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

$$\begin{gathered} p=+10 \\ f=8.0 \end{gathered}$$

The mirror is convex.

The object distance, type of mirror, and focal length are given in the problem. First, find the radius of curvature from the focal length. Then by using the mirror formula, find the image distance. Use the formula for magnification to find the lateral magnification. Using these quantities, determine whether the image is real or virtual and inverted or non-inverted. Also, find the position of the image.

Formulae are as follows:

$$\begin{gathered} r=2f \\ \frac{1}{f}=\frac{1}{p}+\frac{1}{i} \\ m=\frac{-i}{p} \end{gathered}$$

where, $m$is the magnification, $p$is the pole, $f$is the focal length.

The radius of curvature:

Use the following formula to find the radius of curvature:

$r=2\times f$

Since the mirror is convex, the focal length must be negative, i.e.,$f=-8.0$

$\begin{array}{c}r=2\times \left(-8.0\right)\\ r=-16\mathrm{cm}\end{array}$

Therefore, the radius of curvature is $r=-16\mathrm{cm}.$

Image distance:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$

Rearranging for i,

$i=\frac{pf}{\left(p-f\right)}$

Plugging the values,

$\begin{array}{c}i=\frac{10\times \left(-8\right)}{10-\left(-8\right)}\\ i=-4.44\mathrm{cm}\end{array}$

Therefore, the image distance is $i=-4.44\mathrm{cm}.$

Lateral magnification:

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

Therefore,

$\begin{array}{l}m=-\frac{-44444}{10}\\ m=+04\end{array}$

Therefore, the lateral magnification is$m=+0.44$.

Whether the image is real or virtual:

Since the image distance is negative, the image is virtual (V).

Therefore, the image is virtual (V).

Whether the image is inverted or non-inverted:

As the magnification is positive, the image is non-inverted (NI).

Therefore, the image is non-inverted (NI).

Position of image:

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.

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