Question

a particular camera has a focal length of 28 millimeters (mm). a photographer wants to get a closeup of geological formation at 22 inches. a standard equation applies as follows to find image distance (i) involving object distance (o) and focal length (f).
\\( \frac { 1 } { i } = \frac { 1 } { f } - \frac { 1 } { o } \\)
you will need to convert the measurements to the same measurement system, whether u.s. customary or metric. however, to be a rate or decimal applicable to the metric measurement of the camera lens, you will ultimately need to use metric values. so the best approach would be to convert all measures to metric, subtract the fraction, and convert the fraction to a decimal.
the values then are:
\\( f = 15 ( mathrm { mm } ) \\)
\\( o = 22 \\)
\\( \frac { 1 } { f } - \frac { 1 } { o } = \\)
do the math and convert the resulting fraction to a decimal. use 25.4 mm per inch.
round to the nearest hundredth if necessary.

Explanation:

Step1: Convert object distance to mm

Since \(1\) inch \( = 25.4\) mm, for \(o = 22\) inches, then \(o=22\times25.4\) mm.
\(o = 558.8\) mm.

Step2: Calculate \(\frac{1}{f}-\frac{1}{o}\)

Given \(f = 15\) mm and \(o = 558.8\) mm.
\(\frac{1}{f}-\frac{1}{o}=\frac{1}{15}-\frac{1}{558.8}\)
First, find a common denominator. The common denominator of \(15\) and \(558.8\) is \(15\times558.8 = 8382\).
\(\frac{1}{15}-\frac{1}{558.8}=\frac{558.8 - 15}{8382}=\frac{543.8}{8382}\)
\(\frac{543.8}{8382}\approx0.06\)

Answer:

\(0.06\)