Chapter 23, Problem 73P

(a)

To determine

The image distance, whether the image is real or virtual, its orientation and relative size.

Answer to Problem 73P

The object distance is $3.8\text{m}\text{and 2}\text{.6}\text{m}$ for $24\text{mm and 36}\text{mm}$ respectively.

Explanation of Solution

The focal length of the lens is $8.00\text{cm}$ and the image distances are $5.0\text{cm}$, $14.0\text{cm}$, $16.0\text{cm}$ and $20.0\text{cm}$.

Write the thin lens equation

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

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

Rearrange for $q$

  $q=\frac{pf}{p-f}$                                                                                (I)

Write the equation for transverse magnification

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

Here, $m$ is the magnification. A negative magnification implies the image if inverted.

If $\left|m\right|>1$, the image is enlarged, $\left|m\right|<1$, the image is diminished and $\left|m\right|=1$, the image size is same as the object size.

For the object distance of $5.0\text{cm}$,

Substitute $8.00\text{cm}$ for $f$ and $5.0\text{cm}$ for $p$ in (I) to find $q$

  $\begin{array}{c}q=\frac{\left(5.0\text{cm}\right)\left(8.00\text{cm}\right)}{\left(5.0\text{cm}\right)-\left(8.00\text{cm}\right)}\\ =-13.3\text{cm}\end{array}$

Substitute $5.0\text{cm}$ for $p$ and  $-13.3\text{cm}$ for $q$ in (II) to find $m$

  $\begin{array}{c}m=\frac{-\left(-13.3\text{cm}\right)}{5.0\text{cm}}\\ =2.67\end{array}$

Thus, the image distance is $-13.3\text{cm}$, the image is virtual, upright and enlarged since $q$ is negative, $m$ is positive and $\left|m\right|>1$.

For the object distance of $14.0\text{cm}$,

Substitute $8.00\text{cm}$ for $f$ and $14.0\text{cm}$ for $p$ in (I) to find $q$

  $\begin{array}{c}q=\frac{\left(14.0\text{cm}\right)\left(8.00\text{cm}\right)}{\left(14.0\text{cm}\right)-\left(8.00\text{cm}\right)}\\ =18.7\text{cm}\end{array}$

Substitute $14.0\text{cm}$ for $p$ and  $18.7\text{cm}$ for $q$ in (II) to find $m$

  $\begin{array}{c}m=\frac{-\left(18.7\text{cm}\right)}{14.0\text{cm}}\\ =-1.33\end{array}$

Thus, the image distance is $18.7\text{cm}$, the image is real, inverted and enlarged since $q$ is positive, $m$ is negative and $\left|m\right|>1$.

For the object distance of $16.0\text{cm}$,

Substitute $8.00\text{cm}$ for $f$ and $16.0\text{cm}$ for $p$ in (I) to find $q$

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

Substitute $16.0\text{cm}$ for $p$ and  $16.0\text{cm}$ for $q$ in (II) to find $m$

  $\begin{array}{c}m=\frac{-\left(16.0\text{cm}\right)}{16.0\text{cm}}\\ =-1.00\end{array}$

Thus, the image distance is $16.0\text{cm}$, the image is real, inverted and same as the object since $q$ is positive, $m$ is negative and $\left|m\right|=1$.

For the object distance of $20.0\text{cm}$,

Substitute $8.00\text{cm}$ for $f$ and $20.0\text{cm}$ for $p$ in (I) to find $q$

  $\begin{array}{c}q=\frac{\left(20.0\text{cm}\right)\left(8.00\text{cm}\right)}{\left(20.0\text{cm}\right)-\left(8.00\text{cm}\right)}\\ =13.3\text{cm}\end{array}$

Substitute $20.0\text{cm}$ for $p$ and  $13.3\text{cm}$ for $q$ in (II) to find $m$

  $\begin{array}{c}m=\frac{-\left(13.3\text{cm}\right)}{20.0\text{cm}}\\ =-0.667\end{array}$

Thus, the image distance is $13.3\text{cm}$, the image is real, inverted and diminished since $q$ is positive, $m$ is negative and $\left|m\right|<1$.

The results are summarized in the table:

$p$ in $\text{cm}$	$q$ in $\text{cm}$	$m$	Real or Virtual	Orientation	Relative size
$5.00$	$-13.3$	$2.67$	virtual	upright	enlarged
$14.0$	$18.7$	$-1.33$	real	inverted	enlarged
$16.0$	$16.0$	$-1.00$	real	inverted	same
$12.0$	$13.3$	$-0.667$	real	inverted	diminished

(b)

To determine

The image height for different object distance.

Answer to Problem 73P

The image height for $5.00\text{cm and 20}\text{.0}\text{cm}$ are $10.7\text{cm and }-2.67\text{cm}$ respectively.

Explanation of Solution

The image height is $4.00\text{ cm}$.

Write the equation for transverse magnification

  $m=\frac{h^{\prime }}{h}$

Here, $h$ is the object height and $h^{\prime }$ is the image height.

Rearrange for $h^{\prime }$

  $h^{\prime }=mh$                                                                             (III)

For the object distance of $5.0\text{cm}$,

Substitute $2.67$ for $m$ and $4.00\text{ cm}$ for $h$ in (III) to find $h^{\prime }$

  $\begin{array}{c}{h}^{\prime }=\left(2.67\right)\left(4.00\text{ cm}\right)\\ =10.7\text{cm}\end{array}$

For the object distance of $20.0\text{cm}$,

Substitute $-0.667$ for $m$ and $4.00\text{ cm}$ for $h$ in (III) to find $h^{\prime }$

  $\begin{array}{c}{h}^{\prime }=\left(-0.667\right)\left(4.00\text{ cm}\right)\\ =-2.67\text{cm}\end{array}$