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}$$\left(\text{centimeter}\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(including sign), (b) the image distance $\mathbf{i}$, 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 from object O or non-inverted localid="1663055514084" $\left(NI\right)$, and (f) on the same side of the mirror as O or on the opposite side.

The object distance,$p=+15\text{cm}$

Focal length,$f=10\text{cm}$

The mirror is Concave.

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 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 radius of curvature is,

$r=2f$

From the spherical mirror equation:

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

The magnification formula is given by,

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

Here, $m$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.,$f=+10\text{cm}$

Thus, the radius of the curvature will be,

role="math" localid="1663058401501" $$\begin{gathered} r=2\times 10cm \\ =+20cm \end{gathered}$$

Hence, the radius of curvature is$+20\text{cm}$.

Write the spherical mirror equation as below.

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

Rearrange the above equation for the image distance$i$,

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

Plugging the known values in the above equation, and you have

$$\begin{gathered} i=\frac{15\text{cm}\times 10\text{cm}}{\left(15-10\right)\text{cm}} \\ =+30\text{cm} \end{gathered}$$

Hence, the image distance is $+30cm$.

Lateral magnification is define by using following equation.

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

Substitute $+30cm$for $i$and $+15cm$for $p$in the above equation.

$$\begin{gathered} m=-\frac{30\text{cm}}{15\text{cm}} \\ =-2.0 \end{gathered}$$

Hence, the lateral magnification is$-2.0$.

Since the image distance is positive, the image is real $\left(R\right)$. Hence, the image is real$\left(R\right)$.

As the magnification is negative. Hence, the image is inverted$\left(I\right)$.

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 real, it is formed on the same side as the object.

Hence, the image is on the same side as the object.