Chapter 3.2, Problem 85PS

Solution

The optimum pinhole diameter for a camera box with a length of 10 centimeters.

  $0.45$ millimeters

Given:

The optimum diameter d in millimeters of the pinhole in a pinhole camera is given by $d=1.9{\left[\left(5.5\times {10}^{-4}\right)l\right]}^{\frac{1}{2}}$ where l is the length of the camera box in millimeters.

Concept Used:

• Power of a power property: ${\left(x^a\right)}^b=x^{a\cdot b}$
• Product of power property: $a^m\cdot a^n=a^{m+n}$
• Quotient of power property: $\frac{a^m}{a^n}=a^{m-n}$
• Product property of radicals: $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$
• Quotient property of radicals: $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$
• The exponent form of the nth root of a number a is given by $\sqrt[n]{a}=a^{\frac{1}{n}}$

Calculation:

In order to find the optimum pinhole diameter for the camera box with length of 10 centimeters or 100 millimeters, substitute $l=100$ into the given formula and simplify further as shown below,

  $\begin{array}{c}d=1.9{\left[\left(5.5\times {10}^{-4}\right)l\right]}^{\frac{1}{2}}\\ =1.9{\left[\left(5.5\times {10}^{-4}\right)100\right]}^{\frac{1}{2}}\\ =1.9{\left[550\times {10}^{-4}\right]}^{\frac{1}{2}}\\ =1.9{\left[5.5\times {10}^{-2}\right]}^{\frac{1}{2}}\\ =1.9\times {\left(5.5\right)}^{\frac{1}{2}}\times {\left({10}^{-2}\right)}^{\frac{1}{2}}\\ =1.9\times {\left(5.5\right)}^{\frac{1}{2}}\times {10}^{-1}\\ =1.9\times {\left(5.5\right)}^{\frac{1}{2}}\times \frac{1}{10}\\ \approx 0.45\end{array}$

Therefore, the optimum pinhole diameter for the given camera is about $0.45$ millimeters.