9, 11, 13 Spherical mirrors. Object O stands on the central axis of a spherical mirror. For this situation, each problem in Table 34-3 gives object distance (centimeters), the type of mirror, and then the distance $\mathit{p}$(centimeters, without proper sign) between the focal point and the mirror. Find (a) the radius of curvature $\mathit{r}$(including sign), (b) the image distance $\mathit{i}$, and (c) the lateral magnification $\mathit{m}$. Also, determine whether the image is (d) real (R)or virtual (V), (e) inverted (I)from object O or non-inverted (NI), and (f) on the same side of the mirror as O or on the opposite side.

The focal length of the mirror, $f=35\text{cm}$

The object distance from the mirror, $p=+22\text{cm}$

The mirror is convex.

A convex mirror or diverging mirror is a curved mirror in which the reflecting surface bulges towards the light source.

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.

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

Virtual, upright and reduced images are always formed by convex mirrors, regardless of the distance between the object and the mirror.

Formulae:

The radius of curvature of a mirror,

$r=2f$ ….. (i)

The mirror equation is,

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

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.

Since the mirror is convex, the focal length must be negative, i.e.,$f=-35\text{cm}$.

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

$\begin{array}{rcl}r& =& 2f\\ & =& (2\times (-35))\text{cm}\\ & =& -70\text{cm}\end{array}$

Hence, the radius of curvature is $-70\text{cm}$.

Now, the image distance can be calculated by rearranging equation (ii) as follows:

$\begin{array}{rcl}\frac{1}{i}& =& \frac{1}{f}-\frac{1}{p}\\ & =& \frac{p-f}{pf}\end{array}$

$\begin{array}{rcl}i& =& \frac{pf}{(p-f)}\\ & =& \frac{22\text{cm}\times (-35\text{cm})}{22\text{cm}-(-35\text{cm})}\\ & =& -13.5\text{cm}\end{array}$

Therefore, the image distance is $-13.5\text{cm}$.

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

$\begin{array}{rcl}m& =& \frac{-i}{p}\\ & =& -\frac{(-13.5\text{cm})}{22.0\text{cm}}\\ & =& +0.61\end{array}$

Therefore, the lateral magnification is $+0.61$.

As per the calculations done in part (b), 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 $\left(V\right)$.

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

$\begin{array}{rcl}m& =& \frac{h_i}{h_o}\\ & =& \frac{-i}{p}\\ & =& +0.61\end{array}$

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

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

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 as the object.

Hence, the image is on the opposite side of the object O.