Through 29 22 23, 29 more mirror. ObjectOstands 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 localid="1664204659535" $\mathit{m}$. (All distances are in centimeters.) It also refers to whether (g) the image is real localid="1664204665329" $\mathbf{\left(}\mathbf{R}\mathbf{\right)}$or virtual localid="1664204670423" $\mathbf{\left(}\mathbf{V}\mathbf{\right)}$. (h) inverted localid="1664204676312" $\mathbf{\left(}\mathbf{I}\mathbf{\right)}$or non-inverted localid="1664204682860" $\left(\mathrm{NI}\right)$from O, and (i) on the same side of the mirror as the object Oor on the opposite side. Fill in the missing information, where, only a sign is missing, answer with the sign.

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 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.

Object distance refers to the separation between an object and the mirror's pole.

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

Formulae:

The radius of curvature of a mirror,

$r=2f$ ….. (i)

The mirror equation is,

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

The lateral magnification of an object,

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

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

The mirror is a concave type as given in the problem. Hence, the type of mirror is concave.

As the given mirror is a concave mirror, the focal length value of the mirror will be positive in sign.

Thus, using the given data, the focal length of the mirror is given as:

$f=+20\text{cm}$

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

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

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

Therefore, the radius of curvature is $\begin{array}{rcl}& & +40\text{cm}\end{array}$.

The value of the object distance from the mirror is given by the given data in the table as follows:

$p=+10\text{cm}$

Hence, the object distance is $+10\text{cm}$.

Now, the image distance can be calculated using the data in equation (ii) as follows:

$\frac{1}{20\text{cm}}=\frac{1}{i}+\frac{1}{10\text{cm}}$

$\begin{array}{rcl}\frac{1}{i}& =& \frac{1}{20\text{cm}}-\frac{1}{10\text{cm}}\\ & =& \frac{(10-20)\text{cm}}{200{\text{cm}}^2}\\ & =& \frac{-10\text{cm}}{200{\text{cm}}^2}\end{array}$

$\begin{array}{rcl}i& =& -\frac{200{\text{cm}}^2}{10\text{cm}}\\ & =& -20\text{cm}\end{array}$

Hence, the image distance is$\begin{array}{rcl}& & -20\text{cm}\end{array}$.

Thus, the lateral magnification of the mirror can be given using equation (iii) as follows:

$\begin{array}{rcl}m& =& \frac{-i}{p}\\ & =& \frac{-(-20\text{cm})}{10\text{cm}}\\ & =& +2.0\end{array}$

Therefore, the magnification ratio is $\begin{array}{rcl}& & +2.0\end{array}$.

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

Hence, the image formed by the mirror is virtual $\left(V\right)$.

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

$\begin{array}{rcl}m& =& \frac{h_i}{h_o}\\ & =& \frac{-i}{p}\\ & =& +2.0\end{array}$

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

Therefore, the image is non-inverted $\left(NI\right)$.

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

Hence, the image is on the opposite side of the mirror.