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Chapter 24: Problem 12

Kim says that she was less than 10 ft away from the president when she took a picture of him with her \(50-\mathrm{mm}\) focal length camera lens. The picture shows the upper half of the president's body (or \(3.0 \mathrm{ft}\) of his total height). On the negative of the film, this part of his body is $18 \mathrm{mm}$ high. How close was Kim to the president when she took the picture?

Short Answer

Expert verified
Answer: Kim was approximately 1.006 feet (or approximately 1 foot) away from the president when she took the picture.

Step by step solution

01

Write down the known values

We know the following: 1. Focal length of the camera lens, f = \(50 mm\) 2. Object height (upper half of the president's body), h = \(3 ft\) 3. Image height on the film, h' = \(18 mm\) Note: We need to convert the object height to the same unit as the other measurements (millimeters). There are 304.8 millimeters in a foot, so \(3 ft = 914.4 mm\).
02

Write down the lens equation

The lens equation relating the object distance (d), image distance (d'), and focal length (f) is: $$\frac{1}{d}+\frac{1}{d'}=\frac{1}{f}$$
03

Write down the magnification equation

The magnification (M) is the ratio of the image height (h') to the object height (h). It is also equal to the ratio of the image distance (d') to the object distance (d). $$M=\frac{h'}{h}=\frac{d'}{d}$$
04

Solve for the image distance (d')

Now, we can find the image distance (d') using the magnification equation. Rearranging the equation and plugging in the values, we get: $$d' = d \times \frac{h'}{h} = d \times \frac{18\,\text{mm}}{914.4\,\text{mm}}$$ Substitute this expression for d' in the lens equation: $$\frac{1}{d}+\frac{1}{d \times \frac{18}{914.4}}=\frac{1}{50}$$
05

Solve for the object distance (d)

Now, solve the equation for the object distance (d): $$\frac{1}{d}+\frac{1}{d} \times \frac{914.4}{18}=\frac{1}{50}$$ Multiply both sides of the equation by \(50d\) to get rid of the fractions: $$50 + 50 \times \frac{914.4}{18}=d$$ Now, solve for d: $$d=50 + 50 \times \frac{914.4}{18}\approx 306.8\,\text{mm}$$
06

Convert the object distance to feet

Convert the object distance from millimeters to feet: $$d \approx 306.8\,\text{mm} \times \frac{1\,\text{ft}}{304.8\,\text{mm}}\approx 1.006\,\text{ft}$$
07

Conclusion

Kim was approximately 1.006 feet (or approximately 1 foot) away from the president when she took the picture.

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