Object Ostands 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 $\mathbf{\left(}\mathbf{R}\mathbf{\right)}$or virtual $\mathbf{\left(}\mathbf{V}\mathbf{\right)}$, (h) inverted $\mathbf{\left(}\mathbf{I}\mathbf{\right)}$or non-inverted $\mathbf{\left(}\mathbf{NI}\mathbf{\right)}$from O, and (i) on the same side of the mirror as the object Oor the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

The lateral magnification of the mirror, $m=+0.4\text{0cm}$

The object’s distance from the mirror, $p=+\text{30cm}$

The image is inverted which means the image formed is real.

Magnification refers to the ratio of image length to object length measured in planes that are perpendicular to the optical axis.

The focal length is positive if the mirror is a concave mirror. The focal length is negative if the mirror is a convex mirror. The image distance is positive if the image is a real image and is on the mirror side of the object.

A concave mirror is a diverging mirror with its reflective surface bugling opposite of the light source. Thus, as per the properties, the images formed by the concave mirrors can be both real and virtual based on the position of the object.

The distance of an object refers to the separation between an object and the mirror's pole. Image distance refers to the separation between the mirror's pole and the image.

Formulae:

The lateral magnification of an object,

$m=\frac{h_i}{h_o}=\frac{-i}{p}$ ….. (i)

The mirror equation is,

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

The radius of curvature of a mirror,

$r=2f$ ….. (iii)

Here, $f$is the focal length, $p$is the object distance from the mirror, and $i$is the image distance,$h_i$is the height of an image, and$h_o$is the height of an object.

Now, from the lateral magnification value as negative, the image distance can be given using equation (i) as follows.

$\begin{array}{rcl}i& =& -mp\\ & =& 0.4\times (30\text{cm})\\ & =& +12\text{cm}\end{array}$

As in a normal condition, it is difficult to get a real image from the convex mirror, thus, it can be said that the mirror is concave.

Hence, the mirror is concave

Now the focal length value of the concave mirror can be given using equation (i) as follows:

$\begin{array}{rcl}\frac{\text{1}}{f}& =& \frac{\text{1}}{12\text{cm}}+\frac{\text{1}}{30\text{cm}}\\ & =& \frac{30\text{cm}+12\text{cm}}{30\text{cm}\times 12\text{cm}}\\ & =& \frac{42\text{cm}}{360{\text{cm}}^2}\end{array}$

$\begin{array}{rcl}f& =& \frac{360{\text{cm}}^2}{42\text{cm}}\\ & =& +\text{8.6cm}\end{array}$

Hence, the focal length of the mirror is $\begin{array}{rcl}& & +\text{8.6cm}\end{array}$.

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

$\begin{array}{rcl}r& =& \text{2}\times \left(8.\text{6cm}\right)\\ & =& 17.2\text{cm}\\ & \approx & 17\text{cm}\end{array}$

Hence, the radius of curvature is $\begin{array}{rcl}& & 17\text{cm}\end{array}$.

The value of the object’s distance from the mirror is given in the table as:

$p=+30\text{cm}$

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

From the calculations done in part (a), it is found that the value of the image distance can be given as follows:

$i=+\text{12cm}$

Hence, the image distance is $+\text{12cm}$.

As per the given data, the value of lateral magnification can be given as follows:

$m=-\text{0.40}$

Hence, the magnification ratio is $-\text{0.40}$.

From the calculations based in part (e), it is found that the image distance is positive in value. Thus, for an image distance to be positive, the image can be concluded to be real.

Hence, the image formed by the mirror is real $\left(R\right)$.

The lateral magnification of the mirror is given to be negative value. Again, you know that the lateral magnification can be given as:

$\begin{array}{rcl}m& =& \frac{h_i}{h_o}\\ & =& \frac{-i}{p}\\ & =& -0.4\end{array}$

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

Hence, the image is inverted $\left(I\right)$.

As per the given data and calculations based in part (e), the values of image distance and object distance can be seen to be positive in values.

Hence, the image is formed on the same side of the object O.