17 through 29 22 23, 29 More mirrors. Object O stands on the central axis of a spherical or plane mirror. For this situation, each problem in Table 34-4 refers to (a) the type of mirror, (b) the focal distance $\mathit{f}$, (c) the radius of curvature $\mathit{r}$, (d) the object distance $\mathit{p}$, (e) the imagedistance $\mathit{i}$, and (f) the lateral magnification $\mathit{m}$. (All distances are in centimeters.) It also refers to whether (g) the image is real $\left(R\right)$or virtual localid="1662996882725" $\left(V\right)$, (h) inverted $\mathbf{\left(}\mathit{I}\mathbf{\right)}$or noninverted $\mathbf{\left(}\mathit{N}\mathit{I}\mathbf{\right)}$from O, and (i) on the same side of the mirror as object O or on the opposite side. Fill in the missing information. Where only a sign is missing, answer with the sign.

The magnification, $m=-0.50$.

The object’s distance, $p=+60cm$.

The properties of mirror can be used to find the type of mirror. Magnification and object distance can be used to find image distance and focal length, and from focal length, radius of curvature can be found. And from image distance, it is identify whether image is inverted or not.

Formulae:

The radius of curvature is,

$r=\text{2}f$

The spherical mirror equation is,

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

The magnification is,

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

Here, $r$is the radius of curvature, $f$is the focal length, $p$is the object distance from mirror, and $i$is the image distance.

As magnification ratio is less than zero and negative, it means image is smaller than object. Hence, mirror is concave type.

Use following formula to find focal length as,

$\begin{array}{rcl}m& =& \frac{-i}{p}\\ i& =& -mp\end{array}$

$\begin{array}{rcl}i& =& -\left(-0.5\right)\times \left(+60\right)\\ & =& +30cm\end{array}$,

Now focal length is as follows,

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

$\begin{array}{rcl}\frac{\text{1}}{f}& =& \frac{\text{1}}{+30\text{cm}}+\frac{\text{1}}{+60\text{cm}}\\ & =& 0.033{\text{cm}}^{-1}+0.0167{\text{cm}}^{-1}\\ & =& 0.0497{\text{cm}}^{-1}\end{array}$

$\begin{array}{rcl}f& =& \frac{1}{0.0497{\text{cm}}^{-1}}\\ & =& 20.12\text{cm}\\ & \approx & \text{20 cm}\end{array}$

Hence, the focal length is 20 cm.

Use the following formula to find the radius of curvature,

$\begin{array}{rcl}r& =& 2\times f\\ & =& 2\times 20\text{cm}\\ & =& 40\text{cm}\end{array}$

Hence, the radius of curvature of the spherical mirror is 40 cm.

Object distance is$p=+60cm$ as given in table.

According to part (b) the image distance is $i=+30cm$.

Magnification ratio is $m$is $-0.50$as given in the table.

Since image distance is positive, image is real.

As magnification is negative so image is inverted.

An image is formed on same side of mirror from the object.