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 $\mathbf{ps}\mathbf{}$(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 $\mathit{r}$(including sign), (b) the image distance localid="1662986561416" $\mathit{i}$, 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 object distance,$p=+17\mathrm{cm}$

The focal length of the spherical mirror,$f=-14\mathrm{cm}$

The mirror is convex mirror.

Use equations of the radius of curvature and lateral magnification for a spherical mirror. Also, need to use equation 34.4 to calculate the distance of the image. Using these values, determine the nature of the image.If the image distance is negative, then the image is virtual and as the magnification ratio is positive, the image is upright.

Formulae:

The radius of curvature is related to the focal length of the mirror by the flowing formula.

$r=2f$

The spherical mirror equation is,

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

The magnification equation is as below.

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

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

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., .$f=-14\mathrm{cm}$

$\begin{array}{c}r=-2\times 14\mathrm{cm}\\ =-28\mathrm{cm}\end{array}$

Hence, the radius of curvature is $-28\mathrm{cm}$.

Write the spherical mirror equation as below.

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

Rearranging it for the image distance.

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

Substitute known values in the above equation.

$\begin{array}{c}i=\frac{17\mathrm{cm}\times \left(-14\mathrm{cm}\right)}{\left(17-\left(-14\right)\right)\mathrm{cm}}\\ =-7.7\mathrm{cm}\end{array}$

Hence, the image distance is $-7.7\mathrm{cm}$.

The magnification ratio is define as,

$\begin{array}{c}M=\frac{-i}{p}\\ =\frac{7.7\mathrm{cm}}{17\mathrm{cm}}\\ =+0.45\end{array}$

Hence, the magnification ratio is $0.45$.

Since the image distance is negative. Therefore, the image is virtual.

As magnification is positive. Thus, the image is non-inverted.

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