towards the light source. This bulging-out surface reflects light outwards and is not used to focus light. The images formed by the convex mirror are virtual being smaller in size than the actual object's height.

Formulae:

The radius of curvature of a mirror,

$r=\text{2}f$ ......(i)

where, $f$= focal length of the mirror

The mirror equation,

$\frac{\text{1}}{f}=\frac{\text{1}}{i}+\frac{\text{1}}{p}$ ......(ii)

The lateral magnification of an object,

$m=\frac{-i}{p}$ ......(iii)

As the magnification ratio is less than zero and positive, it means the image is smaller than the object and it is located nearer to the mirror that is image distance is lower than the distance of the object, this case is only possible for a convex type mirror.

Hence, the type of mirror is convex.

Since the type of mirror is convex, the focal length value will always be negative for this type of mirror.

Hence, the value of the focal length is$-20\text{cm}$.

Now, the radius of the curvature of the mirror can be given using equation (i) as follows:

$\begin{array}{rcl}r& =& 2\times \left(-20\right)\text{cm}\\ & =& -40\text{cm}\end{array}$

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

Let, the object distance be $p$. Thus, the image distance from the mirror can be given using equation (iii) as follows:

$\begin{array}{rcl}i& =& -mp\\ & =& -\text{0.1}p\end{array}$

Now, using the above value in equation (ii), the value of the object distance can be given as follows:

$\begin{array}{rcl}\frac{\text{1}}{-\text{20 cm}}& =& \frac{\text{1}}{-\text{0.1}p}+\frac{\text{1}}{p}\\ \frac{-9}{p}& =& \frac{\text{1}}{-\text{20 cm}}\\ p& =& +180\text{cm}\end{array}$

Hence, the value of the object distance is$+180\text{cm}$

Now, using the above object distance value in the equation of part (d), the distance of the image from the mirror can be given as follows:

$\begin{array}{rcl}i& =& -0.1\times 180\text{cm}\\ & =& -18\text{cm}\end{array}$

Hence, the image distance is$-18\text{cm}$

As given in the problem, the value of the magnification ratio that is the lateral magnification of the image for this convex mirror is given as: $m=+\text{0.10}$

Hence, the value of the magnification ratio is$+\text{0.10}$

As per the calculations done in part (d), the image distance is found to be negative. This implies that the image formed is opposite to the object placed in front of the mirror (as object distance is positive). Thus the image is virtual in nature.

Therefore, the image is virtual.

The lateral magnification of the mirror is found to be a positive value as per the calculations based in part (c). Again, we know that the lateral magnification can be given as:

$m=\frac{h_i}{h_o}$

Thus, the image height needs to be positive that is possible only in an non-inverted image case.

Hence, the image is non-inverted (I).

For spherical mirrors, real images form on the side of the mirror where the object is located, and virtual images form on the opposite side. Since the image is virtual, it is formed on the opposite side of the object.

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