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 $\mathit{p}\mathit{s}\mathbf{\left(}\mathit{c}\mathit{e}\mathit{n}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{s}\mathbf{\right)}$, 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$\mathit{i}$, and (c) the lateral magnification$\mathit{m}$ . Also, determine whether the image is (d) real$\left(R\right)$or virtual $\mathbf{\left(}\mathit{V}\mathbf{\right)}$, (e) inverted$\mathbf{\left(}\mathit{I}\mathbf{\right)}$from object O or non-inverted $\mathbf{\left(}\mathit{N}\mathit{I}\mathbf{\right)}$, and (f) on the same side of the mirror as O or on the opposite side.

Object distance is,$p=+12\text{cm}$

Focal length of mirror,$f=18\text{cm}$

The mirror is concave.

The object distance, type of the mirror, and focal length is 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 lateralmagnification. Using these quantities, determine whether the image is real or virtual and inverted or non-inverted. Also, find the position of the image.

Formulae:

The relation between the radius of curvature and focal length of the mirror is,

$r=2f$.

The spherical mirror equation is given by,

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

The magnification is define by the following expression.

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

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 concave, the focal length must be positive, i.e., $f=+18\text{cm}$

$$\begin{gathered} r=2\times \left(18\text{cm}\right) \\ =+36\text{cm} \end{gathered}$$

Hence, the radius of curvature is $+36\text{cm}$.

Write the spherical mirror equation as below.

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

Rearranging the above equation foras below.

$$\begin{gathered} \frac{1}{i}=\frac{1}{f}-\frac{1}{p} \\ =\frac{p-f}{pf} \\ i=\frac{pf}{\left(p-f\right)} \end{gathered}$$

Plugging the known values in the above equation, you get

$$\begin{gathered} i=\frac{12\text{cm}\times \left(18\text{cm}\right)}{\left(12-18\right)\text{cm}} \\ =-36\text{cm} \end{gathered}$$

Therefore, the image distanceis$-36\text{cm}$ .

Determine the lateral magnification as below.

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

Substitute role="math" localid="1663072917852" $-36cm$for $i$and $12cm$for $p$in the above equation, and you obtain

$$\begin{gathered} m=-\frac{\left(-36\text{cm}\right)}{12\text{cm}} \\ =+3.0 \end{gathered}$$

Hence, the lateral magnification is$+3.0$.

Since the image distance is negative, the image is virtual$\left(V\right)$ . The image is formed on the other side of the mirror.

As the magnification is positive. Thus, 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 as the object.