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Chapter 24: Problem 75

Veronique is nearsighted; she cannot sce clearly anything more than $6.00 \mathrm{m}$ away without her contacts. One day she doesn't wear her contacts; rather, she wears an old pair of glasses prescribed when she could see clearly up to \(8.00 \mathrm{m}\) away. Assume the glasses are \(2.0 \mathrm{cm}\) from her eyes. What is the greatest distance an object can be placed so that she can see it clearly with these glasses?

Short Answer

Expert verified
Answer: Approximately 10.43 meters.

Step by step solution

01

Find the focal length of the old glasses

We know that with the old glasses, Veronique could see clearly up to 8 meters away (\(d_i = 8~m\)). We also know that without the glasses her far point is 6 meters away (\(d_o = 6~m\)). Using the lens equation: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) \(\frac{1}{f} = \frac{1}{6} + \frac{1}{8}\) \(f = \frac{1}{\frac{1}{6} + \frac{1}{8}} = 24/5~m = 4.8~m\)
02

Calculate the new image distance with the glasses on the face

Since the glasses are 2 cm from her eyes, the new image distance \(d_i'\) will be 2 cm less than the original image distance: \(d_i' = d_i - 0.02 = 8 - 0.02 = 7.98~m\)
03

Calculate the new object distance using the lens equation

Now that we know the focal length and the new image distance, we can find the new object distance \(d_o'\) using the lens equation: \(\frac{1}{f} = \frac{1}{d_o'} + \frac{1}{d_i'}\) \(\frac{1}{4.8} = \frac{1}{d_o'} + \frac{1}{7.98}\) Now, solve for \(d_o'\): \(d_o' = \frac{1}{\frac{1}{4.8} - \frac{1}{7.98}} \approx 10.43 ~m\)
04

Conclusion

The greatest distance an object can be placed so that Veronique can see it clearly with the old glasses is approximately 10.43 meters away.

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