Chapter 23, Problem 72P

(a)

To determine

The image distance and characteristics of the image.

Answer to Problem 72P

The image distance and characteristics of the images is given in table 1.

Explanation of Solution

Write an expression for lens’ law.

    $\frac{1}{f}=\frac{1}{p}+\frac{1}{p}$                                                                                                                (I)

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

Rearrange equation (I) to find $q$.

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

Write an expression for magnification.

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

Here, $m$ is the magnification.

If the magnification is positive, the image is virtual and upright. If the magnification is negative the image is real and inverted. If the magnification is greater than one, the image is larger than the object. If the magnification is less than one, the image is smaller than the object.

Conclusion:

For object distance $5.00\text{cm}$:

Substitute $-8.00\text{cm}$ for $f$ and $5.00\text{cm}$ for $p$ in equation (II) to find $q$.

    $\begin{array}{c}q=\frac{1}{\left(\frac{1}{-8.00\text{cm}}-\frac{1}{5.00\text{cm}}\right)}\\ =\frac{\left(-8.00\text{cm}\right)\left(5.00\text{cm}\right)}{\left(5.00\text{cm}\right)-\left(-8.00\text{cm}\right)}\\ =-3.08\text{cm}\end{array}$

Substitute $-3.08\text{cm}$ for $q$ and $5.00\text{cm}$ for $p$ in equation (III) to find $m$.

    $\begin{array}{c}q=-\frac{-3.08\text{cm}}{5.00\text{cm}}\\ =\frac{3.08\text{cm}}{5.00\text{cm}}\\ =0.615\end{array}$

For object distance $8.00\text{cm}$:

Substitute $-8.00\text{cm}$ for $f$ and $8.00\text{cm}$ for $p$ in equation (II) to find $q$.

    $\begin{array}{c}q=\frac{1}{\left(\frac{1}{-8.00\text{cm}}-\frac{1}{8.00\text{cm}}\right)}\\ =\frac{\left(-8.00\text{cm}\right)\left(8.00\text{cm}\right)}{\left(8.00\text{cm}\right)-\left(-8.00\text{cm}\right)}\\ =-4.00\text{cm}\end{array}$

Substitute $-4.00\text{cm}$ for $q$ and $8.00\text{cm}$ for $p$ in equation (III) to find $m$.

    $\begin{array}{c}q=-\frac{-4.00\text{cm}}{4.00\text{cm}}\\ =\frac{4.00\text{cm}}{8.00\text{cm}}\\ =0.500\end{array}$

For object distance $14.0\text{cm}$:

Substitute $-8.00\text{cm}$ for $f$ and $14.0\text{cm}$ for $p$ in equation (II) to find $q$.

    $\begin{array}{c}q=\frac{1}{\left(\frac{1}{-8.00\text{cm}}-\frac{1}{14.0\text{cm}}\right)}\\ =\frac{\left(-8.00\text{cm}\right)\left(5.00\text{cm}\right)}{\left(14.0\text{cm}\right)-\left(-8.00\text{cm}\right)}\\ =-5.09\text{cm}\end{array}$

Substitute $-5.09\text{cm}$ for $q$ and $14.0\text{cm}$ for $p$ in equation (III) to find $m$.

    $\begin{array}{c}q=-\frac{-5.09\text{cm}}{14.0\text{cm}}\\ =\frac{5.09\text{cm}}{14.0\text{cm}}\\ =0.364\end{array}$

For object distance $16.0\text{cm}$:

Substitute $-8.00\text{cm}$ for $f$ and $16.0\text{cm}$ for $p$ in equation (II) to find $q$.

    $\begin{array}{c}q=\frac{1}{\left(\frac{1}{-8.00\text{cm}}-\frac{1}{16.0\text{cm}}\right)}\\ =\frac{\left(-8.00\text{cm}\right)\left(16.0\text{cm}\right)}{\left(16.0\text{cm}\right)-\left(-8.00\text{cm}\right)}\\ =-5.33\text{cm}\end{array}$

Substitute $-5.33\text{cm}$ for $q$ and $16.0\text{cm}$ for $p$ in equation (III) to find $m$.

    $\begin{array}{c}q=-\frac{-5.33\text{cm}}{16.0\text{cm}}\\ =\frac{5.33\text{cm}}{16.0\text{cm}}\\ =0.333\end{array}$

For object distance $20.0\text{cm}$:

Substitute $-8.00\text{cm}$ for $f$ and $20.0\text{cm}$ for $p$ in equation (II) to find $q$.

    $\begin{array}{c}q=\frac{1}{\left(\frac{1}{-8.00\text{cm}}-\frac{1}{20.0\text{cm}}\right)}\\ =\frac{\left(-8.00\text{cm}\right)\left(20.0\text{cm}\right)}{\left(20.0\text{cm}\right)-\left(-8.00\text{cm}\right)}\\ =-5.71\text{cm}\end{array}$

Substitute $-5.71\text{cm}$ for $q$ and $20.0\text{cm}$ for $p$ in equation (III) to find $m$.

    $\begin{array}{c}q=-\frac{-5.71\text{cm}}{20.0\text{cm}}\\ =\frac{5.71\text{cm}}{20.0\text{cm}}\\ =0.286\end{array}$

Thus, the image distance and characteristics of the images is given in table 1.

$p\left(\text{cm}\right)$	$q\left(\text{cm}\right)$	$m=-q/p$	Real or virtual	Orientation	Relative size
$5.00$	$-3.08$	$0.615$	Virtual	Upright	Diminished
$8.00$	$-4.00$	$0.500$	Virtual	Upright	Diminished
$14.0$	$-5.09$	$0.364$	Virtual	Upright	Diminished
$16.0$	$-5.33$	$0.333$	Virtual	Upright	Diminished
$20.0$	$-5.71$	$0.286$	Virtual	Upright	Diminished

Table 1

(b)

To determine

The height of the image.

Answer to Problem 72P

  The height of the image at $\text{5}\text{.00 cm}$ is $2.46\text{cm}$ and at $20.0\text{cm}$ is $1.14\text{cm}$.

Explanation of Solution

Write an expression for the image height.

    $h\text{'}=mh$                                                                                                             (IV)

Here, $m$ is the magnification, $h$ is the object height and $h\text{'}$ is the image height.

Conclusion:

For object distance $5.00\text{cm}$:

Substitute $4.00\text{cm}$ for $h$ and $0.615$ for $m$ in equation (IV) to find $h\text{'}$.

    $\begin{array}{c}h\text{'}=\left(0.615\right)\left(4.00\text{cm}\right)\\ =2.46\text{cm}\end{array}$

For object distance $20.0\text{cm}$:

Substitute $4.00\text{cm}$ for $h$ and $0.286$ for $m$ in equation (IV) to find $h\text{'}$.

    $\begin{array}{c}h\text{'}=\left(0.286\right)\left(4.00\text{cm}\right)\\ =1.14\text{cm}\end{array}$

Thus, the height of the image at $\text{5}\text{.00 cm}$ is $2.46\text{cm}$ and at $20.0\text{cm}$ is $1.14\text{cm}$.