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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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Most popular questions from this chapter

A person on a safari wants to take a photograph of a hippopotamus from a distance of \(75.0 \mathrm{m} .\) The animal is \(4.00 \mathrm{m}\) long and its image is to be \(1.20 \mathrm{cm}\) long on the film. (a) What focal length lens should be used? (b) What would be the size of the image if a lens of \(50.0-\mathrm{mm}\) focal length were used? (c) How close to the hippo would the person have to be to capture a \(1.20-\mathrm{cm}-\) long image using a 50.0 -mm lens?
A converging lens with a focal length of \(3.00 \mathrm{cm}\) is placed $24.00 \mathrm{cm}\( to the right of a concave mirror with a focal length of \)4.00 \mathrm{cm} .\( An object is placed between the mirror and the lens, \)6.00 \mathrm{cm}\( to the right of the mirror and \)18.00 \mathrm{cm}$ to the left of the lens. Name three places where you could find an image of this object. For each image tell whether it is inverted or upright and give the total magnification.
You plan to project an inverted image \(30.0 \mathrm{cm}\) to the right of an object. You have a diverging lens with focal length \(-4.00 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. Once you put a second lens at \(18.0 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high. what is the image height?
A camera uses a 200.0 -mm focal length telephoto lens to take pictures from a distance of infinity to as close as 2.0 \(\mathrm{m}\). What are the minimum and maximum distances from the lens to the film?
Jordan is building a compound microscope using an eyepiece with a focal length of \(7.50 \mathrm{cm}\) and an objective with a focal length of $1.500 \mathrm{cm} .\( He will place the specimen a distance of \)1.600 \mathrm{cm}$ from the objective. (a) How far apart should Jordan place the lenses? (b) What will be the angular magnification of this microscope?
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