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Chapter 24: Problem 66

A camera with a 50.0 -mm lens can focus on objects located from $1.5 \mathrm{m}$ to an infinite distance away by adjusting the distance between the lens and the film. When the focus is changed from that for a distant mountain range to that for a flower bed at \(1.5 \mathrm{m},\) how far does the lens move with respect to the film?

Short Answer

Expert verified
Answer: The lens moves \(\frac{26}{58}\,\mathrm{m}\) closer to the film when the focus is changed from the distant mountain range to the flower bed at 1.5m away.

Step by step solution

01

Write down the given information and identify the unknowns

We are given the focal length of the lens, \(f = 50.0\,\mathrm{mm}\) or \(0.05\,\mathrm{m}\). The object distances are infinity and \(d_{o1} = 1.5\,\mathrm{m}\). We need to find the difference in the image distances, \(\Delta d_i = d_{i2} - d_{i1}\) when the focus is changed.
02

Calculate the image distance for the distant mountain range

When the focus is on the distant mountain range, the object distance is infinity. Using the lens formula, we can write: \(\frac{1}{f} = \frac{1}{\infty} + \frac{1}{d_{i1}}\). Since any number divided by infinity is considered as zero, the equation becomes: \(\frac{1}{f} = 0 + \frac{1}{d_{i1}}\). Solving for \(d_{i1}\), we get: \(d_{i1} = f\) \(d_{i1} = 0.05 \,\mathrm{m}\)
03

Calculate the image distance for the flower bed 1.5m away

When the focus is on the flower bed, the object distance is \(d_{o2} = 1.5\,\mathrm{m}\). Using the lens formula again, we have: \(\frac{1}{f} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}}\) Now, we need to solve for \(d_{i2}\): \(\frac{1}{d_{i2}} = \frac{1}{f} - \frac{1}{d_{o2}}\) \(\frac{1}{d_{i2}} = \frac{1}{0.05} - \frac{1}{1.5}\) \(\frac{1}{d_{i2}} = 20 - \frac{2}{3} = \frac{58}{3}\) \(d_{i2} = \frac{3}{58} \,\mathrm{m}\)
04

Calculate the distance moved by the lens

Now that we have the image distances for both object distances, we can find the distance moved by the lens by subtracting the two image distances: \(\Delta d_i = d_{i2} - d_{i1}\) \(\Delta d_i = \frac{3}{58} \,\mathrm{m} - 0.05\,\mathrm{m}\) \(\Delta d_i = \frac{3}{58} \,\mathrm{m} - \frac{50}{1000}\,\mathrm{m}\) \(\Delta d_i = \frac{3}{58} \,\mathrm{m} - \frac{29}{58}\,\mathrm{m}\) \(\Delta d_i = \frac{3 - 29}{58}\,\mathrm{m}\) \(\Delta d_i = - \frac{26}{58}\,\mathrm{m}\) Since the negative sign indicates the direction of movement, the lens moves \(\frac{26}{58}\,\mathrm{m}\) closer to the film when the focus is changed from the distant mountain range to the flower bed at 1.5m away.

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Most popular questions from this chapter

Comprehensive Problems Good lenses used in cameras and other optical devices are actually compound lenses, made of five or more lenses put together to minimize distortions, including chromatic aberration. Suppose a converging lens with a focal length of \(4.00 \mathrm{cm}\) is placed right next to a diverging lens with focal length of \(-20.0 \mathrm{cm} .\) An object is placed \(2.50 \mathrm{m}\) to the left of this combination. (a) Where will the image be located? (b) Is the image real or virtual?
A slide projector, using slides of width \(5.08 \mathrm{cm},\) produces an image that is \(2.00 \mathrm{m}\) wide on a screen \(3.50 \mathrm{m}\) away. What is the focal length of the projector lens?
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A refracting telescope has an objective lens with a focal length of $2.20 \mathrm{m}\( and an eyepiece with a focal length of \)1.5 \mathrm{cm} .$ If you look through this telescope the wrong way, that is, with your eye placed at the objective lens, by what factor is the angular size of an observed object reduced?
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