Chapter 4.5, Problem 73PS

Solution

a.

To find: The expression in terms of $s$ .

  $s=2{\log }_{2}f$ .

Given: the $s$ and $f$ are related by the equation is as $s={\log }_2{f}^2$ . Where $s$ to be a measure of the amount of light that strikes the film and $f$ be the $f$ - stops.

Concept used: By the power property of logarithm ${\log }_a{m}^n=n{\log }_{a}m$ ; Where, $a$ is the base of the logarithm.

Calculation:

Now, simplify the above equation as follows:

  $\begin{array}{c}s={\log }_2{f}^2\\ s=2{\log }_{2}f\left[{\log }_a{m}^n=n{\log }_{a}m\right]\end{array}$

Conclusion:

Hence, the required expression is $s=2{\log }_{2}f$ .

b.

To find: The table shows the first eight $f$ - stops on a millimeter camera. Complete the table and also describe the pattern.

$s$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$f$	$1.414$	$2.000$	$2.828$	$4.000$	$5.657$	$8.000$	$11.314$	$16.000$

Given: the $s$ and $f$ are related by the equation is as $s={\log }_2{f}^2$ . Where $s$ to be a measure of the amount of light that strikes the film and $f$ be the $f$ - stops.

Concept used: By the power property of logarithm ${\log }_a{m}^n=n{\log }_{a}m$ ; Where, $a$ is the base of the logarithm.

Calculation:

Now, simplify the above equation as follows:

  $\begin{array}{l}s={\log }_2{f}^2\mathrm{...}\left(1\right)\\ s=2{\log }_{2}f\left[{\log }_a{m}^n=n{\log }_{a}m\right]\end{array}$

Therefore,

substitute the value of $f=1.414$ in the above equation (1);

  $\begin{array}{c}s={\log }_2{\left(1.414\right)}^2\\ s={\log }_{2}2\\ s=1\left[\begin{array}{c}{\left(1.414\right)}^2\approx 2\\ {\log }_{a}a=1\end{array}\right]\end{array}$

And substitute the value of $f=2.000$ in the above equation (1);

  $\begin{array}{c}s={\log }_2{\left(2\right)}^2\\ s=2\cdot {\log }_{2}2\\ s=2\left[\begin{array}{c}{\log }_a{m}^n=n{\log }_{a}m\\ {\log }_{a}a=1\end{array}\right]\end{array}$

Again, substitute the value of $f=2.828$ in the above equation (1);

  $\begin{array}{c}s={\log }_2{\left(2.828\right)}^2\\ s={\log }_{2}8\left[{\left(2.828\right)}^2\approx 8\right]\\ s={\log }_2{\left(2\right)}^3\\ s=3{\log }_{2}2\left[\begin{array}{c}{\log }_a{m}^n=n{\log }_{a}m\\ {\log }_{a}a=1\end{array}\right]\end{array}$

Thus,

  $s=3$

And substitute the value of $f=4.000$ in the above equation (1);

  $\begin{array}{c}s={\log }_2{\left(4\right)}^2\\ s={\log }_{2}16\\ s={\log }_2{\left(2\right)}^4\\ s=4{\log }_{2}2\left[\begin{array}{c}{\log }_a{m}^n=n{\log }_{a}m\\ {\log }_{a}a=1\end{array}\right]\end{array}$

Then,

  $s=4$

Again, substitute the value of $f=5.657$ in the above equation (1);

  $\begin{array}{c}s={\log }_2{\left(5.657\right)}^2\\ s={\log }_{2}32\left[{\left(5.657\right)}^2=32\right]\\ s={\log }_2{\left(2\right)}^5\\ s=5{\log }_{2}2\left[\begin{array}{c}{\log }_a{m}^n=n{\log }_{a}m\\ {\log }_{a}a=1\end{array}\right]\end{array}$

Thus,

  $s=5$

And substitute the value of $f=8.000$ in the above equation (1);

  $\begin{array}{c}s={\log }_2{\left(8\right)}^2\\ s={\log }_{2}64\\ s={\log }_2{\left(2\right)}^6\\ s=6{\log }_{2}2\left[\begin{array}{c}{\log }_a{m}^n=n{\log }_{a}m\\ {\log }_{a}a=1\end{array}\right]\end{array}$

Then,

  $s=6$

Again, substitute the value of $f=11.314$ in the above equation (1);

  $\begin{array}{c}s={\log }_2{\left(11.314\right)}^2\\ s={\log }_{2}128\left[{\left(11.314\right)}^2=128\right]\\ s={\log }_2{\left(2\right)}^7\\ s=7{\log }_{2}2\left[\begin{array}{c}{\log }_a{m}^n=n{\log }_{a}m\\ {\log }_{a}a=1\end{array}\right]\end{array}$

Thus,

  $s=7$

And substitute the value of $f=16$ in the above equation (1);

  $\begin{array}{c}s={\log }_2{\left(16\right)}^2\\ s={\log }_{2}256\\ s={\log }_2{\left(2\right)}^8\\ s=8{\log }_{2}2\left[\begin{array}{c}{\log }_a{m}^n=n{\log }_{a}m\\ {\log }_{a}a=1\end{array}\right]\end{array}$

Then,

  $s=8$

Now, substitute the value of $s$ and $f$ is given table as follows:

$s$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$f$	$1.414$	$2.000$	$2.828$	$4.000$	$5.657$	$8.000$	$11.314$	$16.000$

Conclusion:

Hence, the required table of $s$ , and $f$ is given table as below:

$s$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$f$	$1.414$	$2.000$	$2.828$	$4.000$	$5.657$	$8.000$	$11.314$	$16.000$

c.

To find:the ninth $f$ - stops.

  $f\approx 22.624$ .

Given: The $s$ and $f$ are related by the equation is as $s={\log }_2{f}^2$ . Where $s$ to be a measure of the amount of light that strikes the film and $f$ be the $f$ - stops.

Concept used: By the definition of the logarithm is defined as let $b$ and $y$ be the positive number with $b\ne 1$ . The logarithm of $y$ with base $b$ is denoted by ${\log }_{b}y$ and if defined as follows: ${\log }_{b}y=x$ if and only if $y=b^x$ .

Calculation:

From the table in part (b) the values of $s$ are $1,2,3,4,5,6,7$ and $8$ for the first $8$

  $f$ - stop.

Therefore, the $s$ value for the $9^{th}$ stop is $9$ and we need to solve $f$ .

Now take the given equation as: $s={\log }_2{f}^2$ and substitute $s=9$ it in this equation.

  $\begin{array}{c}9={\log }_2{f}^2\\ {\left(2\right)}^9=f^2\left[{\log }_{b}y=x\text{ if and only if }y=b^x\right]\\ f=\sqrt{{\left(2\right)}^9}\\ f={\left(2\right)}^4\cdot \sqrt{2}\end{array}$

Thus,

  $\begin{array}{c}f\approx 16\cdot 1.414\left[\sqrt{2}\approx 1.414\right]\\ f\approx 22.624\end{array}$

Conclusion:

Hence, the required $9^{th}$

  $f$ - stop is $f\approx 22.624$ .

And also clear that $9^{th}$

  $f$ the stop is under the $35$ meters.