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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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Most popular questions from this chapter

A converging lens with focal length \(3.00 \mathrm{cm}\) is placed $4.00 \mathrm{cm}$ to the right of an object. A diverging lens with focal length \(-5.00 \mathrm{cm}\) is placed \(17.0 \mathrm{cm}\) to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image? (b) Repeat part (a) for the case where the lenses are separated by \(10.0 \mathrm{cm}.\)
You have two lenses of focal length \(25.0 \mathrm{cm}\) (lens 1 ) and $5.0 \mathrm{cm}$ (lens 2 ). (a) To build an astronomical telescope that gives an angular magnification of \(5.0,\) how should you use the lenses (which for objective and which for eyepiece)? Explain. (b) How far apart should they be?
An astronomical telescope provides an angular magnification of 12. The two converging lenses are \(66 \mathrm{cm}\) apart. Find the focal length of each of the lenses.
A photographer wishes to take a photograph of the Eiffel Tower $(300 \mathrm{m}\( tall ) from across the Seine River, a distance of \)300 \mathrm{m}$ from the tower. What focal length lens should she use to get an image that is 20 mm high on the film?
An object is located at \(x=0 .\) At \(x=2.50 \mathrm{cm}\) is a converging lens with a focal length of \(2.00 \mathrm{cm},\) at \(x=16.5 \mathrm{cm}\) is an unknown lens, and at \(x=19.8 \mathrm{cm}\) is another converging lens with focal length \(4.00 \mathrm{cm} .\) An upright image is formed at $x=39.8 \mathrm{cm} .$ For each lens, the object distance exceeds the focal length. The magnification of the system is \(6.84 .\) (a) Is the unknown lens diverging or converging? (b) What is the focal length of the unknown lens? (c) Draw a ray diagram to confirm your answer.
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