Chapter 24, Problem 15E

(a)

To determine

The lens-to-film distance.

Answer to Problem 15E

The lens-to-film distance is $0.0508\text{m}$.

Explanation of Solution

Write the expression for the focal length of the lens.

  $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$                                                                                        

Here, $f$ is the focal length of the lens, $v$ is the image distance and $u$ is the object distance.

Rearrange the above expression for the image distance.

  $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$        (I)

Conclusion:

Substitute $0.05\text{m}$ for $f$ and $3\text{m}$ for $u$ in equation (I).

  $\begin{array}{c}\frac{1}{v}=\frac{1}{\left(0.05\text{m}\right)}-\frac{1}{\left(3\text{m}\right)}\\ =\left(20-\frac{1}{3}\right)\text{m}\\ =\frac{59}{3}\text{m}\end{array}$

Simplify the above expression for the image distance.

  $\begin{array}{c}v=\left(\frac{3}{59}\right)\text{m}\\ =0.0508\text{m}\end{array}$

Thus, the lens-to-film distance is $0.0508\text{m}$.

(b)

To determine

The linear magnification of the lens.

Answer to Problem 15E

The linear magnification of the lens is $-0.0169$.

Explanation of Solution

Write the expression for the focal length of the lens.

  $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$                                                                                        

Here, $f$ is the focal length of the lens, $v$ is the image distance and $u$ is the object distance.

Rearrange the above expression for the image distance.

  $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$        (I)

Write the expression for the magnification of the lens.

  $m=-\frac{v}{u}$        (II)

Here, $m$ is the magnification of the lens.

Conclusion:

Substitute $0.05\text{m}$ for $f$ and $3\text{m}$ for $u$ in equation (I).

  $\begin{array}{c}\frac{1}{v}=\frac{1}{\left(0.05\text{m}\right)}-\frac{1}{\left(3\text{m}\right)}\\ =\left(20-\frac{1}{3}\right)\text{m}\\ =\frac{59}{3}\text{m}\end{array}$

Simplify the above expression for the image distance.

  $\begin{array}{c}v=\left(\frac{3}{59}\right)\text{m}\\ =0.0508\text{m}\end{array}$

Substitute $0.0508\text{m}$ for $v$ and $3\text{m}$ for $u$ in equation (i).

  $\begin{array}{c}m=-\frac{\left(0.0508\text{m}\right)}{\left(3\text{m}\right)}\\ =-0.0169\end{array}$

Thus, the linear magnification of the lens is $-0.0169$.

(c)

To determine

The maximum height of the person.

Answer to Problem 15E

The maximum height of the person is $1.42\text{m}$.

Explanation of Solution

Write the expression for the focal length of the lens.

  $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$                                                                                        

Here, $f$ is the focal length of the lens, $v$ is the image distance and $u$ is the object distance.

Rearrange the above expression for the image distance.

  $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$        (I)

Write the expression for the magnification of the lens.

  $m=-\frac{v}{u}$        (II)

Here, $m$ is the magnification of the lens.

Write the expression for the magnification of the lens.

  $m=\frac{I}{O}$        (III)

Here, $I$ is the image height and $O$ is the object height.

Conclusion:

Substitute $0.05\text{m}$ for $f$ and $3\text{m}$ for $u$ in equation (i).

  $\begin{array}{c}\frac{1}{v}=\frac{1}{0.05\text{m}}-\frac{1}{3\text{m}}\\ =\left(20-\frac{1}{3}\right)\text{m}\\ =\frac{59}{3}\text{m}\end{array}$

Simplify the above expression for the image distance.

  $\begin{array}{c}v=\left(\frac{3}{59}\right)\text{m}\\ =0.0508\text{m}\end{array}$

Substitute $0.0508\text{m}$ for $v$ and $3\text{m}$ for $u$ in equation (ii).

  $\begin{array}{c}m=-\frac{0.0508\text{m}}{3\text{m}}\\ =-0.0169\end{array}$

Substitute $-0.0169$ for $m$ and $0.024\text{m}$ for $O$ in equation (iii).

  $-0.0169=\frac{0.024\text{m}}{O}$

Simplify the above expression for the object height.

  $\begin{array}{c}O=\frac{0.024\text{m}}{-0.0169}\\ =1.42\text{m}\end{array}$

Thus, the maximum height of the person is $1.42\text{m}$.