Chapter 23, Problem 79P

(a)

To determine

The object distance.

Answer to Problem 79P

The object distance is $9.1\text{ cm}$ .

Explanation of Solution

Write the transverse magnification equation in terms of the object and image heights.

  $m=\frac{h^{\prime }}{h}$                                                                                                             (I)

Here, $m$ is the transverse magnification, $h^{\prime }$ is the image length and $h$ is the object length

Write the transverse magnification equation in terms of the object and image distances.

  $m=-\frac{q}{p}$                                                                                                            (II)

Here, $p$ is the object distance and $q$ is the image distance.

Equate equations (I) and (II) and rewrite the equation for $p$ .

  $\begin{array}{c}\frac{h^{\prime }}{h}=-\frac{q}{p}\\ p=-\frac{h}{h^{\prime }}q\end{array}$                                                                                                  (III)

Conclusion:

Given that the height of the object is $8.0\text{ cm}$ , the height of the image is $3.5\text{ cm}$ and the image distance is $-4.0\text{ cm}$ . The negative value of the image distance is due to the fact that the image is virtual.

Substitute $8.0\text{ cm}$ for $h$ , $3.5\text{ cm}$ for $h^{\prime }$ and $-4.0\text{ cm}$ for $q$ in equation (III) to find $p$ .

  $\begin{array}{c}p=-\frac{8.0\text{ cm}}{3.5\text{ cm}}\left(-4.0\text{ cm}\right)\\ =9.1\text{ cm}\end{array}$

Therefore, the object distance is $9.1\text{ cm}$ .

(b)

To determine

The type of the mirror used.

Answer to Problem 79P

The mirror is convex.

Explanation of Solution

The images formed by a plane mirror are upright and virtual and for point objects, the object and the image are equidistant from the mirror and lie on the same normal line. Depending on the location of the object, a concave mirror can form either real or virtual images and the images can be larger or smaller than the object. The images formed by a convex mirror are upright, virtual, smaller than the object and closer to the mirror than the object.

In part (a), it is found that the image is smaller in height than the object and it is given that the image is virtual. Also the image is upright and closer to the mirror than the object. This implies the used mirror is convex.

(c)

To determine

The focal length and the radius of curvature of the mirror.

Answer to Problem 79P

The focal length of the mirror is $-7.1\text{ cm}$ and the radius of curvature is $14\text{ cm}$ .

Explanation of Solution

Write the mirror equation.

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

Here,$f$ is the focal length of the mirror.

Rewrite the above equation for $f$ .

  $f={\left(\frac{1}{p}+\frac{1}{q}\right)}^{-1}$

Put equation (III) in the above equation.

  $\begin{array}{c}f={\left(\frac{-h^{\prime }}{hq}+\frac{1}{q}\right)}^{-1}\\ =q\left(1-\frac{-h^{\prime }}{h}\right)\end{array}$                                                                                                (IV)

Write the equation for the radius of curvature .

  $R=\left|2f\right|$                                                                                                              (V)

Here, $R$ is the radius of curvature.

Conclusion:

Substitute $-4.0\text{ cm}$ for $q$ , $3.5\text{ cm}$ for $h^{\prime }$ and $8.0\text{ cm}$ for $h$  in equation (IV) to find $f$ .

  $\begin{array}{l}f=\left(-4.0\text{ cm}\right)\left(1-\frac{3.5\text{ cm}}{8.0\text{ cm}}\right)\\ =-7.1\text{ cm}\end{array}$

Substitute $-7.1\text{ cm}$ for $f$ in equation (V) to find $R$ .

  $\begin{array}{c}R=\left|2\left(-7.1\text{ cm}\right)\right|\\ =14\text{ cm}\end{array}$

Therefore, the focal length of the mirror is $-7.1\text{ cm}$ and the radius of curvature is $14\text{ cm}$ .