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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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