Question

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.

Short answer

1. Type of mirror is convex.
2. Focal length is $20cm$.
3. Radius of curvature is $40cm$.
4. Object distance is $+60cm$.
5. Image distance is$+30cm$.
6. Magnification ratio is$-0.5$.
7. Image is real.
8. Inverted.
9. Position of image is same side.

Step-by-step solution

Step 1: Given data:

The magnification, $m=-0.50$.

The object’s distance, $p=+60cm$.

Determining the concept:

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.

(a) Determining the type of mirror:

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

(b) Determining the Focal length:

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.

(c) Determining the Radius of curvature:

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.

(d) Determining the Object distance:

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

(e) Determining the Image distance:

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

(f) Determining the Magnification ratio:

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

(g) Determining whether the image is virtual or real:

Since image distance is positive, image is real.

(h) Determining whether the image is inverted or not inverted:

As magnification is negative so image is inverted.

(i) Determining the position of the image:

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