Chapter 24, Problem 14P

(a)

To determine

The focal length of the lens.

Answer to Problem 14P

The focal length of the lens is $\underset{\_}{224\text{mm}}$.

Explanation of Solution

Write the expression for magnification of image

  $M=-\frac{q}{p}=\frac{h^{\prime }}{h}$                                                                                                            (I)

Here, $q$ is the image distance, $p$ is the object distance, $h$ is the height of the object, $h^{\prime }$ is the height of the image

Rearrange equation (I) to obtain an expression for $q$

  $q=\frac{-h^{\prime }p}{h}$                                                                                                                   (II)

Write the expression for focal length

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

Here, $f$ is the focal length

Thus, the expression for focal length is

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

Conclusion:

Substitute, $75.0\text{m}$ for $p$, and $0.225\text{m}$ for $q$ in equation (IV)

  $\begin{array}{c}f=\frac{1}{\left(\frac{1}{75.0\text{m}}+\frac{1}{0.225\text{m}}\right)}\\ =0.224\text{m}\\ \text{=224}\text{mm}\end{array}$

Therefore, the focal length of the lens is $\underset{\_}{224\text{mm}}$.

(b)

To determine

The size of the image.

Answer to Problem 14P

The size of the image is $\underset{\_}{2.67\text{mm}}$.

Explanation of Solution

From equation (III) the equation for $q$ is given below

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

Rewrite equation (II) to obtain an expression for $h^{\prime }$

  $h^{\prime }=\frac{-qh}{p}$                                                                                                                  (VI)

Conclusion:

Substitute, $50.0\text{mm}$ for $f$, and $75\times {10}^3\text{mm}$ for $p$ in equation (V)

  $\begin{array}{c}q=\frac{1}{\left(\frac{1}{50.0\text{mm}}-\frac{1}{75\times {10}^3\text{mm}}\right)}\\ =50.0\text{mm}\end{array}$

Substitute, $50.0\text{mm}$ for $q$, $4.00\text{m}$ for $h$, and $75.0\text{m}$ for $p$ in equation (VI)

  $\begin{array}{c}{h}^{\prime }=\frac{-\left(50.0\text{mm}\right)\left(4.00\text{m}\right)}{75.0\text{m}}\\ =-2.67\text{mm}\end{array}$

Therefore, the image size is $\underset{\_}{2.67\text{mm}}$.

(c)

To determine

The distance at which the man should stand to obtain an image of height $12\text{cm}$ using a $50.0\text{mm}$ lens.

Answer to Problem 14P

The distance at which the man should stand to obtain an image of height $12\text{cm}$ using a $50.0\text{mm}$ lens is $\underset{\_}{16.7\text{m}}$.

Explanation of Solution

Consider equation (II).substitute equation (II) in equation (III) to obtain an expression for $p$

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

Conclusion:

Substitute, $0.050\text{m}$ for $f$, $4.00\text{m}$ for $h$, $-0.0120\text{m}$ for $h^{\prime }$ in equation (VII)

  $\begin{array}{c}p=\left(0.050\text{m}\right)\left(1-\frac{4.00\text{m}}{-0.0120\text{m}}\right)\\ =16.7\text{m}\end{array}$

Therefore, the distance at which the man should stand to obtain an image of height $12\text{cm}$ using a $50.0\text{mm}$ lens is $\underset{\_}{16.7\text{m}}$.